Mechanics 84 6.1 2D Particle in Cartesian Coordinates

A Hostetler Handbook

v 0.82

Contents

Preface v

1 Newton’s Laws1 1.1 Many Particles ...... 2 1.2 Cartesian Coordinates...... 3 1.3 Polar Coordinates ...... 4 1.4 Cylindrical Coordinates ...... 7 1.5 Air Resistance, Friction, and Buoyancy...... 8 1.6 Charged Particle in a Magnetic Field...... 17 1.7 Summary: Newton’s Laws...... 20

2 Momentum and Center of Mass 23 2.1 Rocket with no External Force ...... 24 2.2 Multistage Rockets...... 25 2.3 Rocket with External Force...... 25 2.4 Center of Mass...... 26 2.5 Angular Momentum of a Single Particle...... 28 2.6 Angular Momentum of a System of Particles ...... 29 2.7 Angular Momentum of a Continuous Mass Distribution ...... 30 2.8 Summary: Momentum and Center of Mass ...... 34

3 Energy 36 3.1 General One-dimensional Systems ...... 40 3.2 Atwood Machine...... 43 3.3 Spherically Symmetric Central Forces ...... 44 3.4 The Energy of a System of Particles ...... 45 3.5 Summary: Energy ...... 47

4 Oscillations 49 4.1 Simple Harmonic Oscillators...... 49 4.2 Damped Harmonic Oscillators...... 52 4.3 Driven Oscillators ...... 57 4.4 Summary: Oscillations...... 69

5 Calculus of Variations 72 5.1 The Euler-Lagrange Equation...... 72 5.2 The Brachistochrone Problem...... 77 5.3 General Parametrization...... 80 5.4 Summary: Calculus of Variations...... 83

6 Lagrangian Mechanics 84 6.1 2D Particle in Cartesian Coordinates...... 84 6.2 2D Particle in Polar Coordinates...... 86 iv Contents

6.3 Unconstrained Particles in 3D...... 87 6.4 Constrained Particles in 3D...... 87 6.5 Noether’s Theorem...... 95 6.6 The Hamiltonian...... 96 6.7 Lagrange Multipliers...... 98 6.8 Summary: Lagrangian Mechanics...... 100

7 Orbits 102 7.1 The Kepler Problem...... 105 7.2 Kepler’s Laws...... 109 7.3 Transfer Orbits...... 109 7.4 Summary: Orbits...... 113

8 Noninertial Frames 115 8.1 Frame with Linear Acceleration...... 115 8.2 Rotating Frames ...... 116 8.3 Centrifugal Force...... 118 8.4 Coriolis Force...... 119 8.5 Summary: Noninertial Frames...... 123

9 Rigid Rotations 125 9.1 Rotation About a Fixed Axis...... 126 9.2 The Inertia Tensor...... 128 9.3 Principal Axes ...... 133 9.4 Torque-free Motion...... 136 9.5 Summary: Rigid Rotations ...... 139

10 Coupled Oscillators 141 10.1 Two Masses and Three Springs...... 141 10.2 The General Problem ...... 145 10.3 Double Pendulum...... 147 10.4 Triatomic Molecule...... 149 10.5 Parallel and Series Springs...... 151 10.6 Summary: Coupled Oscillators ...... 152

11 Collision and Scattering 154 11.1 Hard Sphere Scattering ...... 155 11.2 The General Case...... 156 11.3 Rutherford Scattering ...... 158

12 Math Reference 160 12.1 Complex Numbers ...... 160 12.2 Vectors ...... 160 12.3 Curvilinear Coordinates...... 160 12.4 Diﬀerential Equations ...... 163 12.5 Taylor Series ...... 163 12.6 Approximations...... 164

Index 165 Preface

About These Notes

Cover image: NASA (The International Space Station) These are my class notes from two courses on classical mechanics (PHY 3221 and PHY 4222) that I took at Florida State University. Our textbook was Classical Mechanics by John R. Taylor. My class notes can be found at www.leonhostetler.com/classnotes Please bear in mind that these notes will contain errors. Any errors are certainly my own. If you ﬁnd one, please email me at [email protected] with the name of the class notes, the page on which the error is found, and the nature of the error. This work is currently licensed under a Creative Commons Attribution-NonCommercial- NoDerivatives 4.0 International License. That means you are free to copy and distribute this document in whole for noncommercial use, but you are not allowed to distribute derivatives of this document or to copy and distribute it for commercial reasons.

Updates

Last Updated: July 10, 2019 Version 0.82: (Jul. 10, 2019) Updated layout, spell-checked Version 0.81: (Dec. 21, 2017) First upload.

Conventions

I will represent a vector by a bold letter topped with an arrow. For example: F~. I will represent a unit vector (i.e. a direction vector with magnitude 1) by a bold leter topped with a hat. For example: xˆ. In physics, a time derivative is often represented with a dot. For example, instead dx d2x of writing dt or dt2 , one might writex ˙ orx ¨. This convention will generally be used in these notes.

Chapter 1

Newton’s Laws

Mechanics is the study of the motion of material bodies in space as a function of time. We’ll be studying classical mechanics which assumes v << c (i.e. non-relativistic) and Planck’s constant is treated as h = 0 (i.e. non-quantum). First Law: In the absence of forces, a body moves with uniform velocity. It’s the ability to visualize the ideal case that enabled Newton to achieve this break- through since in the real world, moving bodies tend to slow down. Remarkably, a statement about an ideal situation can allow us to make arbitrarily precise predictions. Second Law: F~ = m~a This law does not hold relativistically. Here, m is the inertial mass, which is an attribute of the body. Inertial mass quantiﬁes a body’s resistance to force. F~ is not an attribute of the body, but rather it quantiﬁes the interaction of the body with its environment. Newton also introduced a “quantity of motion” that is known today as momentum, ~p = m~v. Using this, we can write Newton’s law in a more general form that also holds relativistically: d~p F~ = = ~p˙. dt Third law: For every action, there’s an equal but opposite reaction. The notation F~ ij denotes the force exerted on particle i by particle j. Read it as “force on i by j”. According to Newton’s third law, for two particles i and j,

F~ ij = −F~ ji.

ext If a system of two particles is isolated with no outside forces, that is, F~ = 0, then the only force on any of the two particles is the force exerted by the other particle. Since ~ ~ ~ ˙ ~ ˙ F ij = −F ji and F 12 = ~p1 and F 21 = −~p2, we have that d (~p + ~p ) = 0, dt 1 2 or conservation of momentum. This is what the third law is really telling us. The weight force of a body is ~w = m~g. The mass m in this case is gravitational mass. Although gravitational mass and inertial mass are deﬁned diﬀerently, for any object ever measured, the two always had the same value. In fact, Einstein’s principle of equivalence states that

gravitational mass = inertial mass.

Some important concepts implied by Newton’s laws are 2 Newton’s Laws

Inertial frames: An inertial frame is any frame in which Newton’s first law holds. Fic- titious forces arise when not in an inertial frame. Galilean invariance: Any frame in uniform motion with respect to an inertial frame is also an inertial frame. Equation of motion: The equation of motion for bodies is Newton’s second law. The second law can be written as a second order differential equation with respect to position F~ = m~r¨. To get the position equation for a body, we integrate this twice ˙ with respect to time—getting two constants of integration ~r0 and ~r0 = ~v0. These constants are often fixed by initial conditions.

1.1 Many Particles

Suppose you have many particles spread throughout some space. Each particle, in this case, is experiencing an external force from outside the system as well as a force from every other particle. If you define some origin O, then the position of the ith particle can be represented by ~ri, and its momentum by ~pi = mi~ri. Newton’s second law implies d X ext ~p = F~ + F~ . dt i ij i j6=i What this states is that the time derivative of the momentum of the ith particle is equal to the sum of all the forces due to the other particles plus the external force on the particle. The index j 6= i simply means to sum over all j not equal to i. That is, sum over all other particles—ignoring the case where F~ ii since a particle doesn’t exert a force on itself. The total momentum of the system, which we denote with a capital P~ is the sum of the momenta of all the particles ~ X P = ~pi. i Since we can differentiate sums term by term, we know that ~˙ X ˙ P = ~pi. i ˙ Replacing ~pi with our result from above gives us   ~˙ X X ~ ~ ext X X ~ X ~ ext P =  F ij + F i  = F ij + F i . i j6=i i j6=i i Looking at the first term on the right, we can expand this as X X X X X X F~ ij = F~ ij + F~ ij. i j6=i i ji We can write the right side more compactly by combining the two terms and switching the subscripts on one of the force vectors X X X X F~ ij = F~ ij + F~ ji . i j6=i i j

This simpliﬁes to ˙ ext P~ = F~ .

In other words, the rate of change of the total momentum of the system depends only on the external forces on the system—not on the internal forces. This also implies that if the external forces on a system of particles is zero, then the total momentum is a constant (the indeﬁnite integral of 0 is a constant).

1.2 Cartesian Coordinates

A law of motion is the second order diﬀerential vector equation for the position vector ~r as a function of time F~ = m~r¨.

In Cartesian coordinates, the position vector is

~r = xxˆ + yyˆ + zzˆ = hx, y, zi, where x, y, and z are generally functions of t. Diﬀerentiating the position function is clean and easy since the unit vectors do not depend on time—they are just constants when we diﬀerentiate. Since acceleration is the second derivative of the position vector with respect to time, we have that

~r¨ = ~a =x ¨xˆ +y ¨yˆ +z ¨zˆ.

Since the components in the three directions are separable in Cartesian coordinates, we can write a force vector in the form

F~ = Fxxˆ + Fyyˆ + Fzzˆ.

This means we can treat Newton’s second law in 3D as three separate versions of Newton’s second law in 1D  F = mx¨  x ~ F = m~a =⇒ Fy = my¨  Fz = mz¨

These three ordinary diﬀerential equations (ODEs) are separable, so ﬁnding the solutions x(t), y(t), and z(t) is a simple matter of separating variables and integrating twice. For example, integrating Fx = mx¨, gives us the velocity function and a constant—the initial speed vx(t = 0) = vx0 . Integrating the result of that gives us the position function and a second constant—the initial position x(t = 0) = x0

F a =x ¨ = x = constant x m v (t) =x ˙ = x¨ dt = a dt = a t + v x ˆ x ˆ x x0 1 x(t) = (a t + v ) dt = a t2 + v t + x . ˆ x x0 2 x x0 0

We can use the same process in the y and z directions to ﬁnd vy(t), vz(t), y(t), and z(t). 4 Newton’s Laws

1.3 Polar Coordinates

φˆ rˆ If one or more of the forces involved is a central force, it may be easier to use polar • coordinates. Recall the transformation equations when going from polar to Cartesian or vice versa ~r x = r cos φ, y = r sin φ φ p y r = x2 + y2, φ = tan−1 . x Figure 1.1: Polar coordinates where r is the distance to the particle, and φ is the angle between ~r and the positive x-axis. The position vector in 2D polar coordinates, i.e., z = 0 is

~r = hx, y, zi = hr cos φ, r sin φ, 0i.

To find the velocity and acceleration vectors, we differentiate the position vector. This time it’s more complicated because r and φ are both functions of time, so we have to use the product and chain rules to differentiate the position vector to get

~v = ~r˙ = hr˙ cos φ − rφ˙ sin φ, r˙ sin φ + rφ˙ cos φ, 0i =r ˙hcos φ, sin φ, 0i + rφ˙h− sin φ, cos φ, 0i.

Diﬀerentiating again to get the acceleration vector gives us the following after rearranging

~a = (¨r − rφ˙2)hcos φ, sin φ, 0i + (2r ˙φ˙ + rφ¨)h− sin φ, cos φ, 0i.

This is a lot messier than it was with Cartesian coordinates. We can write these in a more condensed manner by noting that hcos φ, sin φ, 0i is the unit vector rˆ in the direction of ~r. Furthermore, h− sin φ, cos φ, 0i, is a unit vector that is at right angles to rˆ and points in the direction of increasing φ. We can verify that they’re at right angles to each other by noting that their dot product is zero. We know it’s pointing in the direction of increasing φ because when φ = 0, it is h0, 1, 0i. Therefore, we call this unit vector φˆ. So our unit vectors in polar coordinates are

rˆ = hcos φ, sin φ, 0i φˆ = h− sin φ, cos φ, 0i.

This allows us to rewrite our position, velocity, and acceleration in terms of the unit vectors rˆ and φˆ

~r = r rˆ ~v = ~r˙ =r ˙ rˆ + rφ˙ φˆ ~a = ~r¨ = r¨ − rφ˙2 rˆ + 2r ˙φ˙ + rφ¨ φˆ.

Notice, that unlike the unit vectors in Cartesian coordinates, these are changing with time. It’s as if the particle is carrying them along. How are the unit vectors changing with time? Diﬀerentiating them gives us d rˆ = φ˙h− sin φ, cos φ, 0i = φ˙φˆ dt d φˆ = φ˙h− cos φ, − sin φ, 0i = −φ˙rˆ. dt 1.3. Polar Coordinates 5

So the time derivatives of the unit vectors are

rˆ˙ = φ˙φˆ φˆ˙ = −φ˙rˆ.

This gives us an alternate way of writing ~v by replacing φ˙φˆ with rˆ˙ d ~v = ~r =r ˙ rˆ + rφ˙ φˆ =r ˙ rˆ + rrˆ˙, dt and an alternate way of writing ~a by diﬀerentiating the new form of ~v d d ~a = ~v = (r ˙ rˆ + rrˆ˙) =r ¨rˆ + 2r ˙rˆ˙ + rrˆ¨. dt dt We can write Newton’s second law in component form as

F~ = Frrˆ + Fφφˆ, where Fr is the force in the direction of rˆ and Fφ is the force in the direction of φˆ. This is not usually convenient since the force may be changing with time, since it depends on φ and r. From the law F~ = m~a and the equation ~a = (¨r − rφ˙2)rˆ + (2r ˙φ˙ + rφ¨)φˆ, we can separate Newton’s law into two components  ˙2 Fr = m r¨ − rφ F~ = m~a =⇒ ˙ ¨ Fφ = m 2r ˙φ + rφ

Example 1.3.1

Using polar coordinates, analyze the motion of a mass m being twirled around on a string of length R. From Newton’s second law in polar coordinates, we have that

 ˙2 ˙ ¨ F~ = Frrˆ + Fφφˆ = m r¨ − rφ rˆ + m 2r ˙φ + rφ φˆ

Since the length of the string R is ﬁxed, we have the r = R is constant. This implies thatr ˙ =r ¨ = 0, giving us F~ = −m Rφ˙2 rˆ + m Rφ¨ φˆ.

We know that the centripetal radial force is being applied by the string tension. Since the tension force points in the direction opposite of rˆ, we have that Fr = −T . We also know that there is no force in the φˆ direction in this idealized situation. So we have that F~ = −T~ T~ = mRφ˙2rˆ. ¨ ¨ ˙ Since Fφ = mRφ = 0, we know that φ = 0, which implies that φ is a constant. We know this constant as the angular velocity ω, so we have that T = mRω2, where Rω2 is the centripetal acceleration associated with uniform circular motion. This tells us that the rφ˙2 from the equation for ~a is just our familiar Rω2. 6 Newton’s Laws

Example 1.3.2

Using polar coordinates, analyze the motion of a simple pendulum with a weight of mass m on a string of length R. From Newton’s second law in polar coordinates, we have that

 ˙2 ˙ ¨ F~ = Frrˆ + Fφφˆ = m r¨ − rφ rˆ + m 2r ˙φ + rφ φˆ

The forces on the mass are T , the tension force in the string and ~w = m~g the weight. Using a free body diagram, we ﬁnd that

Fr = mg cos φ − T

Fφ = −mg sin φ.

Notice that T is negative since the tension force is pointing in the direction opposite of increasing r. Also, g is negative in Fφ since it points in the direction of decreasing φ. Equating these components with the ones we get from Newton’s law gives us

 ˙2 Fr = m r¨ − rφ = mg cos φ − T ˙ ¨ Fφ = m 2r ˙φ + rφ = −mg sin φ.

Since our string length is ﬁxed, we have that r = R is a constant, sor ˙ =r ¨ = 0, giving us

˙2 Fr = −mRφ = mg cos φ − T ¨ Fφ = mRφ = −mg sin φ.

Solving Fr for T gives us T = m Rφ˙2 + g cos φ .

¨ Solving Fφ for φ gives us g φ¨ = − sin φ, R which is a second order nonlinear diﬀerential equation. For |φ| << 1, we can use the small angle approximation sin φ ≈ φ to write g φ¨ ≈ − φ. R or the more familiar φ¨ ≈ −ω2φ, where ω = pg/R is the angular frequency. We know the general solution to this second order diﬀerential equation is

φ(t) = A cos ωt + B sin ωt. 1.4. Cylindrical Coordinates 7

z 1.4 Cylindrical Coordinates ρ In cylindrical coordinates, P 0 is the projection of a point P onto the xy-plane. Then φ is the angle between this projection and the positive x-axis, and ρ is the length of this • projection. Notice that ρ is not the distance from a particle at P to the origin, rather, it P is the distance from the particle P to the z-axis. The transformations between Cartesian and cylindrical coordinates are y x = ρ cos φ, y = ρ sin φ, z = z ρ p y ρ = x2 + y2, φ = tan−1 . φ • 0 x x P Figure 1.2: Cylindrical coordi- We can deﬁne directions in cylindrical coordinates in terms of the unit vectors ρˆ, nates φˆ and zˆ. These vectors are orthonormal (i.e. mutually orthogonal and of unit length). The unit vector ρˆ points in the direction of increasing ρ. The unit vector φˆ is orthogonal to ρˆ and points in the direction of increasing φ. Notice that φˆ and ρˆ are parallel to the xy-plane. The unit vector zˆ is orthogonal to both of the others and points parallel to the z-axis in the direction of increasing z. Their components in terms of Cartesian coordinates are

ρˆ = hcos φ, sin φ, 0i φˆ = h− sin φ, cos φ, 0i zˆ = h0, 0, 1i.

The position vector for a particle in cylindrical coordinates can be written as

~r = ρρˆ + zzˆ.

Diﬀerentiating this twice gives us

~r˙ =ρ ˙ρˆ + ρρˆ˙ +z ˙zˆ ~r¨ =ρ ¨ρˆ + 2ρ ˙ρˆ˙ + ρρˆ¨ +z ¨zˆ.

From 2-dimensional polar coordinates, we know that ρˆ˙ = φ˙φˆ and diﬀerentiating this gives us ρˆ¨ = φ¨φˆ + φ˙φˆ˙, and we know that φˆ˙ = −φ˙ρˆ, so we can rewrite the above two equations as

~r˙ = ~v = ρρˆ + ρφ˙φˆ +z ˙zˆ ~r¨ = ~a = ρ¨ − ρφ˙2 ρˆ + ρφ¨ + 2ρ ˙φ˙ φˆ + (¨z) zˆ.

Example 1.4.1

A particle of mass m is constrained to move at a distance of ρ = R from the vertical z-axis. Using Newton’s second law, ﬁnd the equations of motion for the particle. Newton’s second law in cylindrical components can be broken into components

F~ = Fρρˆ + Fφφˆ + Fzzˆ.

Using F~ = m~a and the equation for ~a in cylindrical coordinates, we can write the 8 Newton’s Laws

force components as

 ˙2 Fρ = maρ = m ρ¨ − ρφ ¨ ˙ Fφ = maφ = m ρφ + 2ρ ˙φ

Fz = maz = m (¨z) .

From the description of the problem, the only forces acting on the particle are the weight force ~w = m~g which is acting entirely in the negative z direction and the centripetal force, which is directed toward the z-axis.

 v2 F = m ρ¨ − ρφ˙2 = −m T = −mρω2 ρ ρ ¨ ˙ Fφ = m ρφ + 2ρ ˙φ = 0

Fz = m (¨z) = −mg.

We know that ρ = R is a constant which implies thatρ ˙ =ρ ¨ = 0. Simplifying the equations gives us

φ˙ = ω φ¨ = 0 z¨ = −g.

Integrating the ﬁrst equation gives us

φ(t) = ωt + φ0.

Integrating the third equation gives us

z˙ = vz(t) = −gt + vz0 .

The constant of integration is C = vz0 since vz(t = 0) = vz0 . Integrating again gives us 1 z(t) = − gt2 + v t + z . 2 z0 0

1.5 Air Resistance, Friction, and Buoyancy

Incorporating air resistance is a good way to introduce diﬀerential equations into our equations of motion. We know from experience that air resistance is a function of velocity, f(v). The faster you go through the air, the more air resistance you experience. Empirically, the magnitude of the force f~(v) of air resistance is

f(v) = bv + cv2,

2 where flin = bv is the linear air resistance and fquad = cv is the quadratic air resistance. The form of this function makes sense since any function f(v) will have a Taylor series expansion of the form f(v) = a + bv + cv2 + ··· , where a, b, and c are constants. In the case of air resistance, the leading constant a is zero since there is no air resistance when an object is not moving. 1.5. Air Resistance, Friction, and Buoyancy 9

We can also assume that the direction of the air resistance is opposite the direction of the object’s velocity. This is exactly true for non-rotating, spherical objects, but is ~v at best an approximation valid for non-rotating or non-spherical objects. Consider an airplane wing for example. A good portion of the air resistance force, the lift, is not in • the direction opposite of the plane’s motion. In our case, we’ll assume that

f~ = −f(v)vˆ. ~ f = −f(v)vˆ ~w = m~g Recall that vˆ = ~v/v. We’ll assume the object is a sphere. The constants b and c depend on the character- Figure 1.3: Air resistance istics of the object and the fluid. Empirically, we find that linear air resistance is due to the viscosity (i.e. viscous drag) of the fluid and the size of the object. ~v b ∝ viscosity · D, where D is the diameter of the sphere. It turns out from Stokes’ law that • b = 3πηD, flin = −b~v ~w = m~g where η is the viscosity of the fluid. We also find that quadratic air resistance is due to density of the fluid and the Figure 1.4: Linear air resistance cross-sectional area of the sphere c ∝ density · D2. We can understand this as meaning that quadratic air resistance is due to the fact that the object must accelerate the mass of the fluid that it collides with. Obviously, then, it will be related to both the density of the fluid and the cross-sectional area (D2) of the object. For a sphere in air at STP, we find that b = 1.6 × 10−4 (kg/ms) D c = 0.25 (kg/m3) D2.

Looking at the ratio of fquad and solving for v, we ﬁnd that if flin 6 × 10−4 m2/s v << , then f dominates D lin 6 × 10−4 m2/s v >> , then f dominates. D quad

For typical everyday objects moving at everyday velocities, fquad dominates.

Linear Air Resistance For a spherical object, the linear air resistance is

~ f lin = −b~v. The sum of forces acting on the particle are

F~ = m~g − b~v.

Linear air resistance is simple because we can still separate the components when using Cartesian coordinates

Fx = −bvx

Fy = −mg − bvy. 10 Newton’s Laws

Horizontal Motion with Linear Air Resistance

~v We can use the same process to consider friction for an object sliding on a plane (as shown in Fig. (1.5)) when the friction depends on the velocity. We will use this idea to f~ = −b~v m investigate the horizontal component of linear air resistance. Applying Newton’s law, we get for the x component, the separable diﬀerential equation

Figure 1.5: 1D horizontal mo- F = mv˙ = −bv tion with friction. x x x dv b x = −kv , k = dt x m 1 1 vx(t) − dvx = dt k ˆ vx ˆ 1 t = − ln v + C. k x

vx0 Then we ﬁx the arbitrary constant of integration C using the initial conditions vx(t =

0) = vx0 . In other words, when t = 0, vx = vx0 . Plugging these values in gives us 1 t 0 = − ln vx0 + C 0 τ k Figure 1.6: The exponential de- 1 C = ln vx , cay of speed for an object sliding k 0 with linear friction. so our complete equation is 1 1 1 vx t = − ln vx + ln vx0 = − ln . k k k vx0 x(t) What we actually want is the velocity in terms of t instead of the other way around. To invert the equation, we solve the right side for the natural logarithm, then raise both sides as powers to e to get

−kt − t v (t) = v e = v e τ , x0 + vx0 τ x x0 x0 where 1 m τ = = , t k b is an example of a characteristic time. x0 τ Figure 1.7: The position func- Notice that our function vx(t), shown in Fig. (1.6), is an exponential function. A tion of an object sliding with related function is the radioactive decay function where the characteristic time is the linear friction. decay constant. Another related function appears in circuits, where for example, the time constant might give the time it takes for a capacitor in a circuit to discharge to 1/e = 36.8% of its initial charge. In our case, τ is the time it takes for the velocity of the object to decay to 36.8% of its initial value. Integrating vx(t) gives us

− t − t x(t) = v e τ dt = −v τe τ + C. x0 ˆ x0

Plugging in the initial condition x(t = 0) = x0, gives us x0 = −vx0 τ +C, so C = x0 +vx0 τ and our complete equation is

 t − τ x(t) = x0 + vx0 τ 1 − e .

Notice that as t → ∞, x − x0 → vx0 τ, that is, vx0 τ is how far the object can travel. This is illustrated in Fig. (1.7). 1.5. Air Resistance, Friction, and Buoyancy 11

Check various limits of x(t) to make sure the solution behaves the way you would expect it to. If vx0 = 0 then x(t) = x0. This makes sense. If the object has no initial velocity, it stays at its starting place. What does the solution tell us if air resistance is negligible? When b → 0, τ = m/b → ∞ and t/τ → 0. Then t t2 x(t) = x + v τ 1 − 1 − + + ··· ' x + v t. 0 x0 τ 2τ 2 0 x0

Notice that we replaced e−t/τ with its Taylor expansion. When the air resistance is negligible, our result reduces to the familiar equation for 1D motion with constant velocity. −t/τ If we look at an infinitely viscous fluid, that is, τ → 0, we see that e → 0, so vx = 0 and x = x0. That is, the object goes from v = vx0 to v = 0 instantaneously. In other words, it comes to a stop instantaneously. No matter how complicated the air resistance function is, the object will eventually slow until flin dominates, and when it does, it has a finite stopping distance—vx0 τ. In general, for 1D motion with only a velocity-dependent force F (v), Newton’s second law gives us dv m v dv0 m = F (v) =⇒ dt = dv =⇒ t = m 0 . dt F (v) ˆv0 F (v ) Doing the integration gives you t as a function of v, t = t(v). You then need to invert the function to get v as a function of t, v = v(t). Once you have v(t), you can calculate the position function by integrating

t 0 0 x − x0 = v(t ) dt . ˆ0

Vertical Motion with Linear Air Resistance We turn now to vertical motion with linear air resistance—as with an object falling through the air. With vertical motion, the forces acting on the body are weight and air resistance. From Newton’s second law, for upward motion, we get

Fy = mv˙y = −mg − bvy. Notice that we chose the upward direction to be positive y. This becomes the first order separable differential equation dv b y = −g − kv , k = . dt y m Separating the differentials and integrating gives us dv dt = − y g + kvy t vy 1 dt0 = − dv0 ˆ ˆ g + kv0 y 0 vy0 y 1 vy 1 g + kv 0 y t = − ln(g + kvy) = − ln k vy0 k g + kvy0

We now have t in terms of v, but we want v in terms of t, so solving for vy gives us

g + kvy −kt − t = e = e τ , g + kvy0 which simpliﬁes to g g − t v (t) = − + + v e τ . y k k y0 12 Newton’s Laws

−t/τ Notice that for t >> τ, e → 0, so vy(t) → −g/k = −vter, the terminal velocity of the −vy(t) body, which can be written mg g v = = = gτ, (linear air resistance). ter b k Another way to ﬁnd the terminal velocity is to realize that it occurs when the acceleration v ter of a falling body is zero, that is, when the weight force is balanced by the air resistance, so bvter = mg =⇒ vter = mg/b = gτ. The expression for terminal velocity allows us to rewrite the equation for vy(t) as

t − τ t vy(t) = −vter + (vter + vy0 ) e . 0 τ If vy0 = 0, then the velocity of the body approaches the terminal velocity in the manner Figure 1.8: Velocity of an ob- shown in Fig. (1.8). ject falling against linear air re- To get the vertical position function of the object, we just integrate vy(t) sistance. t t 0 t 0 0 0 − t y − y = v (t ) dt = −v t − τ (v + v ) e τ . 0 y ter ter y0 ˆ0 0 0 This simpliﬁes to t − τ y(t) = −vtert − τ (vter + vy0 ) e − 1 .

If we calculate the terminal velocity of an object assuming that flin is the dominant resistive force, we should check that our assumption is valid. Our assumption is valid if

fquad << flin, when calculated at the terminal velocity. For an object falling in a dense ﬂuid, such as a liquid, another factor comes into play—that of buoyancy. The buoyant force on such an object is equal to the weight of the displaced ﬂuid

Fb = ρV g, where ρ is the density of the ﬂuid, V is the volume of the object, and g is the local gravitational acceleration. The buoyant force is directed opposite the gravitational force. The net force on an object falling vertically in a dense ﬂuid and experiencing only linear drag, is dv F = m y = 3πηDv + ρV g − mg. y dt y We’ve done horizontal and vertical linear resistance, and since they decouple, we can put them together to get the entire equation of projectile motion with linear air resistance. For the horizontal and vertical motions with x0 = y0 = 0, we have that

 t − τ x(t) = vx0 τ 1 − e

 t − τ y(t) = −vtert + τ (vter + vy0 ) 1 − e .

Recall that the range for a projectile, when not including air resistance, is 2v v R = x0 y0 . vac g What is the range when air resistance is taken into account? To ﬁnd the shape of the

trajectory, we solve x(t) for t to get t = −τ ln(1 − x/(vxo τ)), which we plug in to y(t) and simplify to get x vter + vy0 y(x) = vterτ ln 1 − + x. vxo τ vx0 1.5. Air Resistance, Friction, and Buoyancy 13

Fig. (1.9) shows the trajectory of a body with linear air resistance (solid line) and no air resistance (dashed line). Looking at the equation for x(t), we see that the limit as y x = vx0 τ t → ∞ is x(t) = vx0 τ, so the line x = vx0 τ is our vertical asymptote. Remember that x will never actually reach vx0 τ, but it asymptotically approaches it. The range R of the projectile under linear air resistance is the x-value when y(x) = 0, so R vter + vy0 0 = vterτ ln 1 − + R. x vxo τ vx0 y0 We can’t find a simple closed-form expression for R but we can approximate it by finding the first correction term for the limit of weak air resistance. In the limit of weak air resistance, τ → ∞, so Figure 1.9: The trajectory of a R projectile under linear air resis- << 1. tance. vx0 τ Using the Taylor expansion

1 1 ~v ln(1 − x) = − x + x2 + x3 + ··· , 2 3 • and replacing vter with gτ, we get ~ 2 ! f quad = −cv vˆ ~w = m~g R 1 R 2 1 R 3 gτ + v 0 = −gτ 2 + + + ··· + y0 R vxo τ 2 vxo τ 3 vxo τ vx0 Figure 1.10: Quadratic air resis- gτR gR2 gR3 gτR v R tance = − − − − · · · + + y0 v 2v2 3v3 τ v v xo xo xo x0 x0 gR2 gR3 v R gR2 gR3 v R = − − − · · · + y0 ≈ − − + y0 2v2 3v3 τ v 2v2 3v3 τ v xo xo x0 xo xo x0 2v v 2 R ≈ x0 y0 − R2. g 3vxo τ We now have a quadratic equation in R, and we could solve for R using the quadratic formula, but there’s no need to. We have only an approximation here, and there’s no need to solve an equation exactly if the equation is itself only approximate. We can approximate it by iterating on a calculator. We start by calculating the ﬁrst order approximation of R, that is, we ignore the second term in the equation above and then we improve that approximation by plugging that result in for R

2v v 2 2v v 2 2v v 4v R ≈ x0 y0 − x0 y0 ≈ x0 y0 1 − y0 . g 3vxo τ g g 3vter Note: This kind of approximation has its limitations. A quadratic equation can have two solutions. This method only works when there are two solutions and there is a large diﬀerence between the two solutions. This approximation method will then converge to the solution that is closest to zero.

Quadratic Air Resistance Quadratic air resistance is given by

~ 2 f quad = −cv vˆ = −cv~v.

From Newton’s second law, we get

d~v F = m = m~g − cv~v. dt 14 Newton’s Laws

q 2 2 The complication is the fact that v = vx + vy depends on both the x and y components. Newton’s second law for each component gives us c q v˙ = − v2 + v2v x m x y x c q v˙ = − v2 + v2v − g. y m x y y Now they are nonlinear and coupled, so we can’t solve them separately as we did with linear air resistance. Fortunately, if we focus only on one-dimensional motion, we don’t have to worry about the orthogonal component.

Horizontal Motion with Quadratic Air Resistance From Newton’s second law, we get dv F = m = −cv2. dt We solve this in the usual way, integrating and then solving for v in terms of t to get

v0 m v(t) = t , τ = . 1 + τ cv0

Notice that the characteristic time τ for quadratic air resistance is different from the one for linear air resistance. This one depends on the initial speed. In both cases, though, the characteristic time gives us the time frame in which the object slows appreciably. If we plot v(t), we find that there’s a slower decay than for linear air resistance. That is, an object under quadratic air resistance, doesn’t slow as quickly as an object under linear air resistance. At time τ, the object has a speed exactly half of its initial speed. To find the position function, we integrate v(t) as before to get

 t x(t) = x + v τ ln 1 + . 0 0 τ

Notice that this function grows logarithmically, surpassing any ﬁnite value of x. So under just quadratic air resistance, an object would never slow to a complete stop. In the real world, any object under quadratic air will eventually slow enough that linear air resistance takes over and brings it to a complete stop in a ﬁnite distance.

Vertical Motion with Quadratic Air Resistance For vertical motion, we have two forces—the weight force pointing downward and the quadratic air resistance pointing opposite the direction of motion. From Newton’s second law, for an object going up, we have dv F = m = −cv2 − mg, dt if we take the upward direction to be positive y. For an object going down, we have dv F = m = cv2 − mg, dt Because of the squared term v2, our equation won’t change sign, as it should, when the object goes from upward motion to downward motion. Because of that, we need to treat the two directions separately. In other words, for a single object thrown upwards, we 1.5. Air Resistance, Friction, and Buoyancy 15 need two equations—one for the upward portion of the trajectory and the other for the downward portion. For downward motion, the object reaches terminal speed when the air resistance equals the gravitational force. So we get that terminal velocity under quadratic air resistance is rmg v = (quadratic air resistance). ter c Looking at downward motion, we have that

 2 2 dv c 2 cv v = v − g = −g 1 − = −g 1 − 2 dt m mg vter 1 dt = − dv v2 g 1 − 2 vter t 1 v 1 dt0 = − dv0. v2 ˆ0 g ˆv0 1 − 2 vter

0 0 On the right side, we can make the substitution u = v /vter then du = (1/vter)dv and 0 dv = vter du. v vter vter 1 t(v) = − 2 du. g ˆ v0 1 − u vter We could use partial fraction decomposition, 1/(1 − u2) = 1/(2(1 − u)) + 1/(2(1 + u)), to rewrite and solve it as logarithms, but it gets very messy. From a table of integrals, we ﬁnd that the integral is the inverse of the hyperbolic tangent, so

v v v v v v t = − ter tanh−1 u ter = − ter tanh−1 − tanh−1 0 . g v0 g v v vter ter ter Solving for v gives us −1 v0 gt vd(t) = vter tanh tanh − . vter vter

For v0 = 0, that is, the object is dropped from rest, this simpliﬁes to

 gt vd(t) = −vter tanh . vter

What happens in the limit of weak air resistance? For weak air resistance vter >> gt, since we know that for no air resistance, the terminal velocity is inﬁnite. From the Taylor expansion tanh x = x − x3/3 + ··· , we see that tanh x ≈ x when x is small. So

gt v(t) = −vter + ··· ≈ −gt, vter and we know that v(t) = −gt when there is no air resistance, so this limit checks out. Finally, we can ﬁnd the position function for an object dropped from rest by integrating v(t). The result is

2 vter gt yd(t) = y0 − ln cosh . g vter

For the case of upward motion with quadratic air resistance, we’ll use a slightly diﬀerent trick. For upward motion, the air resistance and the gravitational force are both 16 Newton’s Laws

opposite the direction of the velocity. We start again with Newton’s second law, giving us dv m = −mg − cv2 dt dv c v2 = −g 1 + v2 = −g 1 + . dt mg vter Notice that we are again incorporating the terminal velocity despite this being upward motion. Terminal velocity is the asymptotic velocity for downward motion, but it’s also a useful quantity when dealing with upward motion. It is still deﬁned in terms of downward motion. We want to ﬁnd the velocity v as a function of height y (instead of time t). We could solve for v(t) and y(t), eliminate t and plug one into the other to get v(y), but it’s easier to use the chain rule dv dv dy dv = · = v · . dt dy dt dy The key is in noticing that dy/dt is just the velocity v. Using this, we get dv v2 v · = −g 1 + 2 dy vter 1 v dy = − 2 2 dv g 1 + v /vter y v 0 1 v 0 dy = − 02 2 dv . ˆy0 g ˆv0 1 + v /vter 02 2 We can make the substitution u = 1 + v /vter, then

2 2 2 2 2 1+v /vter 2 2 2 2 v 1 v 1+v /vter v 1 + v /v ter ter ter 0 ter y − y0 = − du = − ln u = ln 2 2 . 2g ˆ 2 2 u 2g 1+v2/v2 2g 1 + v /v 1+v0 /vter 0 ter ter So 2 2 2 vter 1 + v0/vter y(v) = y0 + ln 2 2 . 2g 1 + v /vter To ﬁnd v(y), we just have to solve for v. What is the maximum height of the object if it’s initial upward velocity is v0? Since the maximum height occurs when v = 0, we can just replace v with 0 in the equation above to get 2 2 vter v0 ymax = y(0) = y0 + ln 1 + 2 . 2g vter

For weak air resistance, vter >> v0. The Taylor expansion of ln(1 + x) is 1 1 1 ln(1 + x) = x − x2 + x3 − x4 + ··· 2 3 4 so 2 " 2 2 2 # 2 2 vter v0 1 v0 v0 1 v0 ymax = y0 + 2 − 2 + ··· = y0 + 1 − 2 + ··· . 2g vter 2 vter 2g 2 vter

So for an object starting at y0 = 0, we have that 2 2 v0 1 v0 ymax ≈ 1 − 2 . 2g 2 vter Notice that the ﬁrst term is our familiar result for no air resistance and the second term (i.e. the ﬁrst correction term) is negative, which is what we’d expect since with air resistance, the object will not travel quite as high as without air resistance. 1.6. Charged Particle in a Magnetic Field 17

General Motion with Quadratic Air Resistance In general, for motion that is not just horizontal or vertical, Newton’s second law gives

m~v˙ = m~g − cv2vˆ = m~g − cv~v.

The components are then q ˙ 2 2 m~vx = −c vx + vy vx q ˙ 2 2 m~vy = −mg − c vx + vy vy.

Notice that the x component of the velocity depends also on the y-component and vice versa, i.e., the equations are coupled. These equations can’t be solved analytically, but they can be solved numerically given speciﬁc initial conditions. We know, however, that after a long time, vx → 0 and vy → −vter. Plugging these values into the pair of equations gives us their limit after a long time ˙ m~vx = −cvtervx ˙ m~vy = −mg − cvtervy.

This tells us that if we wait a long time, the coupled equations eventually behave like the uncoupled case of linear air resistance. Motion in the x-direction asymptotically approaches a ﬁnite value.

1.6 Charged Particle in a Magnetic Field

Consider a particle with charge q moving in a magnetic ﬁeld in the positive z-direction, so B~ = Bzˆ. The force on the charged particle is

F~ = q~v × B~ .

Since B~ has only a component in the z direction, the cross product is easily calculated as

xˆ yˆ zˆ ~ F = q vx vy vz = q (vyBxˆ − vxByˆ) .

0 0 B

Newton’s second law gives us

mv˙x = qBvy, mv˙y = −qBvx, mv˙z = 0.

We can immediately see that the acceleration in the z-direction is 0 and therefore, the velocity in that direction is the constant initial velocity in that direction. This gives us the equation of motion in the z direction

z(t) = z0 + vz0 t.

Notice that v˙x = ωvy, v˙y = −ωvx, where ω = qB/m is the cyclotron frequency. If we diﬀerentiate the ﬁrst equation and plug it into the second equation, we get

2 v¨x = −ω vx, 18 Newton’s Laws

a second order diﬀerential equation, which we recognize as the equation for oscillatory motion, and it has the solution

vx = A cos ωt + B sin ωt.

A and B are the two constants of integration, which are determined by the initial conditions. If we let A = a cos δ and B = a sin δ, then we can write it in the form

vx = A cos ωt + B sin ωt = a cos δ cos ωt + a sin δ sin ωt = a cos(ωt − δ), √ where a = A2 + B2 and δ = tan−1(B/A). In this form, we call δ the phase angle. Diﬀerentiating this result for vx and plugging it into the equation we got forv ˙x and solving for vy, gives us vy = −a sin(ωt − δ). All together, the velocity vector for the charged particle in the magnetic ﬁeld is

~v = ha cos(ωt − δ), −a sin(ωt − δ), vz0 i .

A Complex Approach We will now solve the same problem of the charged particle in a magnetic ﬁeld in a more elegant manner involving a use of complex numbers that is a standard approach in physics. Recall that

v˙x = ωvy, v˙y = −ωvx. √ If we multiply the second equation by i = −1 and add it to the ﬁrst equation, we get

d (v + iv ) = ωv − ωiv . dt x y y x If we factor out −i from the right side, we get

d (v + iv ) = −iω(v + iv ). dt x y x y

If we let η = vx + ivy, we can write our ﬁrst order diﬀerential equation as

η˙ = −iωη,

which has the solution η(t) = Ae−iωt. Notice that this has only a single integration constant A. Where’s the second one? The number A is a complex number so there’s a real part and an imaginary part, and that’s where the integration constants lie. Using Euler’s formula eiθ = cos θ + i sin θ, we can write

η(t) = Ae−iωt = A(cos ωt − i sin ωt).

Since the real component of eiθ is cos θ and the imaginary component is sin θ, the complex number eiθ lives on the unit circle of the complex plane. The constant A is a complex number, but we can write it in polar form as

A = aeiδ, 1.6. Charged Particle in a Magnetic Field 19 where a is the magnitude of A (i.e. its distance from the origin in the complex plane) and δ is the angle it makes with the positive x-axis. Using this deﬁnition for A, we can rewrite the solution η(t) as =(η) η(t) = Ae−iωt = aeiδe−iωt = aei(δ−ωt). η(0) = A a This shows us that η(t) lies on the circle with the same radius as A and the angle between δ A and η(t) is ωt. So at t = 0, η(t) is at A, and at some later time t = t, η(t) makes an <(η) angle ωt with A. This is all illustrated in Fig. (1.11). Notice that η(t) moves clockwise ωt around the circle of radius a as time advances. In the complex form of the solution, the constants a and δ are the same as the • constants from our previous solution. Using Euler’s formula again, we can write η(t) = aei(δ−ωt) η(t) = aei(δ−ωt) = a (cos(ωt − δ) − i sin(ωt − δ)) . Figure 1.11: The complex form Recalling that η = v + iv , we can easily substitute between the complex solution and x y of the solution on the complex our earlier solution. plane. q 2 2 Notice that a = vx + vy = v is the magnitude of the transverse velocity of the charged particle. Recall that radial velocity is the velocity toward or away from the origin when dealing with circular motion in a polar coordinate system, and the transverse velocity is the velocity component perpendicular to the radial velocity. The transverse velocity is perpendicular to B, which points in the z direction since we are currently only considering the x and y components of the velocity. To ﬁnd the position function of the particle, we start by introducing another complex number ξ = x + iy, such that ξ = η(t) dt, ˆ since position is the integral of the velocity function. So that we only have to do a single integral (albeit over the complex numbers) we integrate the exponential form of η(t) rather than the expanded form containing sine and cosine. A ξ = η(t) dt = Ae−iωt dt = − e−iωt + C. ˆ ˆ iω Keep in mind that since we integrated over the complex numbers, C is a complex number. If we let D = −A/(iω) and C = X + iY , and we let the point (X,Y ) be our new origin, then our position function becomes ξ(t) = De−iωt. Because of the function being in complex polar form, we know that ξ lives on a circle with radius D in the complex plane, and at t = 0, ξ(0) = D = x0 + iy0. At some later time, t, ξ(t) has moved an angle ω clockwise around the circle. So for the position function, we have that A ia a ξ(t) = De−iωt = − e−iωt = ei(δ−ωt) = (sin(ωt − δ) + i cos(ωt − δ)) . iω ω ω The cyclotron orbit of the particle has a radius

q 2 2 a vx + vy = , ω ω where ω is the familiar angular frequency. Our results tell us that the particle moves in a circle in the complex plane, since we have x+iy, but due to the way we introduced/deﬁned the complex function, (x, y) is also the real position of the particle in a regular rectangular coordinate system. Since x and y describe a circle as t advances, the particle really is moving in a circle. 20 Newton’s Laws

1.7 Summary: Newton’s Laws

Skills to Master • Know Newton’s laws of motion • Be able to write down the position, velocity, and acceleration vectors for a particle in Cartesian, polar, cylindrical, and spherical coordinates • Be able to derive the unit vectors of polar, cylindrical, and spherical coordinates and write them in Cartesian form • Understand the time dependence of the unit vectors in diﬀerent coordinate systems • Obtain the equations of motion for a system by applying Newton’s second law in the most convenient coordinate system • Solve for the equations of motion of a projectile under linear air resistance • Calculate the terminal speed of an object under linear or quadratic air resistance • Solve for the equation of motion of a projectile under quadratic air resistance for the special case of 1D motion (either vertical or horizontal)

Newton’s laws of motion are: Newton’s laws imply that the time rate of change of the total momentum of a system of particles is equal First law: In the absence of forces, a body moves with to the net external force acting on the system uniform velocity. Second law: The acceleration of a body is propor- ~˙ ~ ext tional to the applied force and the proportional- P = F . ity constant is the inertial mass of the body F~ = m~a. To identify the equation of motion for a particle, use this procedure: Third law: For every action, there’s an equal but op- 1. Start with the general form of Newton’s second posite reaction law F~ = m~a 2. Identify the forces acting on the particle ~ ~ F ij = −F ji. 3. Decide which coordinate system to use and write down the acceleration ~a for that coordinate sys- Conservation of momentum is implied by the tem third law. 4. Equate component-wise the net forces from step Newton’s second law is a second order diﬀerential 2 and m~a from step 3 in the form F~ = m~a equation in the position of the moving body 5. Simplify by applying any constraints such as r˙ =r ¨ = 0 ~ ¨ F = m~r. 6. Solve for the coordinate components, integrating This ODE is called the “equation of motion,” and its as needed solution is the position function ~r(t) of the moving body. The total momentum of a system of particles is simply the sum of the momenta of the individual par- Newton’s Laws in Cartesian Coordinates ticles ~ X In Cartesian coordinates, the unit vectors xˆ, yˆ, and zˆ P = ~pi. i do not depend on time, and the position, velocity, and 1.7. Summary: Newton’s Laws 21 acceleration vectors are Newton’s Laws in Cylindrical Coordinates The transformations between Cartesian and cylindrical ~r = xxˆ + yyˆ + zzˆ coordinates are ~r˙ =x ˙xˆ +y ˙yˆ +z ˙zˆ x = ρ cos φ, y = ρ sin φ, z = z ¨ ~r =x ¨xˆ +xy ¨ yˆ +xz ¨ zˆ. p y ρ = x2 + y2, φ = tan−1 . x Then Newton’s second law, aka the equation of motion, In terms of Cartesian coordinates, the unit vectors in separates as cylindrical coordinates are  F = mx¨ ρˆ = hcos φ, sin φ, 0i  x ~ ˆ F = m~a =⇒ Fy = my¨ φ = h− sin φ, cos φ, 0i  Fz = mz¨ zˆ = h0, 0, 1i. The position, velocity, and acceleration of a particle in These can be integrated to get the velocity functions cylindrical coordinates is x˙(t),y ˙(t), andz ˙(t) and integrated a second time to get the position functions x(t), y(t), and z(t). ~r = ρρˆ + zzˆ ~r˙ = ~v = ρρˆ + ρφ˙φˆ +z ˙zˆ Newton’s Laws in Polar Coordinates ~r¨ = ~a = ρ¨ − ρφ˙2 ρˆ + ρφ¨ + 2ρ ˙φ˙ φˆ + (¨z) zˆ. The transformations to switch back and forth between Cartesian and polar coordinates are Air Resistance Empirically, the magnitude of the force f~(v) of air re- x = r cos φ, y = r sin φ sistance is p y 2 r = x2 + y2, φ = tan−1 . f(v) = bv + cv , x where flin = bv is the linear air resistance and fquad = In polar coordinates, the unit vectors are cv2 is the quadratic air resistance.

rˆ = hcos φ, sin φ, 0i Linear Air Resistance φˆ = h− sin φ, cos φ, 0i. For a spherical object, the linear air resistance is ~ The position, velocity, and acceleration can be written f lin = −b~v. as with b = 3πηD, ~r = r rˆ and D is the diameter of the sphere and η is the vis- ~v = ~r˙ =r ˙ rˆ + rφ˙ φˆ cosity of the ﬂuid. For a projectile in the presence of linear air resis- ¨ ˙2 ˙ ¨ ˆ ~a = ~r = r¨ − rφ rˆ + 2r ˙φ + rφ φ. tance, the net force acting on it is F~ = m~g − b~v. In polar coordinates, Newton’s second law, separates as This can be separated into Cartesian components  ˙2 Fr = m r¨ − rφ Fx = −bvx F~ = m~a =⇒ ˙ ¨ Fφ = m 2r ˙φ + rφ Fy = −mg − bvy. 22 Newton’s Laws

For the horizontal component, we can separate which gives us the pair of nonlinear, coupled ODES variables and integrate to get c q t v˙ = − v2 + v2v − τ x x y x vx(t) = vx0 e , m c q v˙ = − v2 + v2v − g. where τ = 1/k = m/b is a characteristic time. Inte- y m x y y grating again gives us the position function These can be solved in the 1D cases of horizontal mo- t − τ x(t) = x0 + vx0 τ 1 − e . tion only or vertical motion only. For horizontal motion, we can integrate to get For the vertical component, we can again separate variables and integrate to get v0 v(t) = t , 1 + τ g g − t v (t) = − + + v e τ . y k k y0 where τ = m/cv0. Integrating again gives us the posi- Integrating gives us the position function tion function

− t y(t) = −v t − τ (v + v ) e τ − 1 . t ter ter y0 x(t) = x + v τ ln 1 + . 0 0 τ Terminal velocity occurs when the downward weight force equals the upward drag force. For linear For vertical motion, we have to treat the upward air resistance, it is and downward cases separately. For an object travel- mg ing downward under quadratic air resistance, Newton’s v = . ter b law is dv F = m = cv2 − mg. For an object falling in a dense fluid, we have to dt include the buoyant force. The buoyant force on such an object is equal to the weight of the displaced fluid For an object dropped from rest, this has solution Fb = ρV g, gt vd(t) = −vter tanh . vter where ρ is the density of the fluid, V is the volume of the object, and g is the local gravitational acceleration. Integrating gives us the position function The buoyant force is directed opposite the gravitational force. The net force on an object falling vertically in a 2 vter gt dense fluid and experiencing only linear drag, is yd(t) = y0 − ln cosh . g vter

dvy Fy = m = 3πηDvy + ρV g − mg. Terminal velocity for an object falling under dt quadratic air resistance is Quadratic Air Resistance rmg v = . Quadratic air resistance is given by ter c f~ = −cv2vˆ = −cv~v. quad For an object traveling upward under quadratic Newton’s second law with quadratic air resistance is air resistance, Newton’s law is d~v dv F = m = m~g − cv~v, F = m = −cv2 − mg. dt dt Chapter 2

Momentum and Center of Mass

In an earlier section, we discussed the conservation of momentum. The momentum of the ith particle is ~pi = mi~vi, and the total momentum of a system of particles is

n ~ X P = ~pi. i=1 From Newton’s third law which states that every action has an equal but opposite reaction, we know that the time derivative of the total momentum is equal to the external force on the system (i.e. the internal forces cancel each other out). If the external force is zero, then the time rate of change of the total momentum is zero, so the total momentum is constant X P~ = mi~vi = constant. i This conservation of momentum greatly simpliﬁes the solutions to many probes. Consider a perfectly inelastic collision between a particle with mass m1 moving at ~v1 and a particle with mass m2 moving at ~v2. Because the particles stick together after the collision, the total mass is m1 + m2, and they move with the same ﬁnal velocity ~vf . Using conservation of momentum before and after the collision, we know that

m1~v1 + m2~v2 = (m1 + m2)~vf , which allows us to calculate the ﬁnal velocity as

m1~v1 + m2~v2 ~vf = . m1 + m2 That we can easily calculate the velocities of the particles after the collision is remarkable given the complex chemistry of an inelastic collision. Another application is that of a person of mass mp standing on a cart with wheels and mass mc, which is itself on a frictionless surface. If the person on the cart begins to run in one direction, the cart beneath him will move in the opposite direction. If the person on the cart runs toward the opposite end of the cart with speed vp relative to the cart, then the cart will move in the opposite direction with some speed vc relative to the ground. Relative to a stationary observer, the person is moving with speed vp − vc. From conservation of momentum, we know that

mcvc = mp(vp − vc).

Solving for vc gives us the speed of the cart

mp vc = vp. mc + mp 24 Momentum and Center of Mass

As always, it is a good idea to check our solution by looking at the limiting cases. If Tip mc → 0, we see that the equation reduces to vc = vp. Since your speed relative to the The tricky part of rocket stationary observer is vp − vc, this shows that you will remain stationary with respect to problems is that the mass the stationary observer. If mc → ∞, then vc → 0, so the cart remains stationary and the depends on time. person runs oﬀ of it at a speed vp.

2.1 Rocket with no External Force

A rocket, which has nothing to push against and nothing to be pushed by, moves by expelling mass. The conservation of linear momentum causes the rocket to move forward with the same but opposite momentum as the mass that is pushed out the back. A similar situation would occur if you were standing on a frictionless surface such as ice. With nothing to push against, how do you move to shore? Your only option is to throw something, such as a shoe, in the direction opposite the shore as hard as you can. To analyze a rocket, we’ll consider the system at a time t and at a later time t + dt. We don’t care what occurred before t or how massive the rocket and its fuel were before t. At time t, the rocket (and its fuel) has mass m, and the rocket is moving forward with speed v. At time t + dt, some amount of mass has been expelled in the form of exhaust. If we say the mass at this time is m + dm, where dm is less than zero, then the mass that has been expelled is −dm. The speed of the rocket (relative to a stationary observer) is now v + dv. If the speed of the exhaust (i.e. the expelled mass) is vex relative to the rocket, then the speed of the exhaust relative to a stationary observer is vex − v − dv. At time t, the total momentum of the system is

P (t) = mv.

At time t + dt, the total momentum of the system is

P (t + dt) = mom. of rocket + mom. of exhaust

= (m + dm)(v + dv) − (−dm)(vex − v − dv) = mv + m dv + vex dm.

Notice that the double infinitesimals dm dv cancel out, but even if they hadn’t we could have neglected them. Because a single infinitesimal is arbitrarily small, two of them multiplied together is effectively zero. So the change in momentum is

dP = P (t + dt) − P (t) = mv + m dv + vex dm − mv = m dv + vex dm.

Given that there are no external forces, the total momentum is conserved, meaning the change in momentum is zero

dP = m dv + vex dm = 0.

Dividing through by dt gives us

P˙ = mv˙ + vexm˙ = 0

mv˙ = −mv˙ ex.

The force −mv˙ ex on the right side is called thrust. Since the time rate of change of the mass is negative, the thrust ends up being a positive quantity. Returning to

m dv + vex dm = 0, 2.2. Multistage Rockets 25 we can ﬁnd the speed as a function of the mass by integrating.

m dv = −vex dm 1 dv = −v dm ex m v m 0 1 0 dv = −vex 0 dm ˆv0 ˆm0 m m v − v = v ln 0 . 0 ex m

Notice that vex is assumed to be constant.

2.2 Multistage Rockets

Notice from the velocity function for a rocket m v = v + v ln 0 , 0 ex m that to maximize v, we need to maximize vex and the ratio m0/m. To do that, we need a multistage rocket. To analyze a two stage rocket (where one stage is emptied of fuel and then jettisoned), we need to analyze it in parts. Let m0 be the initial mass of the rocket and v0 its initial speed. Then let m1 be the mass after the first stage’s fuel has been completely burned and v1 its speed. Then let m2 be the mass after the first stage is jettisoned. Since the first stage is basically disconnected rather than blown off, the rocket’s speed is still v1. Finally, let m3 be the mass after the second stage’s fuel has been completely burned and v2 its speed. After the first stage fuel is burned but before the first stage is jettisoned, we know from the rocket equation and the given quantities that the rocket’s speed is m0 v1 = v0 + vex ln . m1

For the second stage, we begin with an initial mass of m2, a ﬁnal mass of m3, and an initial speed of v1. So the speed after the second stage fuel has been burned is m2 v2 = v1 + vex ln . m3

Plugging in our value for v1 gives us m0 m2 m0 m2 v2 = v0 + vex ln + vex ln = v0 + vex ln . m1 m3 m1 m3

Notice that if the ﬁrst stage is massless, that is, m1 = m2, then we recover the form of the original equation for the single stage rocket.

2.3 Rocket with External Force

Consider a rocket near the surface of the earth where the external force on the rocket is no longer zero. We now have that

ext mv˙ = −mv˙ ex + F . In the case of vertical ascent in the presence of gravity, we have that

mv˙ = −mv˙ ex − mg, 26 Momentum and Center of Mass

where we takev ˙ to be in the positive direction. Assuming a constant burn rate, the rate of mass decrease is constant, som ˙ = −k. Then

mv˙ = kvex − mg dv k = v − g dt m ex v t 0 k 0 dv = vex − g dt ˆv0 ˆ0 m

However, we know that m depends on t according to m = m0 −kt, so we need to substitute this in to get

t k 0 m0 v − v0 = 0 vex − g dt = vex ln − gt. ˆ0 m0 − kt m0 − kt From this, we can find the height of the rocket as a function of time, y(t), by integrating v with respect to time, and we can find the acceleration a(t) by differentiating v with respect to time.

2.4 Center of Mass

For a group of particles mi with positions ~ri, the center of mass is

1 X R~ = m ~r , cm M i i i

P where M = i mi is the total mass of the particles. The center of mass can be decoupled into the components

1 X 1 X 1 X X = m x ,Y = m y ,Z = m z , cm M i i cm M i i cm M i i i i i

then R~ cm = hXcm,Ycm,Zcmi. ~ P ˙ Recall that total momentum is given by P = i mi~ri, which tells us that the total momentum of a system of particles is the total mass times the velocity of the center of mass ˙ MR~ cm = P~ . Also recall that the time derivative of the total momentum is the total external force ~˙ P ~ ext ~ ext P = i F i = F , which tells us that the total external force acting on a system of particles is equal to the total mass times the acceleration of the center of mass

¨ ext MR~ cm = F~ .

In other words, the center of mass of a system of particles obeys Newton’s second law F~ = m~a, as if it were a point particle located at the center of mass. Consider a projectile that is following a typical parabolic projectile path when it explodes into many pieces mi. The only external force being considered is the force of ~ ext P gravity, so F = i mi~g = M~g. From the fact that the total external force equals ¨ the total mass times the acceleration of the center of mass, we have that MR~ cm = M~g 2.4. Center of Mass 27

¨ and so R~ cm = ~g. In other words, the acceleration of the center of mass of the particles following the explosion is still just the acceleration due to gravity. This tells us that after the explosion, where the original projectile would have followed a perfect parabolic trajectory, now the center of mass of all the pieces follows that perfect parabolic trajectory. This allows us to predict the distribution of the pieces after they fall to the ground. Suppose you have two particles m1 and m2 with m1 at the origin of your coordinate system. If R~ cm is the vector from the origin to the center of mass of the two particles, and ~r is the vector from the origin to m2, then the center of mass is given by

m1~0 + m2~r m2 R~ cm = = ~r. m1 + m2 m1 + m2

So the distance between m1 and the center of mass is d1 = |R~ cm| = m2r/(m1 + m2), and the distance between m2 and the center of mass is d2 = |~r| − |R~ cm| = m1r/(m1 + m2), so d m 1 = 2 . d2 m1 We consider now a continuous mass distribution. We start by examining an elemental mass dm at ~r. To ﬁnd the center of mass, we must sum over all elemental masses, each multiplied by their position vector. Our sum becomes the integral

1 R~ = ~r dm. cm M ˆ

The mass density of this mass distribution depends on the position, so we denote it ρ(~r). Recall that density is mass divided by volume, so we can rewrite the elemental mass as dm = ρ(~r) dV , where dV is the volume element. Then

1 R~ = ~rρ(~r) dV, cm M ˆ where the total mass is just

M = ρ(~r) dV. ˆ

Example 2.4.1

Find the center of mass of a hemisphere of radius a and constant density ρ lying on the xy-plane and centered at the origin. The center of mass occurs at some point R~ = hX,Y,Zi. Due to the symmetry we can separate it into components and calculate the x-component, y-component, and z-component of the center of mass separately. However, whenever you do an integral, pause and see if you really have to do the integral. In this case, we can simplify the problem to ﬁnding only the z-component of the center of mass. Because of the symmetry, we know that the x and y components are just zero. So all we have to ﬁnd is 1 R = zρ dV. z M ˆ We’ll do the integral in spherical coordinates. Recall that the transformation relations are

x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ. 28 Momentum and Center of Mass

The volume element, which we should know or be able to derive quickly (via the Jacobian or a geometric picture) is

dV = r2 sin θ dθ dφ dr.

Our integral becomes

π 2π 2 a 1 3 Rz = ρr cos θ sin θ dr dθ dφ. M ˆ0 ˆ0 ˆ0 Since ρ doesn’t depend on any of the integrals, we can pull it outside. We can separate the integral into three, and since the φ integral doesn’t have anything in the integrand, it simpliﬁes to 2π.

π 2 a 1 3 Rz = ρ2π cos θ sin θ dθ r dr. M ˆ0 ˆ0 Evaluating the integral with respect to θ can be done with a straightforward 1 substitution u = sin θ after which it evaluates to 2 . Evaluating the other integral and simplifying, gives us ρπa4 R = . z 4M To calculate the total mass M, we could set up another triple integral, to evaluate ρ dV , but since we’re dealing with a hemisphere with constant density, we know that´ its volume is V = 2πa3/3, and so its mass is M = ρV = 2ρπa3/3. So Rz = 3a/8. The center of mass is then at

 3 R~ = 0, 0, a . cm 8

No matter how this oddly shaped object is tumbling through a vacuum, its center of mass will exactly follow the smooth parabolic path.

2.5 Angular Momentum of a Single Particle

Linear momentum doesn’t care where the origin is, but angular momentum does. When working with angular momentum, we have to specify an origin, since the angular momentum of a particle is about some point. For a particle with position vector ~r and linear momentum ~p, its angular momentum about the origin is

~l = ~r × ~p.

The time rate of change of the angular momentum of the particle is

d~l d = (~r × ~p) = ~r˙ × ~p + ~r × ~p˙. dt dt Notice that we can expand the derivative of the cross product using the regular product rule. The ﬁrst term on the right can be expressed as ~v × m~v, and since the cross product of parallel vectors is zero, that term drops out. Since ~p˙ = F~ we can rewrite the rightmost term as ~r × ~p˙ = ~r × F~, which is equal to the torque, ~Γ, due to F~ about the origin, so

d~l = ~r × F~ = ~Γ. dt 2.6. Angular Momentum of a System of Particles 29

If there is no torque on the particle, that is, ~Γ = 0, then the angular momentum ~l is constant, which gives us conservation of angular momentum. If F~ = 0, then ~v ~ ~ ~ ~ necessarily Γ = 0 and l is conserved. If F is a central force then ~r × F = 0 since the m cross product of parallel vectors is zero. Again, ~l is conserved. A central force is just a force parallel to ~r, and one reason they are special is that they don’t affect angular momentum. F~ Consider a planet of mass m orbiting the sun with mass M. Since the center of mass is close to the center of the system, we assume that the sun is at the origin. Let ~r point ~r from the sun to the planet. The force keeping the planet in orbit is the gravitational force F~ = −GMmrˆ/r2. Since the only force on the planet is a central force, the torque is zero and angular momentum is conserved. From ~l = ~r × ~p = ~r × (m~v), we know that ~r and ~v are both perpendicular to ~l. If M we let ~l point in the z direction, then ~r and ~v, which define the planet’s orbit, both lie in the xy-plane. Figure 2.1: A planet orbiting In a short time dt, the planet moves a small distance ~v dt, and the position vector ~r the sun. sweeps out a small area dA. Recall that the area of a parallelogram defined by two vectors is the magnitude of their cross product. This means that the area of a triangle defined by two vectors is half ~v dt of the magnitude of the cross product of the two vectors. So 1 1 1 dA dA = |~r × ~v dt| = |~r × m~v| dt = |~r × ~p| dt 2 2m 2m dA 1 = ~l . ~r dt 2m Since we know that ~l is constant, we know that dA/dt is constant, which means that the O area swept out by ~r is constant with respect to time. In other words, the radius sweeps out equal areas in equal time. This is Kepler’s second law. Figure 2.2: Kepler’s second law. We could also have used polar coordinates to describe the orbit of the planet with φ being the angle between ~r and some x-axis. Then

~r = rrˆ ~r˙ =r ˙rˆ + rφ˙φˆ

~l = ~r × m~r˙ = (rrˆ) × m(r ˙rˆ + rφ˙φˆ) = mrr˙(rˆ × rˆ) + mr2φ˙(rˆ × φˆ) = mr2φ˙(rˆ × φˆ) = mr2φ˙ zˆ.

2.6 Angular Momentum of a System of Particles

For a system of particles, the angular momentum of the ith particle about the origin is ~ li = ~ri × ~pi. Then the total angular momentum of the system of particles about the origin is ~ X~ X L = li = ~ri × ~pi, i i and the time rate of change of the total angular momentum is

~˙ X ˙ X ˙ X X ˙ X ˙ L = ~ri × ~pi + ~ri × ~pi = ~vi × m~vi + ~ri × ~pi = ~ri × ~pi. i i i i i Notice that the ﬁrst summation is zero since ~v k m~v, and the cross product of parallel vectors is zero. 30 Momentum and Center of Mass

Recall from Newton’s second law that the time rate of change of the momentum of the ith particle is the sum of the forces on i due to all the other particles plus the external force on i ˙ ~ ext X ~ ~pi = F i + F ij. j6=i Using this, we can rewrite the time rate of change of the total momentum as   ~˙ X ~ ext X ~ X ~ ext X X ~ L = ~ri × F i + F ij = ~ri × F i + ~ri × F ij. i j6=i i i j6=i

~ ext Since ~ri × F i is the external torque on the ith particle, the ﬁrst term is just the total external torque on the system about the origin

~ ext X ~ ext Γ = ~ri × F i . i The second term consists of pairs of terms of the form

~ri × F~ ij + ~rj × F~ ji,

where the ﬁrst term is the torque on particle i due to particle j and the second term is the torque on particle j due to particle i. If the internal forces obey Newton’s third law, then F~ ji = −F~ ij and we can write the above as

~ri × F~ ij + ~rj × F~ ji = (~ri − ~rj) × F~ ij.

In a system of particles where the internal forces are all central forces and Newton’s third law holds, we have that (~ri − ~rj) × F~ ij = 0, that is, all the internal torques vanish since ~ri − ~rj is parallel to F~ ij. This would not be the case if the internal forces were not central forces. For example, we could have a system of particles in which the forces between pairs of particles are equal and opposite but not pointing toward each other. In such a case, there would be internal torque and the angular momentum would increase. In our case, the internal torques cancel, so we have that the time rate of change of the total momentum of the system is equal to the total external torque on the system.

˙ ext L~ = ~Γ .

If the external torque is zero, then the angular momentum of the system is conserved.

2.7 Angular Momentum of a Continuous Mass Distribution

Consider a rigid, extended body rotating about the z axis with angular velocity ω. We can think of the rigid body as a system of particles where each particle is frozen into a conﬁguration such that their distances from each other and the angles between them are constant. If we consider the system of rigid particles in cylindrical coordinates, then the ith particle has coordinates (ρi, φi, zi). In rigid rotation about the z axis, ρi and zi are ˙ constant. Only the φi component changes with time. Thenρ ˙i = 0,z ˙i = 0, and φi = ω, and these are the same for all particles. So this is a very restricted motion compared to, say, a gas of particles in which the particles have the freedom to move in all directions. The position of the ith particle is given by the vector

~ri = ρiρˆ + zizˆ. 2.7. Angular Momentum of a Continuous Mass Distribution 31

Diﬀerentiating, gives us the velocity vector Tip Many problems can be ~r˙ =ρ ˙ ρˆ + ρ ρˆ˙ +z ˙ zˆ =ρ ˙ ρˆ + ρ φ˙ φˆ +z ˙ zˆ = 0 + ρ ωφˆ + 0 = ρ ωφˆ. i i i i i i i i i i i solved using conservation of angular momentum of Then the angular momentum of the ith particle about the origin ~l = ~r × (m ~r˙ ) can be i i i a system when the exter- written as nal torque on the system is h i zero. Furthermore, the an- ~l = (ρ ρˆ + z zˆ) × (m ρ ωφˆ) = m ρ ω ρ (ρˆ × φˆ) + z (zˆ × φˆ) = m ρ2ωzˆ − m ρ ωz ρˆ. i i i i i i i i i i i i i i gular momentum of a rigid and symmetric body can be The cross products of the unit vectors are obtained using the right-hand rule. We’re only found easily by ﬁrst calcu- concerned with the z-component of the angular momentum so lating its moment of inertia, then its angular momentum 2 liz = miρi ω, about its axis of rotation is L = Iω. then the z-component of the total angular momentum of the system of rigid particles is

X X L = l = m ρ2ω = Iω, z iz i i Tip i i For the moment of inertia, where I = ρ2 dm you must keep X 2 I = miρi , in mind´ that ρ is the dis- i tance of an elemental mass from the axis of rotation— is called the moment of inertia of the rigid body. The moment of inertia doesn’t depend not from the origin. This on the angular speed ω at which the body is rotating, it is only a parameter associated makes calculating the mo- with the mass distribution of the rigid body. The moment of inertia makes it easy to ment of inertia in spherical calculate the angular momentum of a rigid and symmetric body rotating about a given coordinates, for example, a axis little more tricky. Lz = Iω.

Depending on the shape and mass distribution of a body, a body rotating about the z axis will have angular momentum components in the x and y directions, but we are only considering the component in the z direction. For the continuous case, we consider an elemental mass dm instead of a particle mi, and our summation turns into an integral

I = ρ2 dm = ρ2 ρ dV, ˆ ˆ dens where ρ is the distance from the elemental volume to the axis of rotation, and ρdens is the mass density of the rigid body. Example 2.7.1

Calculate the moment of inertia of a ring with mass M and radius a. This should be immediately obvious as

I = Ma2,

since the moment of inertia is just the distance squared of an elemental mass from the axis of rotation times its mass, summed over all elemental masses. Since all the elemental masses are the same distance, a, from the axis of rotation, I is just Ma2. 32 Momentum and Center of Mass

Example 2.7.2 mi 0 ~ri Calculate the moment of inertia of a uniform disk with mass M and radius a. C.O.M. We know that ~ri I = r2 dm, ˆ R~ where r is the distance from an elemental mass dm to the axis of rotation. We know that the surface density of the disk is σ = M/A where the total mass is M and the surface area is A, so dm = σ dA O

I = σr2 dA. ˆ To convert to 2-dimensional polar coordinates, we make the substitution dA = r dr dθ, then our integral becomes

2π a a 1 I = σr3 dr dθ = σ2π r3 dr = σπa4. ˆ0 ˆ0 ˆ0 2 Since σ = M/A = M/(πa2), we have that 1 I = Ma2. 2

What if you want to know the angular momentum of a system about its center of mass when the center of mass is not at the origin of your coordinate system? Recall that the total linear momentum of a collection of particles is the total mass times the velocity of the center of mass. Is there an analogous thing for angular momentum? If we have a system of particles mi and associated positions ~ri, then their center of ~ 1 P mass is given by Rcm = M i mi~ri. We’ll start by redeﬁning our coordinate system so that the center of mass is at the new origin, then

0 ~ri = R~ cm + ~ri ,

0 0 where ~ri are the new position vectors for the particles. That is, ~ri points from the center of mass to the ith particle. Keep in mind that the center of mass won’t necessarily be in an inertial reference frame, since it may be accelerating relative to the original origin. Diﬀerentiating gives us ˙ ˙ ˙ 0 ~ri = R~ + ~ri . The angular momentum of the system about the original origin is

~ X X ˙ L = ~ri × ~pi = ~ri × mi~ri. i i

Substituting in the equations above gives us

X 0 ˙ ˙ 0 L~ = (R~ + ~ri ) × mi(R~ + ~ri ) i X h ˙ ˙ 0 0 ˙ 0 ˙ 0 i = mi (R~ × R~ ) + (R~ × ~ri ) + (~ri × R~ ) + (~ri × ~ri ) i X ˙ X ˙ 0 X 0 ˙ X 0 ˙ 0 = mi(R~ × R~ ) + mi(R~ × ~ri ) + mi(~ri × R~ ) + mi(~ri × ~ri ). i i i i 2.7. Angular Momentum of a Continuous Mass Distribution 33

P 0 P ˙ 0 The middle two sums drop out because i mi~ri = 0, and so i mi~ri is also zero. P 0 The sum i mi~ri gives the center of mass of the particles with respect to the center of 0 mass since it uses the ~ri vectors, and so the sum is zero. Our equation simpliﬁes to

˙ X 0 ˙ 0 L~ = R~ cm × MR~ cm + ~ri × mi~ri . i Notice that the ﬁrst term on the right is the angular momentum associated with the motion of the center of mass itself, and the second term is the total angular momentum of the system about the center of mass. Solving for the second term on the right, we have that L~ cm = L~ − R~ cm × P~ , so the angular momentum of a system about the center of mass is the angular momentum of the system about the origin minus the angular momentum of the center of mass about the origin. For example, if the sun is taken to be the origin of our coordinate system, but we want to know the angular momentum of Earth about its own center of mass, then we could calculate the angular momentum of the particles of Earth about the sun and subtract from that the angular momentum of Earth’s center of mass about the sun. ~˙ ~ ext P ~ ext ~˙ ~ ext P ~ ext Recall that P = F = i F i and that L = Γ = i Γi . Diﬀerentiating the equation above, gives us

d ˙ ˙ ˙ ext ext L~ = L~ − R~ × P~ − R~ × P~ = ~Γ − R~ × F~ dt cm cm cm cm X ~ ext X ~ ~ ext = ~ri × F − Rcm × F i i i X ~ ~ ext X 0 ~ ext ~ ext = (~ri − Rcm) × F = ~ri × F = Γcm . i i So the time rate of change of the angular momentum of a system about its center of mass is equal to the external torque on the system about the center of mass. This result holds even if the center of mass is accelerating. Recall that Newton’s laws are only valid in inertial frames. 34 Momentum and Center of Mass

2.8 Summary: Momentum and Center of Mass

Skills to Master • Use conservation of linear and angular momentum to solve problems • Calculate the center of mass of objects and systems of objects • Apply Newton’s laws to systems of objects • Calculate the torque acting on an object • Be able to derive Kepler’s second law from angular momentum conservation • Calculate the moment of inertia for various mass distributions

Linear Momentum The time rate of change of the angular momentum of a system about its center of mass is equal to The total momentum of a system of n particles is the external torque on the system about the center of n mass X P~ = ~p . d ~ ~ ext i Lcm = Γcm . i=1 dt The center of mass of a continuous mass distribu- If the external force is zero, then the time rate of change tion with density ρ(~r) is of the total momentum is zero, so the total momentum 1 1 is constant R~ = ~r dm = ~rρ(~r) dV, cm M ˆ M ˆ ~ X P = mi~vi = constant. where the total mass is i M = dm = ρ(~r) dV. This conservation of momentum greatly simpliﬁes the ˆ ˆ solutions to many probes. When integrating over a mass to ﬁnd the center of mass we can often use symmetry and integrate in cylindrical Center of Mass or spherical coordinates to greatly simplify the problem. To do that, we need to remember the volume For a group of particles mi with positions ~ri, the center of mass is elements in those coordinate systems. ~ 1 X Rcm = mi~ri, • Cylindrical coordinates (x, y, z) → (ρ, φ, z), M i dV = ρ dρ dφ dz P • Spherical coordinates (x, y, z) → (r, θ, φ), dV = where M = i mi is the total mass of the particles. The center of mass of a system of particles obeys r2 sin θ dr dθ dφ Newton’s second law as if it were a point particle at ˙ R~ cm. That is, MR~ cm = P~ , where P~ is the total mo- Angular Momentum ~¨ ~ ext ~ ext mentum and MRcm = F where F is the net The angular momentum of a particle about the origin external force. is The angular momentum of a system about its cen- ~l = ~r × ~p. ter of mass is the angular momentum of the system The torque on a particle is the time rate of change of about the origin minus the angular momentum of the its angular momentum center of mass about the origin d~l ~ ~ ~ ~ ~Γ = = ~r × F~. Lcm = L − R × P . dt 2.8. Summary: Momentum and Center of Mass 35

If F~ is a central force, then it applies no torque since where ρ is the distance to the elemental mass from ~r × F~ = 0. If either F~ = 0 or ~Γ = 0, then angular the axis of rotation (not the origin) and ρdens is the momentum is conserved. density. Some common moments of inertia are The total angular momentum of a system of particles is just the sum of the individual angular momenta. • Point mass at distance r: I = mr2 The time rate of change of the total angular momen- • Hoop or cylindrical shell of radius R: I = mR2 tum, provided that all internal forces are central forces, • Rod of length L: I = mR2/12 is just the net external torque on the system • Disk or solid cylinder of radius R: I = mR2/2 • Sphere of radius R: I = 2mR2/5 ˙ ext L~ = ~Γ . Recall that the moment of inertia is the rotational If there’s no net external torque on the system, then analogue of mass and that the rotational analogue of its angular momentum is conserved. Newton’s second law is For a rigid body rotating about the z axis, the z-component of the angular momentum is Γ = Iα,

Lz = Iω, where Γ is the torque and α is the angular acceleration. Likewise, L = Iω is the rotational analogue of ˙ z where ω = φ is the angular velocity of rotation and p = mv.

X 2 When solving problems, determine if angular mo- I = miρi . mentum is conserved. If it is, then you know that, i ~ ~ For a continuous mass distribution, Li = Lf

~ri × ~pi = ~rf × ~pf 2 2 I = ρ dm = ρ ρdens dV, ˆ ˆ ripi sin θi = rf pf sin θf . Chapter 3

Energy

For a particle of mass m traveling at speed v, the kinetic energy is

1 T = mv2. 2

Notice that v2 = ~v · ~v. Substituting this allows us to take the time derivative of the kinetic energy as dT 1 = m ~v˙ · ~v + ~v · ~v˙ = m~v˙ · ~v. dt 2 From Newton’s second law, we know that m~v˙ = F~, and ~v dt = d~r is just the infinitesimal displacement along the particle’s path, which gives us dT = F~ · d~r. This is the work done by the force F~ in the displacement d~r, and we write dW = F~ · d~r. So we can think of work as the differential of the kinetic energy with respect to displacement. For a particle moving from point 1 to point 2 along a path, the total work done by the force is calculated by summing over all the differential d~r’s

2 W1→2 = F~· d~r. ˆ1 Since we are summing over a path that is not necessarily a straight line, the integral above is a line integral. Earlier, we showed that dT = F~ · d~r. This implies that

2 ∆T = T2 − T1 = F~· d~r. ˆ1 So the work done in moving the particle from 1 to 2 is the same as the change in kinetic energy. This is called the work-energy theorem. A force is conservative if it satisﬁes the following two conditions • The force depends only on position and not on velocity or time or any other parameter. This excludes air resistance, for example. • The work done in moving a particle from point 1 to point 2 is independent of the path taken. This excludes friction, for example, since a longer path would obviously mean more work is done. Kinetic friction, for example, fails both of the above since it depends on velocity and the path. Energy 37

Example 3.0.1

Consider a particle being moved from point 1 to point 2 in a uniform gravita- ~ 2 ~ 2 tional ﬁeld F = −mgzˆ. The work done is W1→2 = 1 F · d~r = 1 −mgzˆ· d~r. Since ´ 2 ´ zˆ·~r is just the z component of d~r, this simpliﬁes to 1 −mg d~z = −mg(z2 − z1) = −mg∆z, so the work done depends only on the change´ in the height of the particle and not on its path. Since gravitational force depends only on position, it is a conservative force.

If we have a conservative force, we can define a potential energy for that force. To do that, we first have to define some reference point ~r0 where we define the potential energy to be zero, then the potential energy of the particle at position ~r is the negative of the work done for the force to move the particle from the reference point to the new point

~r ~ ~0 U(~r) = −W~r0→~r = − F · dr . ˆ~r0

Potential energy and work are related in that if a conservative force does work on a particle, the result is an increase in potential energy, and if the force does negative work, the result is a decrease in potential energy. A simple example is that of dropping an object. The force of gravity does negative work when the object drops and there is a resulting decrease in the object’s potential energy. We know that for a particle being taken from point 1 to point 2, the change in kinetic energy is related to the work done via T2 − T1 = W1→2. But for a conservative force, this equals the work done in taking the particle from point 1 to a reference point 0 plus the work done taking the particle from point 0 to point 2, so T2 − T1 = W1→0 + W0→2. The work done in taking the particle from point 0 to point 1 is the negative of the work done taking the particle from point 1 to point 0, so we can write T2 − T1 = −W0→1 + W0→2. Now on the right side we have two potential energy functions—the potential energy at point 1 relative to the reference point and the potential energy of the particle at point 2 relative to the reference point, so we can write T2 − T1 = U1 − U2. This gives is the important result that for conservative forces, mechanical energy, the sum of the kinetic and potential energy, is conserved

E = T + U.

We have related the potential energy function (a scalar function) to the integral of a vector function F~. Now, we want a way to ﬁnd the force F~ given the potential energy function U(~r). We know that dW = F~ · d~r = Fx dx + Fy dy + Fz dz. For a conservative force, we also know that dW = −dU. We can write the diﬀerential dU as dU = ∇~ U · d~r. This tells us that dW = F~ · d~r = −dU = −∇~ U · d~r, which implies that

F~ = −∇~ U.

That is, the force on a particle is the negative of the gradient of the particle’s potential energy function. Example 3.0.2

Calculate the force if the potential energy function is

U(~r) = mgz. 38 Energy

Since there is only a z and no x or y, the gradient is simply

∇~ U = mg zˆ,

so the force is F~ = −∇~ U = −mg zˆ.

Example 3.0.3

Using Cartesian coordinates, ﬁnd the force on a particle if its potential energy varies as the inverse of r. We have that α α U(~r) = = , r px2 + y2 + z2 where α is an arbitrary constant. We know that ∂U ∂U ∂U F~ = −∇~ U = − xˆ − yˆ − zˆ. ∂x ∂y ∂z Taking the partial derivative with respect to x, we have that

∂U 1 3 αx 2 2 2 − 2 = − α(x + y + z ) (2x) = − 3 . ∂x 2 (x2 + y2 + z2) 2

Notice that we can rewrite this as ∂U αx = − . ∂x r3 The other two partials have the same form, so αx αy αz α α α F~ = xˆ + yˆ + zˆ = (xxˆ + yyˆ + zzˆ) = ~r = rˆ. r3 r3 r3 r3 r3 r2 Gravity is an example of this kind of force. The potential function varies with inverse distance and the force varies with inverse distance squared.

Example 3.0.4

Using spherical coordinates, ﬁnd the force on a particle if its potential energy varies as the inverse of r. We have that α U(~r) = , r where α is an arbitrary constant. We know that ∂U 1 ∂U 1 ∂U F~ = −∇~ U = − rˆ − θˆ − φˆ. ∂r r ∂θ r sin θ ∂φ Since our potential function is only a function of r, we only have to worry about the ﬁrst term in the gradient. The rest are zero. ∂U α ∇~ U = = − rˆ. ∂r r2 So α F~ = −∇~ U = rˆ. r2 Energy 39

Given a force, can we deﬁnitely determine if it is conservative or not? The answer is yes, and it derives from Stokes’ theorem. A force F~ is conservative if and only if its curl is zero ∇~ × F~ = ~0.

Example 3.0.5

Determine if F~ = hay, bx, 0i is a conservative force. Taking the curl of F~, we ﬁnd that

∇~ × F~ = (b − a) zˆ,

so the force is conservative if a = b.

Once we know that a force is conservative, it’s straightforward to do the line integral

~r U(~r) = − F~· dr~0. ˆ~r0 Since we know the force is conservative, we can pick any convenient path to integrate over. We can also let the reference point be the origin ~r0 = h0, 0i, and let the ﬁnal point be an arbitrary position ~r = hx, yi. Then the most convenient path would probably be the path from the origin along the x-axis to (x, 0) and from there, vertically to (x, y). Then we can easily split the line integral into a pair of regular integrals—one with respect to x and the other with respect to y. Since y = 0 along the entire path integrated with respect to x, this integral will be 0 if Fx contains a factor of y. Example 3.0.6

Find the potential function for

F~ = hay, ax, 0i.

Earlier, we conﬁrmed that this force is conservative, so we know that we can ﬁnd a unique potential function

~r U(~r) = − F~· dr~0. ˆ~r0 We let our reference point be the origin and the ﬁnal point be (x, y). Then we integrate on a path from the origin to (x, 0), and from there to (x, y), so we can split the integral into

x y x y 0 0 0 0 U = − Fx dx − Fy dy = − ay dx − ax dy . ˆ0 ˆ0 ˆ0 ˆ0 Since y = 0 along the entire path integrated with respect to x, the ﬁrst integral is zero, and we have y U = − ax dy0 = −axy. ˆ0 To conﬁrm, we check that −∇~ U = F~, and it does.

Any force that depends only on x and not on the velocityx ˙ is automatically conservative because there’s only one path. If a force depends only on position, then the work done in moving the particle from point 1 to point 2 is the same even if the particle takes 40 Energy

x2 0 0 a curvy path. The work done in that case is W1→2 = F (x ) dx and the potential x1 Tip x 0 0 energy is U(x) = − F (x ) dx , where x0 is the reference´ point. x0 “One-dimensional” in me- ´ chanics generally means 3.1 General One-dimensional Systems that a system can be completely characterized A second case in which the force is guaranteed to be conservative is in a one-dimensional by a single parameter. system. In this case, 1-dimensional, doesn’t necessarily mean that the particle is moving It doesn’t mean the mo- in a single spatial dimension, but that the particle’s position can be described using a tion is in a single spatial single parameter. For example, a bead sliding on a curvy wire is a 1D system since its dimension. position along the wire can be specified by a single number—the length of wire between it and where it started. Of course there is a normal force exerted by the wire on the bead, but since this force is always perpendicular to the bead’s motion, it does no work. It is U(x) just a constraint force. E Another example of a 1-dimensional system is an Atwood machine with a frictionless and massless pulley. In this case, the tension force is a constraint force and on both strings it does work, but since they cancel, there is no net work done. This is true in general—constraint forces do no net work. x In general, for a one dimensional system (where the force depends only on the particle’s position x) in which the forces are all conservative, the potential energy is x U(x) = − F (x0) dx0, Figure 3.1: A plot of the po- x ˆx0 tential and total energies for a generic 1D system. the force is dU F = − , x dx Tip and the total mechanical energy is conserved and 1 It is often helpful to con- E = mx˙ 2 + U(x) = constant. sider extreme examples 2 when trying to intuit the This is a separable differential. After solving forx ˙ 2 and taking the square root of both behavior of a specific case. sides, we can separate variables and solve: For example, what happens if a box in equilibrium r 2 x˙ = ± pE − U(x) on top of a cylinder is m nudged. We can consider rm 1 the extreme case where dt = ± dx 2 p the box is much smaller E − U(x) r x than the cylinder, and the m 1 t(x) = ± dx0. other extreme where the p 0 2 ˆx0 E − U(x ) box is much larger than the cylinder. Clearly, the The sign on the right side is fixed by the initial velocity of the particle. Is it moving left behavior depends on their or right? The initial position x0 is x(t = 0). To find x(t), the position as a function of relative sizes. t, we only have to invert the result found above. To find the velocity of the particle we would then differentiate x(t). It is often helpful to plot U(x) and the total energy E on the same graph as shown in Fig. (3.1). From a graph like this, there are several things we can say qualitatively about the motion of the object. From the conservation of mechanical energy, we know that the particle will have the largest velocity when U(x) is at its least. We know it will have the lowest velocity when U(x) is at its greatest. This graph shows the potential energy of an oscillator. If the particle is heading to the right of x = 0 its velocity will be decreasing since the potential energy is increasing. Eventually it will reach a stopping point and then accelerate back toward x = 0. Since it is impossible for the potential energy U(x) to be greater than the total energy E, we know that regions on either side of the intersections of U(x) and E are 3.1. General One-dimensional Systems 41 forbidden zones as shown in Fig. (3.2). These intersections are turning points where the U(x) velocity of the particle reaches exactly zero and changes sign. Clearly, this value of x is the amplitude of the oscillator. Consider the potential energy function shown in Fig. (3.3). In Fig. (3.4) is the graph with the turning points noted and the forbidden regions identified. Notice that with this potential energy, a particle with initial position within the first allowed region will oscillate with a large amplitude in that region, and a particle with initial position within the second region will oscillate with small amplitude in that region alone. The behavior −A A of the system is different based on the initial conditions. Figure 3.2: Forbidden regions. dU Recall that the force on the particle is the negative of dx , so we can infer just by looking at any point on the graph, which direction the force is pointing and what its magnitude is. At any point x on the graph, the force is the negative of the first derivative U(x) dU of U(x). This means for points where dx = 0, such as at local minimums and maximums, E the force is zero. A particle placed exactly at any of those points will just sit there at rest. We call these equilibrium points. Notice that there are two kinds of equilibria. x At a local minimum, if a particle is nudged in either direction, the force on it will point d2U back toward the equilibrium. So when dx2 > 0 at that point, it is a stable equilibrium point. At a local maximum, if a particle is nudged in either direction, the force on it will point away from the equilibrium and the particle will fly away from that point. So when d2U Figure 3.3: Potential and total dx2 < 0 at that point, it is an unstable equilibrium point. d2U energy. If we get dx2 = 0, this doesn’t tell us whether the equilibrium is stable or not. In such a case, we may be able to expand U(x) as a Taylor series, differentiate it, then examine the higher terms.

Example 3.1.1

Given the potential function U(x) = 2 cos x + x sin x, determine the stability at x = 0. The ﬁrst derivative is U 0 = − sin x + x cos x. Evaluating the ﬁrst derivative at x = 0 shows that U(x) has an equilibrium point there. To know if it’s a stable or unstable equilibrium point, we evaluate the second derivative at that point. The Figure 3.4: Turning points and second derivative is U 00 = −x sin x. This is 0 at x = 0, the second derivative test forbidden zones. doesn’t tell us if this is a stable or unstable equilibrium point. To answer this, we can expand U(x) about the point 0 to get U(x) x2 x4 x3 U(x) = 2 cos x + x sin x = 2 1 − + − · · · + x x − + ··· E1 2 24 6 x x4 = 2 − + O(x6). 12 E2 From this, we can see that the second derivative of U(x) near x = 0 is negative as zero is approached from either direction since U 00 = −x2 + O(x4). So the equilibrium point is unstable although barely since for x near x = 0, the quantity Figure 3.5: Electron potential x2 will be very small. well.

The plot in Fig. (3.5) represents a more interesting potential energy function. If the particle’s energy is E2 then it is constrained to oscillate in a small allowed region. If its energy is E1 it becomes unbounded and escapes to inﬁnity. Think of the behavior of an electron near its host atom at x = 0. Notice that the electron can never actually reach the nucleus because that region is forbidden. If its energy is low, it is bound to its host atom and will oscillate close to it indeﬁnitely. If its energy is large, however, it will escape this potential well and leave the atom. 42 Energy

Example 3.1.2

Consider an object in free fall, disregarding air resistance. We know that the gravitational force is conservative, and the object’s potential energy function is

U(y) = −mgy,

where +y is taken to be the downward direction. If the mass is dropped from rest, and we take the reference point for the potential energy function to be the y position where the object was dropped from, then the object’s total energy is 0. This is because its potential energy at y = 0 is taken to be zero and its kinetic energy is zero at the moment it is dropped from rest. Then using the formula we derived earlier

rm y 1 r 1 y 1 r 1 y r2y t(y) = √ dy0 = √ dy0 = 2py0 = . 0 0 2 ˆ0 mgy 2g ˆ0 y 2g 0 g Inverting this gives us the familiar result 1 y(t) = gt2. 2

Example 3.1.3

Analyze the energy characteristics of a harmonic oscillator, such as a spring, obeying Hooke’s law. Hooke’s law gives us the force on the object

F (x) = −kx.

Notice that this is a ﬁrst-order (i.e. linear) restoring (notice the negative sign) force. From this, we should already be able to picture the potential energy function. If the object is at its equilibrium position then a nudge in either direction will make it want to return to equilibrium, so the potential energy function should be an upward opening graph with a local minimum at the equilibrium position x = 0. Calculating the potential energy function, we have that

x x 0 0 0 1 2 U(x) = − Fx(x ) dx = k x dx = kx . ˆx0 ˆx0 2 We see that it is an upward opening quadratic centered at the equilibrium position as shown in Fig. (3.6). For a given energy E, we can determine the amplitude A of the object’s oscillations by seeing where E intersects U(x) as shown in Fig. (3.7). When the object reaches x = A, its velocity will be zero, so its kinetic energy will be 0, and so its total energy must equal its potential energy at x = A, that 1 2 is, E = 2 kA . We can now plug E and U(x) into the formula we derived earlier to get its equation of motion

rm x 1 rm x 1 t(x) = ± dx0 = ± √ dx0. p 0 2 02 2 ˆx0 E − U(x ) k ˆx0 A − x

Notice that the time it takes the object to go from x = 0 to x = A can be calculated by integrating this integral from x0 to A. The period (i.e. one cycle), τ, of the oscillator is then 4 times this quantity. In one cycle, the object goes from x = 0 3.2. Atwood Machine 43

to x = A to x = −A and back to x = 0. So U(x)

rm A 1 τ = 4 √ dx0. 2 02 k ˆ0 A − x Making the change of variables u = x0/A lets us express the integral in dimension- x less variables rm 1 1 τ = 4 √ du. 2 k ˆ0 1 − u Figure 3.6: Harmonic oscillator From this, we can see that the period of a harmonic oscillator does not depend potential well. on its amplitude. Noting that the integral above is the arcsin, we can go on to evaluate it to get rm U(x) τ = 2π . k E

If you have an integral, it is always a good idea to try to write it in terms of dimensionless variables. This allows you to see the relations between the diﬀerent variables. In the example above, we made the substitution u = x/A and since both x and A have units of length, the result u is dimensionless. This allowed us to determine that the period of x a harmonic oscillator does not depend on the amplitude without us having to actually evaluate the integral. In general, try to extract dimensionful variables from integrals by making a natural change of variables. Usually then, we don’t even need to know the ac- −A A Figure 3.7: Harmonic oscillator tual value of the integral—it is just some number. What we care about is the physically potential well. interesting relations between the dimensionful variables. This cannot be overemphasized.

3.2 Atwood Machine

Here, we analyze an ideal (frictionless, massless pulley and massless string) Atwood machine using an energy approach. Calculating the potential energy of an Atwood machine can seem tricky given that one mass will be moving down while the other mass will be moving up. However, recall that potential energy is just the negative of the work done, and the work being done is clear. We can ignore the work done by the tension forces since the net work done by the tensions is zero. We can ignore the tension force from the beginning because we’re dealing with a 1-dimensional system, and the tension is just a constraint force. All we have to calculate is the work done by the gravitational force. In the case of an Atwood machine, the work done can be calculated as just the gravitational force times the distance since the angle between the force and the direction of motion is zero. If block M moves downward a distance H, the work done against the gravitational force is −MgH since the object is falling. At the same time, block m moves up a distance H, so the work done on that block is mgH. The total work done is

W = −MgH + mgH = (m − M)gH, and in general, if M moves down a distance x, then

W = (m − M)g(H − x).

Since potential energy is the negative of the work, we have that

U(x) = g(M − m)(H − x).

Notice that if M = m, then U(x) = 0. This makes sense since if the two masses are the same, the system won’t move—it has zero potential energy. If M > m, then in the 44 Energy

Tip first Atwood machine in Fig. (3.8), we have that U(x) = g(M − m)H. This makes sense since the potential energy for an object in a gravitational field is just U = mgH, and in For 1 dimensional systems, this case, the small mass m effectively reduces the total mass. In other words, this is the you can extract an equa- potential energy of a single object at a height H with mass M − m. In the third Atwood tion of motion by differ- machine shown in Fig. (3.8), we have that U(x) = g(M − m)(H − x) = 0 since x = H. entiating the total energy Since we’ve defined M > m, we know that with M resting on the floor, the system is with respect to time. When at rest—its lowest potential energy. The same equations still work if m > M, the only doing this, be careful to difference is that the potential energy instead of going from some positive maximum to use the product rule where zero will go from a maximum at zero to a negative value. you need to... don’t make We can also define the potential energy of the Atwood machine in terms of the length the mistake of differentiat- of the string using the pulley as the reference point rather than the floor. In this case, ing with respect to position the potential energy can be calculated as −Mgx − mg(L − x) where L is the length of when you need to differen- the string. The signs are negative since we’re measuring downward rather than upward. tiate with respect to time. The total energy of the Atwood machine is just the sum of the kinetic and potential Previously, we only knew energies 1 1 how to generate an equa- E = Mx˙ 2 + mx˙ 2 − Mgx − mg(L − x). tion of motion by looking at 2 2 the net forces on an object We know that mechanical energy is conserved, that is, E = constant, so we can differen- and applying Newton’s sec- tiate the energy with respect to time to get us the equation of motion ond law to it. dE 1 1 = M(2x ˙x¨) + m(2x ˙x¨) − Mgx˙ + mgx˙ = 0 dt 2 2 0 = (M + m)x ˙x¨ − (M − m)gx˙ 0 = (M + m)¨x − (M − m)g M − m x¨ = g. M m M + m x Notice that when differentiatingx ˙ 2 with respect to time, we had to use the product rule. If the pulley is massive, the only difference is that you must include the pulley’s H M m rotational kinetic energy Iω2/2 in the total energy.

x m M 3.3 Spherically Symmetric Central Forces Another whole class of forces that are automatically conservative are spherically-symmetric Figure 3.8: Atwood machine. central forces. In spherical coordinates, a general potential energy function has the form U(~r) = U(r, θ, φ), that is, it is a function of r, θ, and φ. A spherically symmetric central force is the special case in which U depends only on the distance r from the origin and not on either of the angles. That is, U = U(r). In this case, the force can be calculated as F~ = −∇~ U(r). Applying the gradient operator for spherical coordinates, we get

∂U F~ = − rˆ, ∂r

∂U ∂U since ∂θ = ∂φ = 0. We can see that it is a central force since it has a component only in the r direction. Any spherically symmetric central force has the form

∂U F~ = f(r) rˆ, f(r) = − . ∂r A force is a central force if it can be written in the form f(~r). It is, further, spherically symmetric if it can be written in the form f(r). A central force is conservative if and only if it is spherically symmetric. 3.4. The Energy of a System of Particles 45

Given U(r), we can calculate F~ using the equation above. Given F~ = f(r) rˆ, we can calculate U(r) as r U(r) = − f(r0) dr0. ˆr0

3.4 The Energy of a System of Particles M m For a system of particles, we can calculate the total kinetic energy by summing the individual kinetic energies. If the external force acting on a system of particles is conservative, H − x x that is, it depends only on the position of a particle, then the external force on particle i is related to the potential energy via

~ ext ~ ext Figure 3.9: Atwood machine. F (~ri) = −∇iUi (~ri).

Consider the case of two interacting particles. If the position of particle 1 is ~r1 and the position of particle 2 is ~r2, then the vector pointing from particle 2 to particle 1 is just ~r = ~r1 − ~r2. The force on particle 1 due to particle 2 is then a function of ~r − ~r , that is F~ = 1 2 12 x L − x F~ 12(~r1 − ~r2). In other words, the force on particle 1 due to particle 2 could depend on the position of both particles, but only in the form ~r − ~r . Since the force between two 1 2 m particles is the same regardless of their position (provided that their relative positions stay M the same), we can conveniently translate the system so that ~r coincides with the origin 2 Figure 3.10: Atwood machine. of our coordinate system. This property is called translational invariance. We can now treat the system as a single particle under an external force, that is F~ 12 = F~ 12(~r1). Provided that the force between the particles is conservative, we can now deﬁne a Tip potential energy function for the pair of particles such that One way to get the equation F~ 12 = −∇~ 1U(~r1), of motion for a 1D system is to diﬀerentiate the total when particle 2 is at the origin, and energy with respect to time.

F~ 12 = −∇~ 1U(~r1 − ~r2), when not at the origin. Note that ∇~ 1 is the gradient with respect to particle 1 ∂ ∂ ∂ ∇~ 1 = xˆ + yˆ + zˆ. ∂x1 ∂y1 ∂z1 In words, the force on particle 1 due to particle 2 is the negative of the gradient (with respect to particle 1) of the potential energy function of the system. For the force on particle 2, we can say that F~ 21 = −∇~ 2U(~r1 − ~r2), that is, the force on particle 2 due to particle 1 is the gradient at particle 2’s position of the same potential energy function. From Newton’s third law, we know that F~ 12 = −F~ 21, so −∇~ 1U(~r1 − ~r2) = ∇~ 2U(~r1 − ~r2). We can now write the total potential energy for a system of two particles as

ext ext U = U1 (~r1) + U2 (~r2) + U12(~r1 − ~r2). It is the sum of the potential energy of particle 1 due to the external force, the potential energy of particle 2 due to the external force, and the internal potential energy of the system. Notice that we don’t have to include a second internal potential energy function 46 Energy

since both particles experience the same internal potential energy. To conﬁrm, we can take the negative gradient of U at particle 1 to get the total force on particle 1

~ ~ ext ~ ~ ext ~ −∇1U = −∇1U1 (~r1) − ∇1U12(~r1 − ~r2) = F 1 + F 12,

which we know is the total force on particle 1 in the system of two particles. Notice that ~ ext the term −∇1U2 (~r2) drops out. The total potential energy for a system of n particles is

n n n X ext X X U = Ui (~ri) + Uij(~ri − ~rj), i i j>i

where the first term is the total potential energy due to external forces and the second term is the total potential energy due to the internal forces of the system. It may seem that we’re ignoring half of the potential energies in the second term, but keep in mind that Uij(~ri − ~rj) is the potential energy of the interaction forces between particles i and j and is experienced by both particle i and particle j. The work-energy theorem also holds for many particles, that is, the total kinetic energy and the total potential energy is constant. The total kinetic energy of a system of particles is just the sum of the individual kinetic energies X 1 T = m~r˙ 2. 2 i i Putting these two results together, we have that the total energy of a system of n particles is n n n X 1 X X X E = m~r˙ 2 + U ext(~r ) + U (~r − ~r ), 2 i i i ij i j i i i j>i where the first term is the total kinetic energy, the second term is the total potential energy due to external forces and the last term is the total potential energy due to the internal forces. An important special case of this is that of a rigid body. For a rigid body, the internal forces (i.e. the interatomic forces) are central forces and so the potential energy does not depend on the direction of (~ri − ~rj). Furthermore, by definition of a rigid body, the quantity |~ri − ~rj| does not change for any given particles i and j. Therefore, for rigid bodies, the potential energy due to the internal forces is just some constant, which we can neglect. For a rigid body, we can also split the total kinetic energy into the translational 2 2 kinetic energy Mvcm/2 and the rotational kinetic energy Icmω /2. So the total energy of a rigid body can be expressed as

1 1 E = Mv2 + I ω2 + U ext + constant. 2 cm 2 cm 3.5. Summary: Energy 47

3.5 Summary: Energy

Skills to Master • Calculate the kinetic energy of a moving object • Identify whether a force is conservative and describe what this implies • Calculate the work done in moving an object against a force • Use conservation of mechanical energy to solve problems • Calculate the force given the potential energy and vice versa • Determine the qualitative behavior of a particle in 1D from a graph of its potential energy

Kinetic energy is Typically, we choose the reference point to be the origin ~r0 = (0, 0), and the final point to be ~r = (x, y), then we 1 1 T = mv2 = m~v · ~v. can break this line integral into a pair of straight line 2 2 integrals, the first from (0, 0) to (x, 0) and the second The work done in moving a particle from 1 to 2 is from (x, 0) to (x, y), then x y 2 0 0 U(~r) = U(x, y) = − Fx dx − Fy dy . W1→2 = F~· d~r. ˆ0 ˆ0 ˆ1 If F~ is in 3-dimensions, we have to add a third in- The work-kinetic energy theorem states that tegral for the path in the z-direction. A force is a spherically-symmetric central force if U = U(r), that ∆T = T2 − T1 = W1→2. is, if the potential energy depends only on r. Then using the gradient operator for spherical coordinates, we To calculate the work done, if given the initial and final get F~ = − ∂U rˆ. Given F~, we can calculate U(r) using speeds, we can just use the work energy theorem. ∂r U(r) = − r f(r0) dr0. A force F~ is conservative if and only if any of the r0 The force´ vector function can be calculated from following are true the scalar potential function by taking the gradient • It depends only on position F~ = −∇~ U. • The work done is independent of the path taken • It’s a one dimensional system, that is, the posi- Given a conservative force field F~, we can learn a tion is described by a single parameter. lot from it: • It is a spherically-symmetric central force 1. Verify that it is conservative. • ∇~ × F~ = ~0 2. Calculate the potential function using U(~r) = − ~r F~· d~r. ~r0 A conservative force implies that 3. For´ 1-dimensional systems, we can go further: • A potential energy function can be defined a) Plot U(x). • Mechanical energy is conserved, that is, E = b) Characterize the equilibrium points of U(x) T + U = constant. by differentiating once to find them and a second time to determine whether they’re Given a conservative force F~, we can define a po- stable or unstable. If d2U/dx2 < 0 it is un- tential energy function stable. If d2U/dx2 > 0 it is stable. If the second derivative test is inconclusive, use ~r Taylor series to expand U(x) and then take U(~r) = − F~· d~r. the second derivative of that. ˆ~r0 48 Energy

c) For small oscillations about a stable equilib- forces and the last term is the total potential energy rium point, d2U/dx2 evaluated at the equi- due to the internal forces. librium point gives the effective spring con- The total energy of a rigid body is stant k. Then the angular frequency of the p oscillations is ω = k/m and the period of 1 2 1 2 ext E = Mv + Icmω + U + constant. the oscillations is τ = 2π/ω. 2 cm 2 d) Write t(x) as an integral by solving E = mx˙ 2/2 + U(x) forx ˙ and integrating. If A is For a one-dimensional system, find the position given, you can calculate E since at x = A, function x(t) by differentiating the total energy and ˙ ˙ v = 0, so E = U(A). Then solve the inte- solving the resulting differential equation T + U = 0. gral for t(x). If given the amplitude A, find When differentiating, remember to differentiate with the period τ by integratingx ˙ from 0 to the respect to t rather than x, using the product rule where amplitude A and multiplying by 4. appropriate. e) Invert t(x) to get x(t). The total energy of an ideal Atwood machine with f) Calculate the velocity functionx ˙(t) by dif- masses M and m and string length L is just the sum ferentiating the position function x(t). of the kinetic and potential energies The total energy of a system of n particles is 1 2 1 2 n n n E = Mx˙ + mx˙ − Mgx − mg(L − x). X 1 X X X 2 2 E = m~r˙ 2 + U ext(~r ) + U (~r − ~r ), 2 i i i ij i j i i i j>i If the pulley is massive, the only difference is that where the first term is the total kinetic energy, the sec- you must include the pulley’s rotational kinetic energy ond term is the total potential energy due to external Iω2/2 in the total energy. Chapter 4

Oscillations

4.1 Simple Harmonic Oscillators

Consider a generic 1-dimensional potential energy due to a conservative force. We are typically interested in the behavior of the system in the small oscillations near a stable equilibrium. If the generic potential energy function U(x) has a stable equilibrium at Tip some point, then we redefine the coordinate system so that this equilibrium point is at x = 0. We can Taylor expand U about the equilibrium point x = 0 as Any system near a stable equilibrium point can be dU 1 d2U approximated as a simple U(x) = U(0) + x + x2 + ··· dx 0 2 dx2 0 harmonic oscillator. We know that dU/dx evaluated at x = 0 is 0 because of the equilibrium point there. The first term, U(0), is just some irrelevant constant that we can neglect by redefining the reference point of the system. If we let

d2U k = , dx2 0 (an eﬀective spring constant), then

1 U(x) = kx2, 2 for small oscillations about the equilibrium point. This is the potential energy of a simple harmonic oscillator. The key point is that we can approximate any system near a stable equilibrium point as a simple harmonic oscillator. Since force is related to potential energy via F (x) = −dU/dx, we get a linear restoring force F (x) = −kx, called Hooke’s law. Applying Newton’s second law gives us the equation of motion for the oscillating object r k x¨ = −ω2x, ω = . 0 0 m

The quantity ω0 is the angular frequency of the oscillation. The period is τ0 = 2π/ω0. The solution of this second order diﬀerential equation is

0 0 x(t) = A cos(ω0t) + B sin(ω0t).

Taking the second derivative of x(t) and plugging it intox ¨ given above, veriﬁes the solution. Since this form of the solution contains both a cosine and a sine term, the 50 Oscillations

overall motion of the oscillator isn’t easy to visualize. However, two particular cases are U(x) easy to visualize. If the oscillator is pulled to x = x0 and released at t = 0, then the sine term vanishes, and the motion is completely described by the cosine term. If the oscillator is at x = 0 and is given a sharp impulse at t = 0, then the cosine term vanishes, a x and the motion is completely described by the sine term. Using U0 1 1 cos x = (eix + e−ix), sin x = (eix + e−ix), 2 2i we can also write the solution in the form Figure 4.1: The Morse poten- iω0t −iω0t tial. x(t) = C1e + C2e ,

0 0 0 0 where C1 = A /2 + B /(2i) and C2 = A /2 − B /(2i). The benefit of this form of the solution is that the exponentials are easy to work with—easy to integrate and differentiate. However, it isn’t easy to visualize the motion of the oscillator in this form. We can write x(t) a third way if we define an angle δ. Consider the right triangle 0 0 with angle δ, hypotenuse A, adjacent√ side A = A cos δ, and opposite side B = A sin δ. Then δ = tan−1(B0/A0) and A = A02 + B02. Then using the trigonometry identity cos(α + β) = cos α cos β − sin α sin β, we can write

0 0 x(t) = A cos(ω0t) + B sin(ω0t) = A cos δ cos(ω0t) + A sin δ sin(ω0t)

= A (cos(ω0t) cos δ + sin(ω0t) sin δ) = A cos(ω0t − δ).

The nice thing about this form of the solution is that A is the real amplitude of the motion and the phase angle δ is the real phase shift of the motion. With this form of the solution, the motion of the oscillator is very easy to visualize—it is simply a cosine curve with amplitude A and shifted to the right by δ. From here, we can go to a third form of the solution by noting that

i(ω0t−δ) Ae = A cos(ω0t − δ) + Ai sin(ω0t − δ).

On the left side, we have Aei(ω0t−δ) = Aeiω0te−iδ. Taking the real part of both sides gives us iω0t Re Ce = A cos(ω0t − δ), where C = Ae−iδ is a complex constant. The signiﬁcance of this solution is that the particle is undergoing uniform circular motion in the complex plane, but its projection on the real axis is undergoing simple harmonic motion. From here, we can go back to any of the other solutions again. For example, using Euler’s formula again and the exponential form of the cosine function, we get

A Re Ceiω0t = A cos(ω t − δ) = ei(ω0t−δ) + e−i(ω0t−δ) 0 2 A A = e−iδeiω0t + eiδe−iω0t = C eiω0t + C e−iω0t. 2 2 1 2

−iδ iδ where C1 = Ae /2 and C2 = Ae /2. So for a simple (i.e. undamped and undriven) harmonic oscillator with angular oscillation frequency ω, we can write the displacement function x(t) in four diﬀerent ways:

x(t) = A0 cos(ωt) + B0 sin(ωt) x(t) = A cos(ωt − δ) iωt −iωt x(t) = C1e + C2e x(t) = Re Ceiωt 4.1. Simple Harmonic Oscillators 51

Example 4.1.1

The Morse potential

2 − x−a U(x) = U0 1 − e δ − U0,

as graphed in Fig. (4.1), is a simple model for 1D motion of a diatomic molecule. The quantity x is the distance between the atoms. The quantity U0 is a convenient constant, U(x = a) = −U0. To ﬁnd the equilibrium point(s), we set the derivative of U equal to zero. Diﬀerentiating gives us

dU 2U0 − x−a − x−a = 1 − e δ e δ . dx δ Setting it equal to zero, we ﬁnd that an equilibrium point occurs at x = a. This tells us that a is the average bond length. That is, if the diatomic molecule is disturbed, it will oscillate and eventually settle to a distance of x = a between the atoms. Recall that F (x) = −dU/dx. By examining, the equation for dU/dx, we can see that for x < a =⇒ x − a < 0 =⇒ −(x − a)/δ > 0 =⇒ a strong repulsive force between the atoms. On the other hand, x > a =⇒ x − a > 0 =⇒ −(x − a)/δ < 0 =⇒ an attractive force between the atoms. Near x = a, the diatomic molecule acts like a harmonic oscillator, and to ﬁnd the proportionality constant k (from Hooke’s law), we start by taking the second derivative of U(x) since we know that

dU d2U F = − = −kx, implies = k. dx dx2 x=0 We can also think of it in terms of a Taylor expansion about x = a. Taylor expanding, we get that 2 − x−a x − a 1 x − a e δ = 1 − + + ··· . δ 2 δ So for |(x − a)/δ| << 1, we have that

 x − a2 x − a2 U U(x) ≈ U 1 − 1 − − U ≈ U − U ≈ 0 (x − a)2 − U . 0 δ 0 0 δ 0 δ2 0

2 2 2 Notice that the term U0(x − a) /δ is analogous to kx /2, so we have that 2U k = 0 . δa Diﬀerentiating U(x) gives us the force dU 2U F (x) = − = − 0 (x − a), dx δ2 which is just a translated version of Hooke’s law. If we pretend that we’re dealing with a single mass now (e.g. one atom might be much larger than the other), then we can calculate an equation of motion by applying Newton’s second law. We have that 2U mx¨ = − 0 (x − a). δ2 52 Oscillations

If we make the substitution z = x − a, then we get the straightforward diﬀerential equation 2 z¨ = −ω0z, p where ω0 = k/m is the angular frequency of the oscillation—something which can be measured with a spectrometer, and k is the value given above. The solution to this ODE is z(t) = A cos(ω0t − δ), and when we back-substitute, we get

x(t) = A cos(ω0t − δ) + a.

Diﬀerentiating x(t) = A cos(ω0t − δ), gives us the velocity of the oscillator as a function of time v =x ˙ = −ω0A sin(ω0t − δ). Then its kinetic energy as a function of time is 1 1 T = mv2 = mω2A2 sin2(ω t − δ). 2 2 0 0 Its potential energy is 1 1 U = kx2 = kA2 cos2(ω t − δ). 2 2 0 Its total mechanical energy is 1 1 T + U = mω2A2 sin2(ω t − δ) + kA2 cos2(ω t − δ) 2 0 0 2 0 1 1 = kA2 sin2(ω t − δ) + kA2 cos2(ω t − δ). 2 0 2 0 This simpliﬁes to 1 T + U = kA2. 2 So the total energy of a harmonic oscillator depends only on its amplitude. An interesting and perhaps counter-intuitive note on oscillators is that an impulse applied to an oscillator does not change the oscillator’s frequency. It only changes its amplitude. For example, if you kick an oscillating mass attached to a spring, the mass won’t oscillate faster or slower, it will just oscillate with a larger amplitude. This is because an impulse changes the kinetic energy of the oscillator, therefore, it changes the amplitude, but not the frequency. The frequency depends only on the mass and the spring constant, and these are system characteristics rather than initial conditions.

4.2 Damped Harmonic Oscillators

All simple harmonic oscillators (i.e. undamped) have a natural frequency ω0. No matter how far such an oscillator is displaced from the equilibrium position, it will oscillate p at its natural frequency. This is because ω0 = k/m depends only the physical properties of the oscillator, k and m, and not on the amplitude A or the initial position or velocity of the oscillator. We will now deal with damped harmonic oscillators, which will no longer be oscillating at their natural frequencies. In the real world, oscillators are subject to friction forces, which damp the oscillations. The oscillations of a block on a spring and a pendulum will both die out in the long-term due to friction forces. The sound of a ringing bell (produced by the oscillations of the metal) will also die out after a time. Damping removes energy from the system. 4.2. Damped Harmonic Oscillators 53

A harmonic oscillator subject to a velocity dependent friction force is a damped oscillator. The equation of motion becomes

mx¨ = −kx − bx.˙ This is usually written in the form

2 x¨ + 2βx˙ + ω0x = 0, (4.1) p where ω0 = k/m is the natural frequency or the regular “undamped” angular frequency, and β = b/2m is the damping parameter. We can also write this more compactly in operator notation as Dx = 0, where D is the linear differential operator d2 d + 2β + ω2. dt2 dt 0 The solution to Eq. (4.1) is rich with multiple interesting regimes. Since it is a second order differential equation, there will be two independent solutions. If we have two solutions, then by the principle of superposition, the linear combination of the two solutions will also be a solution. Since it is a second order ODE, we need two linearly independent solutions, which we combine to get the general solution. The following method will work for any second order linear homogeneous differential equation. We start by letting x(t) = ert, thenx ˙ = rert andx ¨ = r2ert. Substituting these into the equation of motion gives us

2 x¨ + 2βx˙ + ω0x = 0 2 rt rt 2 rt r e + 2βre + ω0e = 0 rt 2 2 e r + 2βr + ω0 = 0 2 2 r + 2βr + ω0 = 0. This last equation is called the auxiliary equation. In words, x(t) = ert is a solution if 2 rt rt 2 rt 2 2 r e + 2βre + ω0e = 0, and this is true if the quadratic equation r + 2βr + ω0 = 0. Using the quadratic formula, we see that this equation is zero in two cases: q 2 2 r1 = −β + β − ω0 q 2 2 r2 = −β − β − ω0. The general solution has the form

r1t r2t x(t) = A1e + A2e . In our case, the general solution is √ √ −βt β2−ω2t − β2−ω2t x(t) = e A1e 0 + A2e 0 .

There are three distinct cases: 2 2 1. β − ω < 0, that is, β < ω0. In this case, the exponents on e are imaginary. This is the regime of underdamping. 2 2 2. β − ω = 0, that is, β = ω0. In this case, the exponents on e are zero. This is the regime of critical damping. 2 2 3. β − ω > 0, that is, β > ω0. In this case, the exponents on e are real. This is the regime of overdamping. Notice that if β = 0, we recover the undamped solution, as expected. 54 Oscillations

Underdamped Oscillator x(t) A The general solution for damped oscillations is √ √ −βt β2−ω2t − β2−ω2t x(t) = e A1e 0 + A2e 0 . t 2 2 In the underdamped case, β − ω0 < 0, which means the exponents in the solution are p 2 2 p 2 2 imaginary. By factoring a negative out of the radical, we get β − ω0 = i ω0 − β , −A and now the radical is a real quantity. We deﬁne this quantity, which is a frequency lower p 2 2 than the natural frequency ω0, as ω1 = ω0 − β . Now our solution can be written as x(t) = e−βt A eiω1t + A e−iω1t . Figure 4.2: Position function of 1 2 an underdamped oscillator. This is a complex quantity, but in the end, this doesn’t matter because we can always just take the real part, which is also a solution. A complex function of time is real for all time if it equals its complex conjugate. The complex conjugate of x(t) is

∗ −βt ∗ −iω1t ∗ iω1t x(t) = e A1e + A2e . ∗ ∗ In order for x(t) = x(t) , it must be that A1 = A2. That is, one constant must be the ∗ ∗ complex conjugate of the other. So we can deﬁne the new constants C = A1 = A2 and ∗ C = A2 = A1, so that x(t) = e−βt C∗eiω1t + Ce−iω1t . The pair of terms in parentheses are now complex conjugates, and we know that when complex conjugates are added, the imaginary parts disappear. However, we can write this in a more convenient form. Any complex number can be written in exponential form containing a real amplitude and a real angle. If we write C = Aeiδ/2, then C∗ = Ae−iδ/2. Substituting these values gives us A x(t) = e−βt ei(ω1t−δ) + e−i(ω1t−δ) , 2 and so the general solution for underdamped oscillations can be written as q −βt 2 2 x(t) = Ae cos(ω1t − δ), ω1 = ω0 − β (underdamped). (4.2)

This tells us that in the underdamped case, the oscillator is oscillating with a magnitude −βt Ae and a frequency ω1, which is less than the natural frequency ω0. Notice that the amplitude Ae−βt is no longer constant in time. In fact, it is decaying exponentially, and we call Ae−βt the envelope function since the oscillations stay between Ae−βt and −Ae−βt. This can be seen in Fig. (4.2). Strictly speaking, this motion is non-periodic, but we can deﬁne the period as the time to go from peak to peak. This period is 2π τ1 = , ω1 and can be found either by diﬀerentiating Eq. (4.2) and looking at twice the time intervals between extrema or by looking at twice the time period between consecutive zeros of Eq. (4.2). The decrement of motion refers to the rate at which the amplitudes decrease and can be found as the ratio of the amplitude of a peak to the amplitude of the next peak Ae−βt . Ae−β(t+τ1)

Notice that the envelope function Ae−βt drops to 1/e of its initial value in time 1/β. This time is called the decay time for the oscillator. 1 decay time = . β 4.2. Damped Harmonic Oscillators 55

Critically Damped Oscillator

2 2 x(t) When β − ω0 = 0, that is, β = ω0, the system is said to be critically damped. In this A case, the general solution for damped oscillations √ √ −βt β2−ω2t − β2−ω2t t x(t) = e A1e 0 + A2e 0 , reduces to the single solution x(t) = e−βt. −A However, a general solution to a second order differential equation is the sum of two solutions, so we have to find a second solution. It turns out that Figure 4.3: Position function of critically damped oscillator. x(t) = te−βt, is also a solution in the special case that β = ω0. This can be verified by taking the first and second derivatives and plugging them intox ¨ + 2βx˙ + β2x = 0. Adding the two solutions together, we get the general solution for the critically damped case

x(t) = (A + Bt)e−βt (critically damped).

This is shown in Fig. (4.3). Critical damping brings the object to the equilibrium position as fast as possible without going past the equilibrium position. Critical damping is usually optimal if you’re designing something in which you want the oscillations damped as quickly as possible.

Overdamped Oscillator Perhaps unexpectedly, the oscillations of a system that is more than critically damped will decay more slowly than if it was critically damped. In a way, overdamping gets in the way of itself. 2 2 When β − ω0 > 0, that is, β > ω0, the system is said to be overdamped. In this case, the general solution for damped oscillations √ √ −βt β2−ω2t − β2−ω2t x(t) = e A1e 0 + A2e 0 , becomes q −βt ω2t −ω2t 2 2 x(t) = e A1e + A2e , ω2 = β − ω0 (overdamped).

Note that ω2 is a new frequency. After a long time,

−(β−ω2)t x(t) ≈ A1e .

An overdamped system returns to equilibrium more slowly than an underdamped or a critically damped system. Unlike an underdamped system, there are no oscillations in an overdamped system—the system is so strongly damped that if displaced from equilibrium it returns to equilibrium only after inﬁnite time.

Energy Dissipation Recall that damping is removing energy from the oscillating system. Since damping isn’t conservative, the mechanical energy is not conserved, so dE/dt is no longer zero. With damping, the equation of motion for a harmonic oscillator is

mx¨ = −kx − bx,˙ 56 Oscillations

x(t) the energy is 1 1 A E = mx˙ 2 + kx2, 2 2 and the time derivative is t dE 1 2 3 =x ˙(mx¨ + kx). dt We applied the chain rule here. Plugging in mx¨ from the equation of motion gives us −A dE = −bx˙ 2, dt Figure 4.4: Undamped oscillator from the example. which is the rate of energy dissipation. The negative sign indicates that damping is always taking energy out of the system. We can also look at the rate of work done by the resistive force F = −bx˙ x(t) A dW = −bx˙ dx. Dividing by dt gives us t dW dx = −bx˙ = −bx˙ 2. 1 2 3 dt dt

−A Example 4.2.1 Consider a typical harmonic oscillator in the form of a block connected to a Figure 4.5: Damped oscillator spring all resting on a frictionless surface. The other end of the spring is connected from the example. to the wall and the whole experiment is being done in a vacuum. The system is motionless when you pull the block (at the end of the spring) 1 meter past its equilibrium point (motionless resting position) and release it. It then begins −1 oscillating with an angular frequency ω0 = 2π s with a period of τ0 = 2π/ω0 = 1 second. That is, the block is going from equilibrium to 1 meter past equilibrium, back through equilibrium to 1 meter past in the other direction, and returning to equilibrium once every second. The position equation for an undamped harmonic oscillator is

x(t) = A cos(ω0t − δ).

−1 In our case, the amplitude is A = 1m, the natural frequency is ω0 = 2π s , and the phase shift is δ = 0. The position equation is simply

x(t) = 1 m cos(2π s−1 t).

A plot of the displacement is shown in Fig. (4.4). Let’s consider now a more realistic version of the experiment. We let air back into our experiment and account for friction between the tabletop and the oscillating block. The general solution for the underdamped harmonic oscillator is q −βt 2 2 x(t) = Ae cos(ω1t − δ), ω1 = ω0 − β .

−1 p 2 2 −1 If we have β = 0.5 s , then ω1 = 4π − (0.5) ≈ 6.26 s . Then our position function is −1 x(t) = (1 m) e−0.5 s t cos(6.26 s−1t). As we can see from the graph in Fig. (4.5), after only 3 seconds, the block is reaching only about a fourth as far from the equilibrium point of x = 0 as it did when it started. −1 What happens if the friction is adjusted perfectly such that β = ω0 = 2π s ? That is, when the system becomes critically damped? For a critically damped 4.3. Driven Oscillators 57

system, the position equation is x(t) A x(t) = (A + Bt)e−βt.

We know that at t = 0, x = 1m, andx ˙ = 0 m/s. This allows us to compute t A = 1m and B = 2π m/s. Our position function is 1 2 3

−1 x(t) = (1 m + 2π m/s t)e−2π s t. −A

Notice from the graph in Fig. (4.6) that the block is practically at rest at the Figure 4.6: Critically damped equilibrium position in a very short time—just a little over 1 second. At t = τ0 = 1 s, the block’s position is oscillator from the example.

−1 x(t = 1 s) = (1 m + 2π m/s 1 s)e−2π s 1 s = (1 + 2π)e−2πm ≈ 0.014 m. x(t) So in the time it takes the undamped oscillator to complete one period, the crit- A ically damped oscillator has decayed to 0.014 m. This is true in general. If you wait the period of the undamped oscillator, the critically damped one will have t decayed to 0.014x0, and you can’t get better performance than this. 1 2 3 Finally, what happens if we increase the friction beyond the point of critical damping? For overdamping, the position function is −A

−βt ω2t −ω2t x(t) = e A1e + A2e , Figure 4.7: Overdamped oscil- p 2 2 lator from the example. where ω2 = β − ω0. Setting x(t = 0) andx ˙(t = 0) to zero, we can get a system of equations in A1 and A2. Using the fact that x(t = 0) = 1m and x˙(t = 0) = 1 m/s, we ﬁnd that A1 = 1/2 + β/(2ω2) and A2 = 1/2 − β/(2ω2). If −1 β = 20, then ω2 = 19.0 s , A1 = 1.026 and A2 = 0.0263, so our position function becomes x(t) = e−20t 1.026e19t + 0.0263e−19t . In the graph in Fig. (4.7), we can see that the damping (i.e. friction) is so high that it takes the block a lot longer to move back to the equilibrium position than when it was critically damped.

4.3 Driven Oscillators

The equation of motion for a damped harmonic oscillator with a sinusoidal driving force is

mx¨ = −kx − bx˙ + F0 cos ωt, where −bx˙ is the damping force and F0 cos ωt is the driving force. Frequently, we write the equation of motion in the form

2 x¨ + 2βx˙ + ω0x = f0 cos ωt,

p where β = b/2m is the damping parameter, ω0 = k/m is the natural frequency (i.e. the frequency in the absence of damping), ω is the frequency of the driving force, and f0 = F0/m. This is now an inhomogeneous second order diﬀerential equation. Inhomogeneous since it has a term f0 cos ωt that does not depend on x or any of its derivatives. In operator notation, we can write the equation of motion as

Dx = f0 cos ωt, 58 Oscillations

d2 d 2 where D = dt2 + 2β dt + ω0 as before. After some time, the motion associated with the natural frequency ω0 will die down x(t) because of the damping force, and all that will be left will be motion due to the driving frequency ω.

A t Driven Undamped Oscillators −0.5 0.5 1 1.5 2 −A In the undamped case, β = 0, and our equation becomes 2 x¨ + ω0x = f0 cos ωt. 2 In this case, if ω << ω0, then the ω0x term dominates. If ω >> ω0, then the driving Figure 4.8: Driven undamped term dominates the motion. oscillator. For the particular, or steady-state solution, we guess xs(t) = A1 cos ωt since both 2 x andx ¨ will contain cosine. Taking the second derivative, we getx ¨ = −A1ω cos ωt. Plugging this in gives us 2 x¨ + ω0x = f0 cos ωt 2 2 −A1ω cos ωt + ω0A1 cos ωt = f0 cos ωt f0 A1 = 2 2 . ω0 − ω 2 2 So xs(t) = A1 cos ωt is the unique steady-state solution and A1 = f0/(ω0 − ω ). Notice 2 2 that the sign of A depends on which of ω0 or ω is larger. We can write the steady-state solution as xs(t) = A1 cos(ωt − δ1),

if we redeﬁne A1 as ( f0 0 if ω < ω0 A1 = 2 2 , δ1 = |ω0 − ω | π if ω > ω0.

The δ1 serves to change the sign of the function. The key thing to remember is that in the driven undamped case, the amplitude is 1 A ∝ 2 2 . |ω0 − ω |

We know that the homogeneous solution is xh(t) = A2 cos(ω0t − δ2) where A2 is the p amplitude in the undriven and undamped case, ω0 = k/m is the natural frequency of the oscillator, and δ2 is the phase shift in the undriven and undamped case. The general solution is the sum of the homogeneous and steady-state solutions

x(t) = xs(t) + xh(t) = A1 cos(ωt − δ1) + A2 cos(ω0t − δ2). Consider a simple harmonic oscillator, say a block at the end of a spring on a frictionless table. The oscillator is released/driven from rest, has a natural frequency of ω0 = 10π, and the driving force has frequency ω = 2π and amplitude f0 = 1000. Since ω < ω0, we have that δ1 = 0. From the amplitude of the driving force and the two frequencies, we calculate A1 = 1.055. We chose such a large value for f0 in order to get a convenient unit for A1. Plugging in what we have so far, gives us

x(t) = 1.055 cos(2πt) + A2 cos(10πt − δ2).

By solving x(t) andx ˙(t) for the initial conditions, we get δ2 = 0, and A2 = −1.055. So the position function is x(t) = 1.055 cos(2πt) − 1.055 cos(10πt). The dashed line in the graph shown in Fig. (4.8) shows the driving force cos(2πt), and the solid line shows the resulting motion x(t). 4.3. Driven Oscillators 59

Driven Damped Oscillators We now consider the damped case of the driven oscillator. The equation of motion is

2 x¨ + 2βx˙ + ω0x = f0 cos ωt, where the ﬁrst term is the acceleration of the oscillator, the second term is the damping force, the third term is the restoring force, and the right side is the driving force. Using the operator notation deﬁned earlier, we can also write this as

Dx = f0 cos ωt. This is a second order inhomogeneous linear diﬀerential equation, so we expect two constants. We only have to ﬁnd one particular solution of Dx = f0 cos ωt, then we can add any solution of the homogeneous equation Dx = 0. We know that the solution for r1t r2t the homogeneous case is xh(t) = A1e + A2e because we solved it when studying the undriven damped oscillator. The general solution of Dx = f0 cos ωt is just the sum of the particular solution and the homogeneous solution

x(t) = xp(t) + xh(t).

r1t r2t Remember that in the long-term, the homogeneous solution xh(t) = A1e + A2e goes to zero because of the damping. So in the long-term the particular solution xp(t) is the only surviving part of x(t), so we call it the steady state solution. The steady state part of the solution survives in the long-term because of the driving force. The terms from the homogeneous part of x(t) go down to zero in the long-term, so we call them the transients. 2 To ﬁnd the particular solution forx ¨ + 2βx˙ + ω0x = f0 cos ωt, it is convenient to consider 2 y¨ + 2βy˙ + ω0y = f0 sin ωt. We do this because we know that for any solution x(t), there is a corresponding solution y(t) with the cosine replaced by the sine term because they are only shifted in time. If we multiply the above equation by i and add it to the one containing x, we get

2 iωt x¨ + iy¨ + 2β (x ˙ + iy˙) + ω0 (x + iy) = f0 cos ωt + if0 sin ωt = f0e . If we let z = x + iy, we can write this as

2 iωt z¨ + 2βz˙ + ω0z = f0e . Now we solve for z(t) and then x(t) = Re (z(t)). If we try z(t) = Ceiωt where C is a complex constant, thenz ˙ = iωCeiωt andz ¨ = −ω2Ceiωt. Plugging them in gives us

2 iωt iωt 2 iωt iωt (−ω Ce ) + 2β(iωCe ) + ω0(Ce ) = f0e f0 C = 2 2 . ω0 − ω + i2ωβ Any complex number can be written in exponential form with a real-valued amplitude A and a real-valued phase angle δ. We write C in exponential form as

f0 −iδ C = 2 2 = Ae . ω0 − ω + i2ωβ Recall that multiplying a complex number by its complex conjugate gives the magnitude squared of the complex number, so 2 ∗ −iδ iδ f0 f0 A = CC = (Ae )(Ae ) = 2 2 2 2 ω0 − ω + i2ωβ ω0 − ω − i2ωβ 2 f0 = 2 2 2 2 2 . (ω0 − ω ) + 4ω β 60 Oscillations

Remember that taking the complex conjugate of even a complicated expression is very easy—just change the sign of every i that appears in the expression. From C = Ae−iδ, x(t) we see that A/C = eiδ, and we can rewrite eiδ as cos δ + i sin δ, so plugging C back in gives us A A 2 2 A cos δ + i sin δ = = (ω0 − ω + i2ωβ). t C f0 −0.5 0.5 1 1.5 2 −A Separating the right side into real and imaginary parts and equating the real and imaginary parts implies that A sin δ = 2ωβ f0 Figure 4.9: Driven damped har- A 2 2 monic oscillator. cos δ = (ω0 − ω ). f0 Dividing the ﬁrst equation by the second gives us 2ωβ tan δ = 2 2 ω0 − ω −1 2ωβ δ = tan 2 2 . ω0 − ω Recall that our goal was to ﬁnd the solution z(t) and then take the real part of it. Our solution is z(t) = Ceiωt. Taking the real part gives us our particular solution

iωt −δt iωt i(ωt−δ) xp(t) = Re (z(t)) = Re Ce = Re Ae e = Re Ae = Re (A(cos(ωt − δ) + i sin(ωt − δ)) .

So our particular solution is xp(t) = A cos(ωt − δ), where the amplitude and phase angle are

f 2ωβ A = 0 , δ = tan−1 . p 2 2 2 2 2 ω2 − ω2 (ω0 − ω ) + 4ω β 0 An important thing to note is that A and δ are determined by the system—not by the initial conditions. Putting it all together, we have that the general solution for the equation of motion

2 x¨ + 2βx˙ + ω0x = f0 cos ωt,

of a damped, driven harmonic oscillator is x(t) = xp(t) + xh(t) or

−βt x(t) = A cos(ωt − δ) + Ate cos(ω1t − δt).

where the amplitude and frequency of the non-transient term is

f 2ωβ A = 0 , δ = tan−1 . p 2 2 2 2 2 ω2 − ω2 (ω0 − ω ) + 4ω β 0

These are determined by the system, and At and δt are determined by the initial conditions. The subscript t indicates “transient” since that part of the solution disappears in the long-term. In Fig. (4.9) we shown an example of a strongly damped system. The dashed line shows the result of the driving force, and the solid line gives the total amplitude. Notice 4.3. Driven Oscillators 61 that the transients die out quickly, and the system is soon displaying only the driven behavior. If we wait for t >> 1/β, the homogeneous part has gone to zero, and we are left with the particular or “steady-state” solution 1 x(t) ≈ A cos(ωt − δ), after t >> . β

Remember from a previous section that we can express the homogeneous solution in a number of diﬀerent ways. For example, we can also write the general solution as

−βt x(t) = A cos(ωt − δ) + e (B1 cos ω1t + B2 sin ω1t), which might be easier to use when evaluating the constants for the initial conditions x(0) = x0 andx ˙(0) = v0.

Resonance Notice that the amplitude A of the steady-state solution is a function of the frequency ω of the driving force. That is,

f A(ω) = 0 . p 2 2 2 2 2 (ω0 − ω ) + 4ω β

2 Let us look at some limits. As ω → 0 we see that A → f0/w . We defined f0 = F0/m p 0 and ω0 = k/m, so as ω → 0, A → F0/k. Recall that the driving force is F0 cos ωt. At ω = 0, F0 cos ωt = F0, that is, the driving force is just a constant force F0. The steady-state solution reduces to x(t) = F0/k, so the oscillator is just sitting there at a distance x = F0/k from equilibrium. Let us look at another limit. As ω → ∞, A → 0. If 2 β is very small (i.e. a weakly damped system), then for large ω, A ≈ f0/ω . We’ve seen that the amplitude varies with the driving frequency, so to see if there is a maximum of A(ω), we differentiate it with respect to ω, and set it equal to zero. We p 2 2 find that it is zero when ω = ω0 − 2β , a frequency that we call resonance frequency

q 2 2 ωR = ω0 − 2β .

The most interesting case is when β is small, that is, the system is weakly damped. Then the second term in the denominator of A(ω) is eﬀectively zero and

f A(ω) ≈ 0 . p 2 2 2 (ω0 − ω )

Then as ω approaches ω0, the amplitude becomes very large. This is the phenomenon of resonance. That is, the amplitude of a driven, damped oscillator, after the transients have died down peaks when the driving frequency is near the natural frequency of the system. When the driving frequency equals the natural frequency (i.e. ω = ω0), the system is “at resonance”. p 2 2 We can calculate the amplitude at resonance by plugging ω = ωR = ω0 − 2β into A(ω) and simplifying. We ﬁnd that

f0 f0 Amax = = . p 2 2 2βω 2β ω0 − β 1 √ p 2 2 From ωR = ω0 − 2β we can see that when β = ω0/ 2 =√ 0.707ω0, then ωR =√ 0. That means that there is no maximum in A(ω) when β ≥ ω0/ 2. When β = ω0/ 2 62 Oscillations

we’re still technically in the underdamped regime, but there’s no longer a resonance phenomenon. In Fig. (4.10) is a graph of the amplitude A(ω) as a function of the driving frequency ω with β = 0.2 ω0. Notice that the resonance frequency is less than the natural frequency. The two coincide when β = 0, i.e. when there is no damping.

A(ω)

f0 Amax = 2βω1

f0 F0 2 = ω0 k

ω ω R 0 ω

Figure 4.10: A plot of the amplitude A(ω) of a driven, damped oscillator with damping parameter β = 0.2 ω0.

In Fig. (4.11) we show the same kind of graph but with various values of the damping parameter β. Notice that at β = 0.707 ω0, there’s no resonance peak. As β is reduced, a resonance peak appears at ω < ω0. As β → 0, the resonance peak centers on ω = ω0 and blows up to inﬁnity.

A(ω) β = 0 β = 0.050 ω0 β = 0.200 ω0 β = 0.707 ω0

f0 F0 2 = ω0 k

ω0 ω

Figure 4.11: A plot of the amplitude A(ω) of a driven, damped oscillator with various damping parameters ranging from β = 0 to β = 0.707 ω0.

Resonance peaks are characterized by their quality factor, Q, where

ω Q = R . 2β 4.3. Driven Oscillators 63

A high quality factor means a sharp peak. For weak damping (weaker than underdamping), we have that β << ω0, which p 2 2 p 2 2 implies that ω1 = ω0 − β ≈ ω0, and ωR = ω0 − 2β ≈ ω0. Furthermore, near 2 2 resonance, ω ≈ ω0. We also have that ω0 − ω = (ω0 + ω)(ω0 − ω) ≈ 2ω0(ω0 − ω). This approximation is sort of like a Taylor expansion of ω about ω0. It is remarkable how close this approximation is if you test it with a pair of large numbers that are close to each 2 2 2 2 other. Finally, we have that 4β ω ≈ 4β ω0. Using these approximations, we have that f f A(ω) = 0 = 0 p 2 2 2 2 2 p 2 2 2 (ω0 − ω ) + 4ω β (2ω0(ω0 − ω)) + 4β ω0 f f β = 0 = 0 . p 2 2 p 2 2 2ω0 (ω0 − ω) + β 2ω0β (ω0 − ω) + β

p 2 2 Notice that exactly at resonance, β/ (ω0 − ω) + β = 1 and A(ω) = f0/2ω0β. We know that Amax = f0/(2βω1), and near resonance, Amax ≈ f0/(2βω0), so we 2 can replace the ﬁrst part of the right side above with Amax. Since A ∝ E, we are often more interested in the amplitude squared. Squaring both sides gives us

2 2 2 β A (ω) = Amax 2 2 . (ω0 − ω) + β

2 2 2 The expression β /((ω0 − ω) + β ) is called a Lorentzian function and is often used 2 2 to describe resonance peaks. At resonance A (ω0) = Amax. At a distance (angular 2 2 frequency) β from resonance, the amplitude squared drops by half; A (ω ± β) = Amax/2, as shown in Fig. (4.12).

A(ω)

2 Amax

1 2 2 Amax ∆ω = 2β

ω0 − β ω0 Figure 4.12: A plot of the amplitude A(ω) of a driven, damped oscillator showing the full width at half maximum.

The width ∆ω = 2β of the resonance peak at half of its maximum value is called the full width at half maximum FWHM = 2β. This is a natural measure of the resonance width. It is easy to see the range of frequencies over which the system displays a strong response. For weak damping at resonance, ω ω E Q = R ≈ 0 = 2π , 2β ∆ω ∆Edis 64 Oscillations

where E is the energy of the oscillator and ∆Edis is the energy dissipated in one cycle at resonance. We can also plot the phase angle δ as a function of the driving frequency ω, −1 2ωβ δ = tan 2 2 , ω0 − ω as shown in Fig. (4.13). In the undamped β = 0 case, δ is a step function that is 0 until ω = ω0 where it jumps to π. As the damping parameter β is turned up, the step function turns into an arctangent function, ﬂattening out more and more as β is increased. No matter what β is, as ω → 0, δ → 0 and as ω → ∞, δ → π. When ω = ω0, δ = π/2. Just as ∆ω = 2β gives the width of the resonance peak at half maximum, it also characterizes the horizontal distance between the sharpest points of the δ curve in the graph below. When β = 0, δ is the step function and this horizontal distance is zero. Note: Plotting δ on a calculator will give a graph that looks diﬀerent. Here, the arctangent value is taken to be between −π and 0 so that it represents a phase lag for both positive and negative values of the arctangent’s argument.

δ(ω) β = 0 β = 0.05 ω0 β = 0.2 ω0

π 2

ω0 Figure 4.13: A plot of the phase angle δ of a driven, damped oscillator as a function of the driving frequency ω for various values of the damping parameter β.

Power Dissipation Recall that the general solution for the driven damped oscillator with equation of motion 2 x¨ + 2βx˙ + ω0x = f0 cos ωt is

−βt x(t) = A cos(ωt − δ) + Ate cos(ω1t − δt). where f 2ωβ A = 0 , δ = tan−1 . p 2 2 2 2 2 ω2 − ω2 (ω0 − ω ) + 4ω β 0

For weak damping at resonance, ω = ω0 and δ = π/2, and after the transients have died down, our solution simplifies to π x(t) = A cos ω t − = A sin (ω t) . max 0 2 max 0 Differentiating, we find that the velocity is

x˙(t) = ω0Amax cos(ω0t). 4.3. Driven Oscillators 65

This shows us that the velocity is in phase with the driving force F (t) = F0 cos(ω0t), which means the driving force is constantly pouring energy into the system. The power input is the rate of work done by the driving force. Given the driving force and the velocity, what is the rate of work being done? Since work is force times distance, the rate of work done is force times velocity.

2 P = F (t)x ˙(t) = (F0 cos(ω0t)) (ω0Amax cos(ω0t)) = F0Amaxω0 cos (ω0t) ≥ 0. This shows us that the power input is always positive at resonance. The average power is τ 1 2 hP i = F0Amaxω0 cos (ω0t) dt, τ ˆ0 where the period is τ = 2π/ω0. Note, the angle brackets on P mean “time average”. We know that the time average of cosine squared over a full period is 1/2, so the average power is 1 hP i = F A ω . 2 0 max 0 The dissipative (i.e. damping) force is

Fd = −bx.˙ Recall that β = b/2m. The rate of work done by the damping force, that is, the rate of energy dissipation, is given by

2 2 2 2 2 Pdis = Fdx˙ = −bx˙ = −b (ω0Amax cos(ω0t)) = −bω0Amax cos (ω0t). The time average of the power of the damping force is 1 hP i = − bω2A2 . dis 2 0 max By substituting f0 Amax = , 2βω0 and f0 = F0/m and β = b/2m, we can verify that the rate of work done by the damping force hPdisi equals the rate of work done by the driving force hP i. This is the expected result.

Fourier Series Solution for Driven Oscillators

Up to this point, we’ve only been considering a driving force of the form f0 cos ωt. What if we had a more general driving force f(t), such that the equation of motion became

2 x¨ + 2βx˙ + ω0x = f(t)? Consider a driving force that is the sum of diﬀerent driving forces X f(t) = fn(t). n

If xn(t) is the steady state solution to the equation

2 x¨n + 2βx˙ n + ω0xn = fn(t), then by the principle of superposition, X x(t) = xn(t). n 66 Oscillations

Since we already know the general solution for a sinusoidal driving force, and since any periodic function f(t) can be expressed as the sum of sinusoidal functions in a Fourier series, we can express the general solution for any periodic function as a sum of solutions for sinusoidal driving forces. Using Fourier series, we can ﬁnd the solution to any equation of motion of the form

2 x¨ + 2βx˙ + ω0x = f(t),

provided that f(t) is periodic. We can do this because any periodic function f(t) can be expressed as an infinite sum of sines and cosines (i.e. Fourier series), so if f(t) can be expressed as a series of sines and cosines, then for each term in the Fourier expansion, we can find a solution xn(t) to the equation of motion, then the general solution is just the sum over n of all these solutions. The Fourier series of a function f(t) defined with a period τ is

∞ X f(t) = [an cos(nωt) + bn sin(nωt)] n=0 τ 1 2 a0 = f(t) dt τ ˆ τ − 2 τ 2 2 an = f(t) cos(nωt) dt τ ˆ τ − 2 τ 2 2 bn = f(t) sin(nωt) dt. τ ˆ τ − 2

Remember that τ = 2π/ω. For example, if a damped oscillator is driven by the function

∞ X f(t) = an cos(nωt), n=1

2 then we know that the steady state solutions tox ¨n + 2βx˙ n + ω0xn = an cos(nωt) are xn(t) = An cos(nωt − δn), so the overall steady-state solution is

∞ X x(t) = An cos(nωt − δn), n=0

where an −1 2ωnβ An = , δn = tan . p 2 2 2 2 2 2 2 ω2 − n2ω2 (ω0 − n ω ) + 4n ω β 0

Example 4.3.1

Consider a weakly damped, β << ω0 oscillator with a high quality factor Q = 5000, where the driving force is the rectangular function deﬁned as

 τ τ 0 if − 2 < t < − 4  τ τ f(t) = f(0) if − 4 < t < 4 .  τ τ 0 if 4 < t < 2 4.3. Driven Oscillators 67

The Fourier series for this function is ∞ f0 2f0 X 1 nπ f(t) = + sin cos(nωt). 2 π n 2 n=1 2 We know that the steady state solution tox ¨ + 2βx˙ + ω0x = f0 cos ωt is xp(t) = A cos(ωt − δ), where f 2ωβ A = 0 , δ = tan−1 . p 2 2 2 2 2 ω2 − ω2 (ω0 − ω ) + 4ω β 0

In our case, the driving force isn’t just f0 cos ωt, but rather, f(t), which is the sum of infinite terms of the form an cos(nωt). Due to the principle of superposition, our general steady-state solution is the sum of the steady-state solutions tox ¨n + 2 2βx˙ n + ω0xn = an cos(nωt). Each of these has the steady-state solution xn(t) = An cos(nωt − δn). So our overall steady-state solution is ∞ X x(t) = An cos(nωt − δn), n=0 where an −1 2ωnβ An = , δn = tan . p 2 2 2 2 2 2 2 ω2 − n2ω2 (ω0 − n ω ) + 4n ω β 0 We can now examine the physics of our solution. We’ll consider a driving force that has 3 beats per beat of natural frequency, that is, ω0 = 3ω. Now we can look at the first few Fourier coefficients.

f0 a0 2 1 f0 A0 = = = · p 2 2 ((3ω)2) 18 ω2 (ω0) 2f0 2f0 a1 π π 1 f0 A1 = ≈ = = · p 2 2 2 2 2 |ω2 − ω2| |(3ω)2 − ω2)| 4π ω2 (ω0 − ω ) + 4ω β 0

A2 = 0

2f0 a3 − 3π f0 2Q f0 A3 = = = − = − · . p((3ω)2 − 32ω2)2 + 4 · 32ω2β2 p4 · 32ω2β2 9πωβ 27π ω2

In the last one, we made the substitution β = ω0/(2Q) = 3ω/(2Q). Notice that A3 contains Q making it a thousand times larger than the previous coeﬃcients. 2 2 2 This is because at n = 3, ω0 = n ω , so that part of the denominator is zero. This is the phenomenon of resonance.

A4 = 0

2f0 a5 5π 1 f0 A5 = ≈ = · . p 2 2 2 2 2 2 2 p 2 2 2 2 40π ω2 (ω0 − 5 ω ) + 4 · 5 ω β ((3ω) − 5 ω )

Notice that the magnitude of A5 is much less than the magnitude of A3. As we continue calculating coeﬃcients, they will continue getting smaller and smaller now that we’re past resonance. Since the A3 coeﬃcient is a thousand times larger than the others, it completely dominates and the entire Fourier series is well approximated by the n = 3 term. ∞ X 2Q f0 π x(t) = A cos (nωt − δ ) ≈ − · cos 3ωt − . n n 27π ω2 2 n=0 68 Oscillations

We know that δ3 = π/2 since that is always the value of the phase angle at resonance. In this case, we had ω0 = 3ω and found that the n = 3 term was exactly at resonance causing a huge spike in the amplitude of the oscillator. Similarly, if we would have had ω0 = ω, we would have found the n = 1 term to be at resonance giving us an even bigger spike in amplitude, and if we had ω0 = 2ω, we would have found the n = 2 term to be at resonance with again a huge spike in amplitude.

Root Mean Square Displacement We want a single number that tells us how strongly the system is responding to our driving force. The mean square displacement is deﬁned as the time average of the displacement squared over a single period.

τ 1 2 hx2i = x2 dt. τ ˆ τ − 2

Keep in mind that once you know the time average of a periodic function over one period, then you know its time average over all time. The root mean square displacement is then the square root of the mean square displacement

p 2 xrms = hx i,

which gives us a number with units of distance and can be used to characterize how strongly an oscillator is responding to a driving force. For an oscillator with a periodic driving force whose Fourier series has only the cosine terms (like the square wave example), we have that

∞ X x(t) = An cos(nωt − δn), n=0 then τ " ∞ # 2 2 1 X hx i = An cos(nωt − δn)Am cos(mωt − δm) dt. τ ˆ τ − 2 n,m=0 Interchanging summation and integration gives us

∞ τ 2 2 X 1 hx i = AnAm [cos(nωt − δn) cos(mωt − δm)] dt. τ ˆ τ n,m=0 − 2

2 1 The integrand vanishes when n 6= m, and for n = m, it becomes hcos i = 2 , so it reduces to Parseval’s theorem ∞ 1 X hx2i = A2 + A2 . 0 2 n n=1 4.4. Summary: Oscillations 69

4.4 Summary: Oscillations

Skills to Master • Solve differential equations and Fourier series • Convert between different forms of the solution for the simple harmonic oscillator • Given a potential, find any equilibria, characterize their stability, calculate the force, velocity, kinetic energy, effective spring constant, and its natural frequency of oscillation • Write down the ODE for a harmonic oscillator with a velocity-dependent damping force and solve it in the under-, over-, and critical-damping cases • Solve a driven, damped harmonic oscillator • Given a damped, driven oscillating system, calculate the resonance frequency, maximum amplitude, quality factor, and FWHM • Calculate the root mean square displacement of an oscillator

Simple Harmonic Oscillators This is usually written in the form

Small motion about any stable equilibrium point can r be approximated as a simple harmonic oscillator. 2 k b x¨ + 2βx˙ + ω0x = 0, ω0 = , β = . The force on a simple harmonic oscillator is m 2m F = mx¨ = −kx, The general solution is which is generally written as √ √ −βt β2−ω2t − β2−ω2t x(t) = e A e 0 + A e 0 , r 1 2 2 k x¨ = −ω0x, ω0 = . m and it can be found by letting x = ert, then factoring rt 2 2 The solution of this diﬀerential equation can be written out e to get the auxiliary equation r + 2βr + ω0 = 0, in several ways: which you solve for r1 and r2 using the quadratic for- x(t) = A0 cos(ω t) + B0 sin(ω t) mula and plug the results into the general form of the 0 0 r1t r2t solution x(t) = A1e + A2e . = C eiω0t + C e−iω0t 1 2 In the underdamped case, β < ω0 and the gen- = A cos(ω0t − δ) eral solution simpliﬁes to

iω0t = Re Ce . q x(t) = Ae−βt cos(ω t − δ), ω = ω2 − β2. Remember that the potential energy is 1 1 0 1 U(x) = kx2, This can be obtained from the general solution by fac- 2 ∗ toring i out of the exponents and noting that A1 = A2. and the velocity can be obtained by differentiating x(t), Next, we write the constants in exponential form in so it is easy to calculate the kinetic energy and the total terms of a real amplitude and a real angle. energy. For the underdamped oscillator, we can define a period as the peak to peak time and find it as twice Damped Oscillators the time periods between the extrema of x(t). In the critically damped case, β = ω0 and the The equation of motion for a damped oscillator is solution is F = mx¨ = −kx − bx.˙ x(t) = (A + Bt)e−βt. 70 Oscillations

This is found by plugging β = ω0 into the general so- The general solution for the driven damped oscil- lution to get a single solution, and then adding to it a lator in the underdamped case is −βt second solution x = te . −βt x(t) = A cos(ωt − δ) + e (B1 cos ω1t + B2 sin ω1t), In the overdamped case, β > ω0, and the general solution doesn’t simplify, so we write where q f 2ωβ −βt ω2t −ω2t 2 2 0 −1 x(t) = e A1e + A2e , ω2 = β − ω . A = , δ = tan . 0 p 2 2 2 2 2 ω2 − ω2 (ω0 − ω ) + 4ω β 0 To remember the form of ω1 and ω2, remember A and δ are determined by the system and B and that it is the square root of the discriminant formed 1 2 B2 are determined by the initial conditions. The first by the auxiliary equation, so β is the first term in the expression in x(t) is the particular (or steady-state) so- general form. Then in the specific solutions for the dif- lution and the second expression is the homogeneous ferent damping regimes, the first term is the larger of 2 2 solution. β and ω0. After some time, the homogeneous part of the so- When asked to obtain x(t) given β and initial con- lution will have gone to zero, and all that will be left ditions, the first thing to do is identify the regime. How is the steady-state part of the general solution. β relates to ω0 determines whether we’re in the under- Since the amplitude A depends on ω, we can have damped, critically damped, or overdamped regime. To resonance, which occurs when there is a sudden in- find the value of the constants, plug the initial condi- crease in A for a small change in ω. The frequency at tions into x(t) andx ˙(t) and solve for the constants. To which resonance occurs is do this, it is generally easier to use the form of x(t) q that contains both sine and cosine. 2 2 ωR = ω0 − 2β . The rate of energy dissipation is It can be calculated by differentiating A with respect dE = −bx˙ 2, to ω and setting it equal to zero to find its maximum. dt Resonance occurs in the underdamped case, and it and can be found by differentiating the total energy occurs when the driving frequency equals the natural E = T + U and plugging inx ¨ from the equation of frequency, ω = ω0. The amplitude at resonance is motion. f0 Amax = , 2βω1 Driven Damped Oscillators and it can be calculated by plugging ω = ωR into A(ω). If we drive the damped oscillator with the force If the driving force equals the natural frequency F0 cos ωt, then our equation of motion becomes (ω = ω0), the oscillator is being driven at resonance and the steady state equation of motion becomes 2 F0 x¨ + 2βx˙ + ω0x = f0 cos ωt, f0 = . m x(t) = A cos(ωt − δ) = Amax sin ω0t, This is now an inhomogeneous second order ODE, because one term in the denominator of A cancels and which has a general solution that is the sum of a par- π δ = 2 . ticular solution and any solution to the homogeneous To calculate the work done by the driving force in form of the equation. We can find a particular solution one period, use the fact that the rate of work done is by replacing the x’s with y’s in the equation of motion the driving force times the velocity, so and the cosine by sine. Then multiply this equation by dW i and add it to the original to get a complex equation = F x.˙ of motion in z. We then try z(t) = Ceiωt, plug it in dt and solve for C. Finally, we write C in exponential Then to compute the work done, separate variables and form, then take the real part of z(t) to get x(t). integrate from 0 to τ. Use the fact that the integral of 4.4. Summary: Oscillations 71 cosine squared from 0 to τ is τ/2. Simplify the final The mean square displacement is defined as answer using τ = 2π/ω. the time average of the displacement squared. For pe- Resonance peaks are characterized by their qual- riodic functions, we only have to integrate over a single ity factor, Q, where period. τ ω 1 2 Q = R . hx2i = x2 dt. τ ˆ τ 2β − 2

Keep in mind that for weak damping (β << ω0), The root mean square displacement is then the ωR ≈ ω0. square root of the mean square displacement The full width at half max, used to characterize p 2 resonance peaks, is simply 2β. xrms = hx i, Chapter 5

Calculus of Variations

The calculus of variations is used to ﬁnd a path that minimizes certain integral functions. For example, proving that the path y(x) = x is the shortest path between (0, 0) and (1, 1) is a calculus of variations problem. Fermat’s principle tells us that light going from one point to another point in a medium will follow the path that minimizes travel time. So if we have a medium in which the index of refraction is a function of position, then calculus of variations is needed to determine the path that the light takes through the medium. To use the calculus of variations to ﬁnd the path that minimizes a certain integral function, we need to be able to describe any path between two points, and we need to be able to calculate its length. The calculus of variations is the mathematical basis of Lagrangian mechanics. This is our primary motivation for familiarizing ourselves with this method.

5.1 The Euler-Lagrange Equation

In the calculus of variations, we seek to extremize (i.e. ﬁnd the maximum or minimum of) a function of the form

x2 dy S = f y(x), , x dx, ˆx1 dx where y = y(x) is an unknown curve that joins the ﬁxed endpoints (x1, y1) and (x2, y2). We can express this a little more compactly as

x2 S = f (y, y0, x) dx. ˆx1

To be more precise, we want to find the function y(x) describing an as yet unknown curve, for which S is a minimum or maximum. Note, another way of stating that S is extremized is to say that it is stationary. Recall from elementary calculus, that the extrema of a function y(x) occur where dy/dx = 0. To find the minimum or maximum of a function, we simply had to differentiate it and set it equal to zero. What we want to do now is a step up from that in difficulty since we aren’t minimizing a simple function, but rather, a functional which is a function (in this case S) that takes in a function (y(x)) and returns a number. If we plug the right function y(x) into S and evaluate S, we obtain a number that is smaller than what we would get if we plugged any other function into S. The only constraint on this “any other function” is that it must also pass through the endpoints (x1, y1) and (x2, y2). 5.1. The Euler-Lagrange Equation 73

If we let y = y(x) denote the “right” path—the one which results in the smallest value for S, then any neighboring paths (including the right one) can be represented by

Y = y(x) + αη(x).

The function αη(x) is simply the diﬀerence between the right curve y = y(x) and any wrong curve. The function η(x) can be any function provided that Y goes through the endpoints (x1, y1) and (x2, y2). That is, η(x1) = η(x2) = 0 so that y(x1) = y(x1) + αη(x1) = y1 and y(x2) = y(x2) + αη(x2) = y2. So in order for y = y(x) to be the “right” curve, α = 0 must result in the smallest possible value for S. We can now express S as a function of α

x2 x2 S(α) = f (Y,Y 0, x) dx = f (y + αη, y0 + αη0, x) dx, ˆx1 ˆx1 with the condition that α = 0 is an extremum of S(α). We have now reduced the problem to the elementary calculus problem of ﬁnding an extremum of a function S(α) at the point α = 0. We know that an extremum of S occurs at α = 0 if dS = 0. dα α=0

Example 5.1.1

Show that the quantity

2π dY 2 S(α) = dx, ˆ0 dx is minimized for the family of functions Y = x + α sin x if α = 0. First, notice that this integral has the form described above with f(Y,Y 0, x) = Y 02 and with endpoints (0, 0) and (2π, 2π). The function η(x) = sin x works since it vanishes at the endpoints, that is, at x = 0 and x = 2π, x + α sin x = x. We ﬁnd that

2π dY 2 2π S(α) = dx = (1 + α cos x)2 dx ˆ0 dx ˆ0 2π = 1 + 2α cos x + α2 cos2 x dx = 2π + α2π. ˆ0 This shows us that S(α) is an upward opening quadratic function centered on α = 0, so obviously, it will have a minimum value at α = 0. To show this explicitly, we diﬀerentiate S and set it equal to zero dS = 2πα = 0 =⇒ α = 0. dα In this case, we’ve shown that if η(x) = sin x, then for the family of functions 2π dY 2 Y = x + αη(x), the quantity S(α) = 0 dx dx is minimized when α = 0. Our goal is to be able to show that this is´ the case for any η(x).

Since S(α) is just a regular function of α, we can diﬀerentiate it with respect to α with the knowledge that its minimum occurs at α = 0. In other words, the derivative of the integral must equal 0 when α = 0. dS x2 ∂ = f (y + αη, y0 + αη0, x) dx = 0. dα ˆx1 ∂α 74 Calculus of Variations

To diﬀerentiate the integral with respect to α, we can move the derivative inside the integral. Applying the chain rule gives us

x2 dS ∂f ∂f 0 = η + 0 η dx = 0, dα ˆx1 ∂y ∂y which we can separate as

x2 x2 dS ∂f ∂f 0 = η dx + 0 η dx = 0. dα ˆx1 ∂y ˆx1 ∂y Consider now the formula for integration by parts

u dv = uv − v du, ˆ ˆ which we can also write as dv du u dx = uv − v dx. ˆ dx ˆ dx In other words, if the integral of a function contains a function and a derivative of another function, then integration by parts allows us to switch the function being diﬀerentiated. This is a common application of integration by parts in physics. Using integration by parts, we can write

x x x 2 ∂f 2 ∂f d ∂f x2 2 d ∂f η0 dx = η dx = η(x) − η dx. 0 0 0 x 0 ˆx1 ∂y ˆx1 ∂y dx ∂y 1 ˆx1 dx ∂y Since η(x) is deﬁned to be zero at the endpoints, the ﬁrst term on the right is zero, and we have that x2 x2 ∂f 0 d ∂f 0 η dx = − η 0 dx. ˆx1 ∂y ˆx1 dx ∂y Putting it back together again, we can now factor η out to get

dS x2 ∂f d ∂f x2 ∂f d ∂f = η − η 0 dx = η(x) − 0 dx = 0. dα ˆx1 ∂y dx ∂y ˆx1 ∂y dx ∂y Given our condition that S is an extremum, we need dS/dα = 0 for all possible η(x). Therefore, the quantity in parentheses must be zero for all values of x between x1 and x2. This is the Euler-Lagrange equation

∂f d ∂f − = 0. ∂y dx ∂y0

Example 5.1.2

Consider a general path connecting the two points 1, (x1, y1) and 2, (x2, y2) that can be described as a function y of x. Show that the shortest path between the two points is a straight line. We want to minimize the path length

2 L = ds, ˆ1 where ds is an inﬁnitesimal displacement along the path. If we zoom in on the 5.1. The Euler-Lagrange Equation 75

path, and the inﬁnitesimal segment ds, we see by the Pythagorean theorem, that ds2 = dx2 + dy2, so ds = pdx2 + dy2. Factoring dx out of the square root gives us ds = p1 + y02dx where y0 = dy/dx, so our integral is

2 p L = 1 + y02 dx. ˆ1 To find the shortest path between points 1 and 2, we need to minimize this integral. Notice that this is a calculus of variations problem where f (y, y0, x) = p1 + y02 depends only on y0. Our goal is to find the specific function y(x) that minimizes L. To do that, we just have to apply the Euler-Lagrange equation

∂f d ∂f − = 0. ∂y dx ∂y0

Calculating the partial derivatives, we get ∂f = 0 ∂y ∂f y0 = . ∂y0 p1 + y02

So the Euler-Lagrange equation gives us ! d y0 0 − = 0. dx p1 + y02

We are left with a second order differential equation that looks messy once we differentiate the quantity in parentheses with respect to x. However, there’s no need to do that. Since the derivative of whatever is inside the parentheses is 0, we know that that quantity is constant. There’s no need to perform the differentiation with respect to x, since we already know the first integral, it is just

y0 = C, p1 + y02

where C is a constant. Solving for y02 gives us

C2 y02 = , 1 − C2 which is just a constant, which implies that y0 is just a constant dy y0 = = m. dx Integrating this gives us y(x) = mx + b, which proves that the shortest path between two points is a straight line. ∂f d ∂f In general, if ∂y = 0, as it was in this case, then we will get dx ∂y0 = 0 ∂f ∂f from the Euler-Lagrange equation, which implies that ∂y0 = constant. Then, ∂y0 will be separable, which allows us to ﬁnd the second arbitrary constant easily by integrating. 76 Calculus of Variations

Example 5.1.3

Do the same problem as above but in polar coordinates. We’ll define the path between points 1 and 2 as a function φ of r. In this case, r is the independent variable and φ is the dependent variable. We could also define it as a function r of φ, but it would make it a little more difficult at the end. Again we want to minimize the path length

2 L = ds, ˆ1

where ds is an inﬁnitesimal displacement√ along the path. Recall that d~r = dr rˆ+ r dφ φˆ, and we have that ds = d~r 2 = pdr2 + r2dφ2. Factoring out dr gives us ds = p1 + r2φ02dr, where φ0 = dφ/dr so the path length is given by

r 2 p L = 1 + r2φ02 dr. ˆr1

In this case, we have f(φ, φ0, r) = p1 + r2φ02. Here φ is playing the role of y and r is playing the role of x. In our case, the Euler-Lagrange equation is

∂f d ∂f − = 0. ∂φ dr ∂φ0

Taking the partial derivatives, we ﬁnd that ∂f = 0 ∂φ ∂f r2φ0 = . ∂φ0 p1 + r2φ02

So from the Euler-Lagrange equation, we get ! d r2φ0 = 0, dr p1 + r2φ02

which tells us that r2φ0 = C, p1 + r2φ02 where C is a constant. We now have a ﬁrst order separable diﬀerential equation. Solving for φ0 gives us dφ C φ0 = = √ , dr r r2 − C2 so 1 φ(r) = C √ dr. ˆ r r2 − C2 If we substitute r = C/ cos u then u = cos−1(C/r) and dr = (C/ cos2 u) sin u du. Then 1 C sin u φ(r) = C sin u du = √ du = du = u+δ, q 2 2 ˆ C C2 2 cos u ˆ 1 − cos u ˆ cos u cos2 u − C 5.2. The Brachistochrone Problem 77

where δ is the second constant. Back-substituting u with cos−1(C/r) gives us ~r(t) r cos(φ − δ) = C.

Using a trig identity, we can write this as dr

~r r cos φ cos δ + r sin φ sin δ = C d cos δ C dφ x cos δ + y sin δ = C =⇒ y = − x + , sin δ sin δ which is clearly the equation of a line since C and δ are constants. Figure 5.1: An infinitesimal segment of the path. In the example done in polar coordinates, we used the fact that d~r = dr rˆ + r dφ φˆ. We can understand this by means of Fig. (5.1). We consider first an elemental segment of the line denoted d~r. Since we are dealing with an infinitesimal element, we can treat d~r as a straight vector rather than a curvy vector function. Notice that in going from one end of d~r to the other, we go through the angle dφ and the distance from the origin changes by dr. rˆ dr In Fig. (5.2), we zoom in on one of the triangles in Fig. (5.1). In polar coordinates, we can break a vector d~r into the sum of a pair of perpendicular basis vectors—one in the direction rˆ and the other in the direction φˆ. Notice that the perpendicular side lengths d~r of this triangle are dr and r dφ (the angle change times the distance from the origin), so the vector can be written as r dφ d~r = dr rˆ + r dφ φˆ. φˆ

5.2 The Brachistochrone Problem Figure 5.2: Breaking the diﬀer- ential into component vectors. This is a classic calculus of variations problem. The brachistochrone problem asks, if a bead sliding on a frictionless wire in the presence of gravity is released from rest at t = 0 at the origin, what shape of the wire will minimize the travel time for the bead to go from O • x the origin to some lower point (x2, y2)? To gain some intuition, it helps to consider the extreme cases. If the second point is directly beneath the origin, what will be the path? The bead will just drop straight down. If the second point is directly to the right of the origin, what will be the path? • The only path that will work is a curve that drops down from the origin (so that the (x2, y2) bead can gain velocity) then curves up to the second point. Since the bead on the wire is y frictionless, the bead will be able to return to its initial height if it follows such a curve. This is a calculus of variations problem in that we want to minimize the transit time Figure 5.3: The brachistochrone for the bead. To do that, we integrate the inverse velocity over the path traveled by the problem. bead. Think of d = rt which implies t = d/r.

2 1 t = ds. ˆ1 v

The only force on the bead is gravity, so its velocity depends on its height (i.e. distance from the x-axis). Using conservation√ of energy, and the fact that the bead starts from rest, we easily ﬁnd that v = 2gy. Since v is a function of y, we choose to treat the path as a function x of y. That is, x = x(√y) where y is the independent variable. Then we know that ds = pdx2 + dy2, so ds = 1 + x02dy, and our integral becomes

√ s y2 1 + x02 1 y2 1 + x02 t = √ dy = √ dy. ˆy1=0 2gy 2g ˆ0 y 78 Calculus of Variations

So we have that f(x, x0, y) = p1 + x02/y. The Euler-Lagrange equation in this case is O ∂f d ∂f • x − = 0. ∂x dy ∂x0 Calculating the partial derivatives, we get (x2, y2) • ∂f = 0 ∂x y ∂f x0 = √ √ . ∂x0 y 1 + x02 Figure 5.4: The solution to the brachistochrone problem is a cy- The Euler-Lagrange equation implies that cloid. x0 √ √ = C, y 1 + x02 √ where C is a constant. A convenient choice for the value of this constant is C = 1/ 2a as we will see later. Squaring both sides allows us to solve for x0 and then integrate to obtain x = x(y).

x02 1 = y(1 + x02) 2a dx r y y x0 = = = dy 2a − y p2ay − y2 y x(y) = dy. ˆ p2ay − y2

If we make the substitution y = a(1−cos θ), then dy = a sin θ dθ and our integral becomes

a(1 − cos θ) x(y) = a sin θ dθ ˆ p2a2(1 − cos θ) − a2(1 − cos θ)2 (1 − cos θ) sin θ = a √ dθ = a(θ − sin θ) + D, ˆ 1 − cos2 θ where D is our second constant. The path of the bead starts at the origin (0, 0), so when x = 0, we know that y = 0, which tells us that θ = 0. Plugging these into y = a(1 − cos θ) and x = a(θ − sin θ) + D, we ﬁnd that D = 0. So our solution, in parametric form is ( x = a(θ − sin θ) . y = a(1 − cos θ)

The curve described by these equations is a cycloid, which is the curve that a point on the circumference of a circle of radius a traces out as it rolls without slipping (along the underside of the x axis in this case). If the “wheel” starts with a point on its rim touching the origin and then it rolls to the right (in Fig. (5.4)) along the underside of the x-axis, the point on the rim will trace out a cycloid. For any point (x2, y2), there is a unique cycloid that goes through it and the origin. To ﬁnd the right cycloid, just adjust the radius a of the wheel until the cycloid goes through the point. If x2 < πy2, the path is monotonic. That is, the portion of the cycloid between point 1 (the origin) and point 2 is strictly decreasing. If x2 > πy2, the portion of the cycloid between points 1 and 2 is not monotonic. The tangents to the cycloid at the endpoints (where the cycloid touches the x-axis) are vertical, but the cycloid is not a half-circle. For a complete loop, the distance between 5.2. The Brachistochrone Problem 79 the endpoints is 2πa, the circumference of the wheel, and the deepest part of the cycloid reaches a depth of 2a as shown in Fig. (5.5). 2πa An interesting feature of the cycloid path is that it is also a tautochrone. This means that for a given cycloid, multiple beads released at diﬀerent points along the 2a cycloid will reach the bottom at the same time. In other words, the period of oscillation is independent of the amplitude. To prove that the cycloid is a tautochrone, we start by calculating the time it takes for the bead to travel from an arbitrary point P (instead of Figure 5.5: One “loop” of the 0 cycloid. the origin) to the bottom of the cycloid. When it starts from the origin, the initial angle is θ = 0. This is the angle that the wheel has rotated through at that point. In this case, the initial angle is some θ . 0 O Given θ0 at P0, we get y0 = a(1 − cos θ0), and h = y − y0 = a(cos θ0 − cos θ). From • x 1 2 conservation of mechanical energy, we get 2 mv = mgh = mga(cos θ0 − cos θ), so the y0 speed of the bead is p p • y v = 2ga cos θ0 − cos θ. P0 h We want to calculate the time for the bead to go from P to P (the bottom of the cycloid), 0 • so we use P 1 t = ds. ˆ v P0 y We know that ds = pdx2 + dy2 and from the parametric equation of the cycloid, we get dx = a(1 − cos θ) dθ and dy = a sin θ dθ. Plugging these in and simplifying, we get Figure 5.6: The time for the √ √ bead to reach the bottom is in- ds = 2a 1 − cos θ dθ. So we get dependent of starting position. √ √ π 2a 1 − cos θ ra π r 1 − cos θ t = √ √ dθ = dθ. ˆθ0 2ga cos θ0 − cos θ g ˆθ0 cos θ0 − cos θ When we change from integrating over the path to integrating over the change in θ, keep in mind that the angle is the angle through which the generating wheel has rotated through. At the bottom of the cycloid, the generating wheel has rotated through an angle√ of π. If we√ now make the change of variables u = cos θ, then du = − sin θ dθ = − 1 − cos2 θ = − 1 − u2 dθ, and

ra −1 r 1 − u 1 ra u0 1 t = − √ du = √ √ du, 2 g ˆu0 u0 − u 1 − u g ˆ−1 u0 − u 1 + u where u0 = cos u0. We now make the second change of variables w = (u + 1)/(u0 + 1) then u = (u0 + 1)w − 1, and our integral simpliﬁes to

ra 1 1 t = p dw. g ˆ0 w(1 − w) Notice that the integral is now dimensionless, so it is just a number. This shows that the time for the bead to reach the bottom is independent of u0 and therefore independent of the initial position given by θ0. For the special case that the initial position P0 is the origin (0, 0), we can use an earlier integral to evaluate the time it takes the bead to reach the bottom.

ra π r 1 − cos θ ra π r1 − cos θ ra t = dθ = dθ = π. g ˆθ0 cos θ0 − cos θ g ˆ0 1 − cos θ g The period of the oscillations, is then

ra r4a τ = 4t = 4π = 2π , g g which we recognize as the period of a simple pendulum of length 4a. 80 Calculus of Variations

In principle, one can construct a truly isochronal pendulum by having it trace out a cycloid. This can be accomplished if the pendulum is a mass on a ﬂexible string that is constrained to wind around two cycloids. 1

5.3 General Parametrization

In this section, we calculate the shortest path between two points using a general parametrization of the path. We can introduce an independent parameter u, then any path can be represented as

x = x(u) y = y(u).

The length of the path from point 1 to point 2 is given by

2 L = ds, ˆ1 where " # dx2 dy 2 ds2 = dx2 + dy2 = + du2. du du Then the path length becomes

u 2 p L = x02 + y02 du. ˆu1 Then the more general case of the quantity to be minimized is given by

u2 S = f(x, x0, y, y0, u) du. ˆu1 To extremize S, we consider the families of paths

X = x(u) + αηx(u)

Y = y(u) + αηy(u),

where ηx(u1) = ηy(u1) = ηx(u2) = ηy(u2) = 0. Then we perform the same analysis as before. We diﬀerentiate S with respect to α to get

u2 d ∂ 0 0 0 0 S(α) = f(x + αηx, x + αηx, y + αηy, y + αηy, u) du dα ˆu1 ∂α u2 ∂f d ∂f ∂f d ∂f = − 0 ηx(u) + − 0 ηy(u) du. ˆu1 ∂x du ∂x ∂y du ∂y

For this to hold for all ηx, ηy this implies that both quantities in parentheses must be zero independently. In other words, we get two Euler-Lagrange equations. In general, for each dependent variable, you get an Euler-Lagrange equation, so we end up with a system of diﬀerential equations. In this case, we have one for x and one for y. The solution to this pair of Euler-Lagrange equations is the parametrized form of the shortest path. In this case, we have that

u2 L = f(x, x0, y, y0, u) du, ˆu1 1http://mathworld.wolfram.com/TautochroneProblem.html 5.3. General Parametrization 81 where f(x, x0, y, y0, u) = px02 + y02. We know that

∂f d ∂f − = 0 ∂x du ∂x0 ∂f d ∂f − = 0, ∂y du ∂y0

∂f ∂f and since ∂x = ∂y = 0, we have that d ∂f = 0 du ∂x0 d ∂f = 0, du ∂y0 which implies that ∂f x0 = = a ∂x0 px02 + y02 ∂f y0 = = b. ∂y0 px02 + y02 where a and b are constants. Now we have a system of two nonlinear second order diﬀerential equations. We can simplify by dividing the ﬁrst equation by the second

√ x0 x02+y02 a = √ y0 b x02+y02 x0 dx dx = du = . y0 dy dy du So we have that dy b = = m dx a dy = m dx y = mx + c.

In the general case of one independent variable and multiple dependent variables, we want to extremize x2 S = f dx, ˆx1 where 0 f = f (yi(x), yi(x), x) , and i = 1, 2, . . . , n. Here we have one independent variable x and 2n dependent variables. To study variations, we introduce n functions ηi(x) and families of functions

Yi = yi(x) + αηi(x).

The same α can be used for every family of functions Yi because it is absorbed by ηi(x). From there, we follow the same procedure as before, by diﬀerentiating S with respect to α and evaluating it at α = 0

n dS x2 ∂ x2 ∂f d ∂f 0 X = f (Yi,Yi , x) dx = − 0 ηi(x) dx = 0. dα α=0 ˆ ∂α ˆ ∂y dx ∂y x1 x1 i=1 i i 82 Calculus of Variations

The η are all independent of each other, and this implies that each term in the parentheses, not just the entire sum, must vanish separately. That is, you get a total of n Euler- Lagrange equations ∂f d ∂f − = 0, ∂yi dx ∂yi where i = 1, 2, . . . , n. 5.4. Summary: Calculus of Variations 83

5.4 Summary: Calculus of Variations

Skills to Master

• Understand the idea of calculus of variations and be able to derive the Euler-Lagrange equation(s) • Solve the brachistochrone problem using calculus of variations • Understand calculus of variations for multiple functions of a parameter

Fermat’s principle tells us that light going from path between point 1 and point 2, you want to mini- one point to another point in a medium will follow the mize the path length, which is given by path that minimizes travel time. 2 In the calculus of variations, we seek to extremize L = ds, ˆ1 x2 S = f (y, y0, x) dx. where ds = p1 + y02dx. Remember, you don’t have ˆx1 to evaluate the integral—you just apply the Euler- To be more precise, we want to ﬁnd the function y(x) Lagrange equation to the integrand. describing an as yet unknown curve, for which S is a Instead of having the function to be minimized de- minimum or maximum. scribed by y(x) or x(y), we could also parametrize it The condition that S is extremized is given by the as Euler-Lagrange equation x = x(u) ∂f d ∂f y = y(u). − = 0. ∂y dx ∂y0 Then the integral to be minimized has the form

Solving the resulting differential equation gives y(x) a u2 function y of x. The function y(x) is the curve that S = f(x, x0, y, y0, u) du, ˆ minimizes S. u1 When the first term in the Euler-Lagrange equa- and we now have an Euler-Lagrange equation for each tion is zero, the result is a separable differential equa- dependent variable tion that is easy to solve. When solving a calculus of variations problem, the ∂f d ∂f − = 0 first step is to identify the quantity that is to be min- ∂x du ∂x0 imized and express that quantity as an integral of the ∂f d ∂f − = 0. form given above. For example, to find the shortest ∂y du ∂y0 Chapter 6

Lagrangian Mechanics

So far, we’ve only looked at classical Newtonian mechanics. Newtonian mechanics is mathematically complete, meaning there is no mechanics problem that we cannot solve given only Newton’s laws. However, Newton’s laws are not always the most convenient to use. Newton’s laws are convenient when working in Cartesian coordinates because F = ma is separable into components. These laws prefer Cartesian coordinates. However, Cartesian coordinates are often not a natural choice when one could make use of spherical or axial symmetry. Lagrangian mechanics is an equivalent (and maybe deeper) formulation of mechanics, that doesn’t have a preferred coordinate system, and so, is often useful where Newtonian mechanics would be messy. Hamilton’s principle or the principle of least action states that the path ~r(t) a particle takes from point 1 to point 2 in the time interval t1 to t2 is that which minimizes the action which is deﬁned as t2 S = L ~r(t), ~r˙(t), t dt, ˆt1 where the Lagrangian is L = T − U. Note that T is the kinetic energy and U is the potential energy. In other words, the actual path ~r(t) that the particle follows is the path that makes the integral S stationary. To say that S is stationary, is like saying the derivative is zero in ordinary calculus. The function could be a minimum, a maximum, or a saddle at that point. The solution is obtained by solving the Euler-Lagrange equations with the Lagrangian L.

6.1 2D Particle in Cartesian Coordinates

Find the Lagrangian and the path a particle follows in two dimensions using Cartesian coordinates. The Lagrangian is 1 1 L = T − U = m|~v|2 − U = m(x ˙ 2 +y ˙2) − U(x, y, t), 2 2 where U(x, y, t) is the potential energy. Often the potential energy does not depend on time, but it could. By Hamilton’s principle, we know that the actual path the particle takes is the one that minimizes the action S, so we solve the pair of Lagrange equations ∂L d ∂L ∂L d ∂L − = 0, − = 0. ∂x dt ∂x˙ ∂y dt ∂y˙ 6.1. 2D Particle in Cartesian Coordinates 85

Taking the partial derivatives, the Lagrange equations simplify to Tip ∂U d ∂U d Remember when diﬀerenti- − − (mx˙) = 0, − − (my˙) = 0. ∂x dt ∂y dt ating the Lagrangian, the partial derivatives only de- Taking the relevant time derivative and simplifying, gives us pend on what is explicitly ∂U ∂U shown in the Lagrangian, mx¨ = − , my¨ = − . ∂x ∂y whereas the time derivative is a total derivative, Recall that in Cartesian coordinates which often requires using ∂U ∂U the chain rule. F~ = −∇~ U = − xˆ − yˆ = F xˆ + F yˆ. ∂x ∂y x y In other words, the Lagrange equations tell us that

mx¨ = Fx, my¨ = Fy. That is, the actual path that the particle takes (the path that makes L stationary) is the one that satisﬁes Newton’s law F~ = m~a everywhere. Example 6.1.1

Given a particle in 3D under the inﬂuence of only the gravitational force, what is the path the particle takes from t1 to t2 if it is at the same height z at both times? Does it ﬂy up in the air, make a few somersaults and return to its starting position? Does it just sit there levitating? According to Hamilton’s principle, the actual path the particle follows is the one that minimizes the action t2 S = L ~r(t), ~r˙(t), t dt. ˆt1 The Lagrangian in our case is 1 1 L = T − U = m|~v|2 − U = m(x ˙ 2 +y ˙2 +z ˙2) − mgz. 2 2 Notice that the potential energy function is U = mgz since there is only the gravitational force. From this, we get three Euler-Lagrange equations ∂L d ∂L − = 0 ∂x dt ∂x˙ ∂L d ∂L − = 0 ∂y dt ∂y˙ ∂L d ∂L − = 0. ∂z dt ∂z˙ Taking the relevant partial derivatives gives us the simpler equations d 0 − (mx˙) = 0 dt d 0 − (my˙) = 0 dt d −mg − (mz˙) = 0. dt Taking the time derivatives and simplifying gives us x¨ = 0, y¨ = 0, z¨ = −g. 86 Lagrangian Mechanics

Solving these three ordinary diﬀerential equations gives us

x(t) = x0 + v0xt

y(t) = y0 + v0yt 1 z(t) = z + v t − gt2. 0 0z 2 These are the ordinary equations of motion for a projectile, and we can see that the path is a parabola.

6.2 2D Particle in Polar Coordinates

Recall that in polar coordinates ~r = rrˆ ~v = ~r˙ =r ˙rˆ + rφ˙φˆ ˙2 ¨ ˙ ~α = (¨r − rφ )rˆ + (rφ + 2r ˙φ)φˆ = αrrˆ + αφφˆ. The Lagrangian in polar coordinates (if U does not depend on t) is 1 1 L = T − U = m|~v|2 − U = m(r ˙2 + r2φ˙2) − U(r, φ). 2 2 Remember, |~v|2 = ~v · ~v. The Euler-Lagrange equations are then ∂L d ∂L − = 0 ∂r dt ∂r˙ ∂L d ∂L − = 0. ∂φ dt ∂φ˙ Plugging in the partial derivatives gives us ∂U d mrφ˙2 − − (mr˙) = 0 ∂r dt ∂U d − − mr2φ˙ = 0. ∂φ dt Taking the time derivatives, being careful to apply the product rule in the second equation, gives us ∂U mrφ˙2 − − mr¨ = 0 ∂r ∂U − − m 2rr˙φ˙ + r2φ¨ = 0. ∂φ Recall that in polar coordinates, ∂U 1 ∂U F~ = −∇~ U = − rˆ − φˆ = F rˆ + F φˆ. ∂r r ∂φ r φ Rearranging the Lagrange equations as ∂U − = m r¨ − rφ˙2 ∂r 1 ∂U − = m 2r ˙φ˙ + rφ¨ , r ∂φ 6.3. Unconstrained Particles in 3D 87 we see that they are the same as the Newtonian versions Tip F = mα r r The more coordinate sys- Fφ = mαφ. tems in which you can write the kinetic energy 6.3 Unconstrained Particles in 3D from the beginning, the faster it will be to solve La- For a particle in 3D, we can specify its position ~r with three numbers. For example, in grangian mechanics prob- Cartesian coordinates it is (x, y, z), in spherical coordinates it is (r, θ, φ), and in cylindrical lems. However, you can always fall back to the Carte- coordinates it is (ρ, φ, z). There are infinite possible coordinate systems we could use. 2 2 2 Generalized coordinates q , q , and q must have the following properties sian L = m(x ˙ +y ˙ +z ˙ )/2− 1 2 3 U(x, y, z). 1. Knowing the position ~r, we can find the coordinates qi 2. Knowing the coordinates qi, we can uniquely determine the position ~r. Note that ~r is a function only of q1, q2 and q3. That is, ~r = ~r(q1, q2, q3). The velocity ˙ ˙ ˙ ~r can be a function of qi and its first time derivatives. That is, ~r = ~r(q1, q2, q3, q˙1, q˙2, q˙3). The velocity function will never contain something likeq ¨1. For example, in polar coordinates, ~r˙ = ~r˙(r, r,˙ φ˙) =r ˙rˆ + rφ˙φˆ. The Lagrangian in generalized coordinates is

1 2 L = T − U = m~r˙ − U(~r) = L(q , q , q , q˙ , q˙ , q˙ , t). 2 1 2 3 1 2 3

For n unconstrained particles in 3D, we have the position vectors ~r1, ~r2,..., ~rn. To describe the position of n particles in 3D, we need 3n numbers (i.e. three numbers for each particle since we are in 3D). Since the system can be speciﬁed by 3n numbers, it is said to have 3n degrees of freedom. Our generalized coordinates are represented by qi where i = 1, 2,..., 3n, and our Lagrangian is

L = T − U = L(q1, q2, . . . , q3n, q˙1, q˙2,..., q˙3n, t).

We will get a total of 3n Lagrange equations

∂L d ∂L − = 0, ∂qi dt ∂q˙1 where i = 1, 2,..., 3n. So we will end up with 3n equations of motion—one for each generalized coordinate.

6.4 Constrained Particles in 3D

Constraints reduce the number of generalized coordinates that we need. For example, if a particle moving in 3D is constrained to move on some surface given by f(x, y, z) = constant, we only need two generalized coordinates to describe the particle’s position since we can always solve for one variable in terms of the others. Note that the force constraining a particle to a surface does no work on the particle since it is perpendicular to the surface. In general, with k constraints, we need 3n − k generalized coordinates. In general, we have k constraints of the form f`(q1, q2, . . . , q3n, t) = 0, where ` = 1, 2, . . . , k. Notice that the constraints can depend on the generalized coordinates qi and on time t (i.e. the constraint surface could be moving in time). The Euler-Lagrange equation

∂L d ∂L − = 0, ∂q dt ∂q˙ 88 Lagrangian Mechanics

Tip can be interpreted as follows To remember which deriva- ∂L = generalized force tives of the Lagrangian ∂q are generalized forces and ∂L = generalized momentum which are generalized mo- ∂q˙ menta, use dimensional d analysis, remembering that (gen. mom.) = generalized force. dt the Lagrangian has units of energy. In other words, for each generalized coordinate qi, there is a generalized momentum ∂L pi = , ∂q˙i y and a generalized force ∂L dpi Fi = = . ∂qi dt x If L does not depend on qi, then qi is a cyclic coordinate (also called “ignorable” φ coordinate). For any cyclic coordinate ` ∂L d Fi = = 0 =⇒ pi = 0, ∂qi dt and this implies that ∂L pi = = constant. Figure 6.1: Simple pendulum ∂q˙i In other words, there is a conservation law for each coordinate missing from the La- grangian. y Example 6.4.1: Simple Pendulum 1 2 2 at x Consider a simple pendulum of length `, mass m, and angle φ as shown in Fig. (6.1). The rod provides a constraint force. That is, the bob is constrained to φ move on the circle x2 + y2 = `2. Where we would normally need two coordinates ` to deﬁne the position of a particle in a 2D, we only need a single generalized coordinate here. A convenient generalized coordinate in this case is the angle φ. The tangential velocity of the bob is v = `φ˙, and its height measured from its lowest point is h = `(1 − cos φ). So our Lagrangian is Figure 6.2: A pendulum with a moving pivot. 1 1 L = T − U = mv2 − mgh = m`2φ˙2 − mg`(1 − cos φ). 2 2 From the Euler-Lagrange equation, we get ∂L d ∂L d − = −mg` sin φ − m`2φ˙ = −mg` sin φ − m`2φ¨ = 0 ∂φ dt ∂φ˙ dt g φ¨ = − sin φ. ` This is the same well-known result that we get for a simple pendulum by using other methods. Notice that in ∂L d ∂L d − = −mg` sin φ − (m`2φ˙), ∂φ dt ∂φ˙ dt the ﬁrst term is the generalized force (torque in this case), and the quantity in parentheses in the second term is the generalized (angular in this case) momentum. 6.4. Constrained Particles in 3D 89

Example 6.4.2: Moving Pendulum Tip

Consider a pendulum as in the above example, but now the pivot point is Any constant in the La- accelerating to the right with a constant acceleration a. Imagine the pendulum grangian can be imme- swinging in a car that is accelerating with a constant acceleration. diately discarded because Our generalized coordinate is again φ. See Fig. (6.2). We don’t need the it will not contribute to position of the pivot to serve as a generalized coordinate since the position of the the Euler-Lagrange equa- top of the pendulum is constrained. That is, the bob is constrained to exist on a tion(s). time-dependent circle now. To compute the Lagrangian, we need to compute T = mv2/2, but what is the speed v of the bob? In the worst case scenario, we can always write v in terms of Cartesian coordinates. In this case, we know that the position of the pivot at time t is x = at2/2 since it is moving to the right with constant acceleration a. 1 x = ` sin φ + at2 2 y = −` cos φ.

Then the speed squared of the bob is

 2 2 v2 =x ˙ 2 +y ˙2 = `φ˙ cos φ + at + `φ˙ sin φ = `2φ˙2 + 2at`φ˙ cos φ + a2t2.

So our Lagrangian is 1 1 L = T − U = mv2 − mgh = m `2φ˙2 + 2at`φ˙ cos φ + a2t2 + mg` cos φ. 2 2 Note there’s an additional constant in L from U, but we can ignore it because any constant in the Lagrangian does not enter the equation of motion. Taking the relevant derivatives gives us ∂L = −mat`φ˙ sin φ − mg` sin φ ∂φ d ∂L d = mat` cos φ + m`2φ˙ = ma` cos φ − mat`φ˙ sin φ + m`2φ.¨ dt ∂φ˙ dt

Note that φ and φ˙ both depend on t, so diﬀerentiating functions containing these requires us to apply the chain rule. Plugging these into the Euler-Lagrange equation ∂L d ∂L − = 0, ∂φ dt ∂φ˙ gives us −mg` sin φ − ma` cos φ − m`2φ¨ = 0. Solving for φ¨ gives us the equation of motion g a φ¨ = − sin φ − cos φ. ` ` A quick check conﬁrms that a = 0 gives us the equation of motion for a simple pendulum as it should. 90 Lagrangian Mechanics

Example 6.4.3: Equilibria

What is the equilibrium angle of the pendulum with a moving pivot point ¨ (see previous example). There will be some equilibrium angle φe given by φ = 0. Setting φ¨ = 0 in the equation of motion gives us a tan φ = − . e g Whenever you have a complicated system with an equilibrium point, you should determine 1. What is the equilibrium point? 2. Is it stable? 3. What is the frequency of small oscillations about the equilibrium point? To determine if the equilibrium point φe is stable, we let

φ(t) = φe + ε(t),

where ε(t) << 1 is a small angle. Plugging this back into the equation of motion, we simplify and keep only the linear terms. We get g a ε¨ = − sin(φ + ε) − cos(φ + ε). ` e ` e We can linearize it by Taylor expanding the sine and cosine or by applying trig identities to get g a ε¨ ≈ − (sin φ cos ε + cos φ sin ε) − (cos φ cos ε − sin φ sin ε) . ` e e ` e e Since ε is small, we use the small angle approximations cos ε = 1 and sin ε = ε to get g a ε¨ ≈ − (sin φ + ε cos φ ) − (cos φ − ε sin φ ) . ` e e ` e e However, we know that at equilibrium, the equation of motion reduces to g a 0 = − sin φ − cos φ . ` e ` e This allows us to simplify the equation of motion in ε to g a ε¨ ≈ − cos φ + sin φ ε. ` e ` e From the triangle shown in Fig. (6.3) relating a, g, and the equilibrium angle, we note that a g sin φe = , cos φe = . pa2 + g2 pa2 + g2 Using these, we can rewrite our equation as

pa2 + g2 ε¨ = − ε. ` We see that the equilibrium point is stable since the minus sign implies a restoring force. The angular frequency of small oscillations about this stable equilibrium point is s pa2 + g2 ω = . ` 6.4. Constrained Particles in 3D 91

Example 6.4.4: Atwood Machine 2 g Consider the Atwood machine shown in Fig. (6.4). Because of the constraint + φe g 2 force supplied by the rope, we can describe the system with a single generalized a p coordinate x. d a The speed of mass one is v1 =x ˙, and the speed of mass two is v2 = dt (` − x) = −x˙. Then the kinetic energy of the system is Figure 6.3 1 1 1 T = m v2 + m v2 = (m + m )x ˙ 2. 2 1 1 2 2 2 2 1 2 The potential energy of the system if we use the center of the pulley (i.e. the dashed line) as the reference point is

U = −m1gx − m2g(` − x) = (m2 − m1)gx.

We have discarded the irrelevant constant −m2g` since it won’t appear in the x equation of motion anyway. Our Lagrangian is

1 2 m2 L = T − U = (m1 + m2)x ˙ − (m2 − m1)gx. 2 m1 Then the Euler-Lagrange equation gives us Figure 6.4: Atwood machine. ∂L d ∂L d − = −(m2 − m1)g − ((m1 + m2)x ˙) = 0 ∂x dt ∂x˙ dt Tip = (m1 − m2)g − (m1 + m2)¨x = 0 For a system with multiple m1 − m2 x¨ = g. moving “particles,” your m1 + m2 Lagrangian will contain a kinetic and a potential energy term for each particle.

Example 6.4.5: Multiple Atwood Machines

With a pair of coupled Atwood machines as shown in Fig. (6.5), the process of applying Lagrangian mechanics is similar to that for the regular Atwood machine, but now we need two generalized coordinates x and y. We will take `1 to be the length of the rope over the ﬁrst pulley, and `2 to be the length of the rope over the second pulley. The velocities of the three masses are

v1 =x ˙ d v = (` − x + y) = −x˙ +y ˙ 2 dt 1 d v = (` − x + ` − y) = −x˙ − y.˙ 3 dt 1 2 The kinetic energy of the system is then 1 1 1 T = m x˙ 2 + m (−x˙ +y ˙)2 + m (−x˙ − y˙)2. 2 1 2 2 2 3 The potential energy of the system is

U = −m1gx − m2g(`1 − x + y) − m3g(`1 − x + `2 − y). 92 Lagrangian Mechanics

Here, we have two Euler-Lagrange equations

∂L d ∂L − = 0 x ∂x dt ∂x˙ `1 − x ∂L d ∂L m − = 0. 1 ∂y dt ∂y˙

The ﬁrst one gives us y `2 − y m1x¨ + m2(¨x − y¨) + m3(¨x +y ¨) = (m1 − m2 − m3)g.

m2 The second one gives us m 3 −m2(¨x − y¨) + m3(¨x +y ¨) = (m2 − m3)g.

Figure 6.5: Coupled Atwood As this example demonstrates, once you have done a bunch of Lagrangian machines. mechanics problems, you can easily crank out equations of motion even for complicated systems.

Example 6.4.6: Particle on a Cone z0 Consider a particle conﬁned to move on the surface of a cone of half-angle α in the presence of gravity as shown in Fig. (6.6). The particle is constrained to move on the surface of this cone, a constraint α which is holonomic. A natural coordinate system to use here is cylindrical coordinates (ρ, φ, z). y0 Normally we would need three coordinates to specify the position of a particle in 3D, but due to the constraint, we can eliminate one of those coordinates. We can eliminate z everywhere since on the surface of the cone, we have z = ρ cot α. In cylindrical coordinates, velocity is given by x0 ~v =ρ ˙ρˆ + ρφ˙φˆ +z ˙zˆ. Figure 6.6: A particle constrained to move on a cone. This is the same as with polar coordinates, but with the added termz ˙zˆ. If you don’t remember this, you can always write ~v =x ˙xˆ +y ˙yˆ+z ˙zˆ and rewrite in terms of ρ, φ, and z. The kinetic energy in our case is 1 1 T = mv2 = m(ρ ˙2 + ρ2φ˙2 +z ˙2). 2 2 Substituting in the constraint z = ρ cot α gives us 1 1 T = m(ρ ˙2 + ρ2φ˙2 +ρ ˙2 cot2 α) = m(ρ ˙2 csc2 α + ρ2φ˙2). 2 2 The potential energy is given by

U = mgz = mgρ cot α.

Our Lagrangian is then 1 L = T − U = m(ρ ˙2 csc2 α + ρ2φ˙2) − mgρ cot α. 2 The Lagrangian includes ρ, but not φ. So φ is a cyclic coordinate implying that there is a conserved quantity. 6.4. Constrained Particles in 3D 93

For this system, we have two Euler-Lagrange equations ∂L d ∂L − = 0 ∂φ dt ∂φ˙ ∂L d ∂L − = 0. ∂ρ dt ∂ρ˙

From the ﬁrst one, we have that ∂L = 0, which implies ∂L = mρ2φ˙ = constant. So ∂φ ∂φ˙ 2 ˙ 2 we have a conserved quantity. Note that mρ φ = mρ ω = Iω = Lz. That is, the z-component of angular momentum is conserved. So from the φ Euler-Lagrange equation, we get an ODE with two unknowns ρ and φ˙ 2 ˙ mρ φ = Lz. From the second Euler-Lagrange equation, we get ρ¨ − ρφ˙2 sin2 α + g sin α cos α = 0.

˙ 2 Noting that φ = Lz/(mρ ), we have that L2 sin2 α ρ¨ − z + g sin α cos α = 0. (6.1) m2ρ3

Suppose the particle is undergoing a circular “orbit” then ρ = ρ0 = constant, ρ˙ = 0, andρ ¨ = 0, and our equation of motion reduces to

2 2 Lz sin α 2 3 = g sin α cos α. m ρ0

Solving for ρ0 gives us 1 L2 tan α 3 ρ = z . 0 m2g To linearize Eq. (6.1), to obtain the frequency of small oscillations about the equilibrium motion (i.e. about the orbit), we plug in ρ(t) = ρ0 + ε(t) where |ε| << ρ0 to get

2 2 −3 Lz 2 Lz 2 ε ε¨ = 2 3 sin α − g sin α cos α = 2 3 sin α 1 + − g sin α cos α. m (ρ0 + ε) m ρ0 ρ0 Next we make use of the Taylor expansion (1 + x)n = 1 + nx + O(x2), and note that keeping only the linear terms gives us

(1 + x)n ≈ 1 + nx.

This is an approximation that we need to remember. Using this approximation gives us 2 3Lz 2 ε¨ ≈ − 2 4 sin (α) ε. m ρ0 2 2 3 Note that Lz and ρ0 are not independent since Lz = m gρ0 cot α. This allows us to rewrite the result as 3g ε¨ ≈ − sin(α) cos(α) ε. ρ0 94 Lagrangian Mechanics

Tip This is the equation of a simple harmonic oscillator with a restoring force. So the If a generalized coordinate circular orbit is stable. A minor tap on the rotating particle, and it comes back to does not appear in the La- the stable orbit. The frequency of small oscillations about this equilibrium orbit grangian, then the asso- is r3g ciated generalized momen- ω = sin(α) cos(α). tum is conserved. ρ0

Example 6.4.7: Bead on a Spinning Hoop

Consider a circular hoop of radius R as illustrated in Fig. (6.7). The hoop is spinning with a constant angular speed ω about the vertical dashed line. On the hoop is a bead of mass m that slides without friction. Given the presence of gravity, obtain the equation of motion for the bead. We could move to a reference frame that is rotating with the hoop, but we R θ will use Lagrangian mechanics instead. The constraint force in our case is the R sin θ normal force of the hoop pointing inward. Since ω is constant, we need a single generalized coordinate θ to specify the bead’s position. Spherical coordinates are useful in this example. We have φ˙ = ω = constant. The velocity of the bead is Figure 6.7: Bead on a spinning hoop. ~v = Rθ˙θˆ + R sin θφ˙φˆ, so our kinetic energy is 1 T = mR2 θ˙2 + ω2 sin2 θ . 2 Our potential energy is U = mgR(1 − cos θ). Our Lagrangian is then 1 L = T − U = mR2 θ˙2 + ω2 sin2 θ − mgR(1 − cos θ). 2 The Euler-Lagrange equation tells us that ∂L d ∂L − = 0 ∂θ dt ∂θ˙ d mR2ω2 sin θ cos θ − mgR sin θ − mR2θ˙ = 0 dt mR2ω2 sin θ cos θ − mgR sin θ − mR2θ¨ = 0.

So our equation of motion is g θ¨ = ω2 cos θ − sin θ. R As a simple check, we see that ω = 0 reduces the equation of motion to that of the simple pendulum. Any equilibriums that exist occur when θ is constant. This means θ = θ0 and θ˙ = θ¨ = 0. Plugging this condition in, we have that g 0 = ω2 cos θ − sin θ. 0 R 0 Since it is already factored, we can easily see that we have two solutions. When sin θ0 = 0, we have θ0 = 0, π. These equilibrium points occur at the very top and 6.5. Noether’s Theorem 95

2 very bottom of the hoop. The other solution is when ω cos θ0 − g/R = 0, which Tip 2 implies cos θ0 = g/(Rω ). We know that | cos θ0| ≤ 1, so this is only a solution if ω2 ≥ g/R. When writing down the La- To test the stability of these equilibrium points, we consider the small os- grangian, it is crucial that you do it in an inertial cillations about them. For the equilibrium point θ0 = 0, we use the small angle approximations cos θ ≈ 1 and sin θ ≈ θ, to give us frame of reference. g θ¨ ≈ ω2 − θ. R If ω2 −g/R < 0, then there is a restoring force, and the equilibrium is stable. Else, 2 it is unstable. So θ0 = 0 is a stable equilibrium when ω < g/R. 2 −1 2 For ω > g/R, we have the equilibrium point θ0 = cos (g/ω R). These are actually two equilibrium positions, but they are mathematically equivalent. To check their stability, we study the small oscillations by letting θ(t) = θ0 + ε(t). g θ¨ = ω2 cos θ − sin θ R g ε¨ = ω2 cos(θ + ε) − sin(θ + ε) 0 R 0 g = ω2[cos θ + ε sin θ ]) − [sin θ + ε cos θ ]. 0 0 R 0 0 Here we used the trig identities

cos(θ0 + ε) = cos θ0 cos ε − sin θ0 sin ε

sin(θ0 + ε) = sin θ0 cos ε + cos θ0 sin ε,

and the small-angle approximations cos ε ≈ 1 and sin ε ≈ ε. Next, we use the fact 2 2 that at this equilibrium, ω cos θ0 − g/R = 0 and ε ≈ 0, to get

2 2 ε¨ ≈ −ω sin (θ0)ε.

This equilibrium (when it exists, that is, if ω is large enough) is always stable since 2 2 2 2 the right side is always negative. We can write sin θ0 = 1−cos θ0 = 1−(g/Rω ) , then ε ≈ −Ω2ε, where r g2 Ω = ω2 − , R2ω4 is the frequency of the small oscillations about the equilibrium point.

6.5 Noether’s Theorem

If we apply a transformation to a mathematical object and that object looks the same before and after the transformation, we say the object has symmetry with respect to that transformation. For example, a sphere has rotational symmetry because it looks the same before and after we apply a rotation. Symmetries in physics lead to conservation laws. Suppose you have an isolated system of n particles, and you translate the entire system (i.e. every particle) by a ﬁxed displacement ~ε. Nothing physical should change— this is an empirical observation of the laws of physics. That is, the laws of physics have translational symmetry. How do we say this mathematically? 96 Lagrangian Mechanics

If the n particles in your original system have positions ~ri with respect to some Tip ˙ origin, then under the translation, ~ri goes to ~ri + ~ε. Since ~ε is constant in time, ~ri is Noether’s theorem tells us the same before and after the translation. In other words, under translation, the kinetic that each symmetry implies energy of the system X 1 a conservation law. T = m~r˙ 2, 2 i i is unchanged. That is, ∆T = 0. It is a feature of isolated systems that displacing every bit of the system by ~ε does not change the potential energy of the system. In other words,

U = U(~r1 + ~ε,..., ~rn + ~ε) = U(~r1,..., ~rn),

so ∆U = 0. This fundamental assumption will lead to Newton’s third law. Putting these two conclusions together, we have that

∆L = ∆T − ∆U = 0.

That is, the Lagrangian does not change under translation. This is the mathematical statement of translational symmetry. Without loss of generalization, we can specify our coordinate system such that the translation is in the x-direction. That is, ~ε = εxˆ. Then by the chain rule, ∂L ∂L ∂L ∆L = ε + ε + ··· + ε = 0. ∂x1 ∂x2 ∂xn The Euler-Lagrange equations for our system are ∂L d ∂L = , i = 1, 2, . . . , n. ∂xi dt ∂x˙ i This allows us to write n X d ∂L ∆L = ε = 0. dt ∂x˙ i=1 i We can divide out the ε, and we know that ∂L = mx˙ i = pix , ∂x˙ i

where pix is the x-component of the momentum of the ith particle. So

n d X p = 0, dt ix i=1

which implies that the x-component of the total momentum Px of the system is constant (i.e. conserved). So translational symmetry in the x-direction implies conservation of the x-component of the total momentum. That is, conservation of momentum is a mathematical consequence of translational invariance. Similarly, rotational symmetry implies conservation of angular momentum.

6.6 The Hamiltonian

Consider a system with n degrees of freedom, that is, it requires n generalized coordinates. Then the Lagrangian is

L = T − U = L(q1, . . . , qn, q˙1,..., q˙n, t), 6.6. The Hamiltonian 97

where each coordinate is a function of time, qi = qi(t). How does the Lagrangian change in time? To ﬁnd out, we look at the total derivative, which we obtain by applying the chain rule n n d X ∂L X ∂L ∂L L = q˙ + q¨ + . dt ∂q i ∂q˙ i ∂t i=1 i i=1 i

Now, assume that qi(t) are the physical trajectories of the particles in the system. We know these must satisfy Euler-Lagrange equations,

∂L d ∂L = . ∂qi dt ∂q˙i

Using this substitution, we can rewrite the total derivative as

n n d X d ∂L X ∂L ∂L L = q˙ + q¨ + dt dt ∂q˙ i ∂q˙ i ∂t i=1 i i=1 i n X d ∂L ∂L d ∂L = q˙ + (q ˙ ) + dt ∂q˙ i ∂q˙ dt i ∂t i=1 i i n X d ∂L ∂L = q˙ + . dt ∂q˙ i ∂t i=1 i

Rearranging gives us

n d X ∂L ∂L q˙ − L = − dt i ∂q˙ ∂t i=1 i d ∂L H = − , dt ∂t where n X ∂L H = q˙ − L, (6.2) i ∂q˙ i=1 i

∂L is the Hamiltonian of the system. Since = pi, we can also write the Hamiltonian as ∂q˙i

n X H = q˙ipi − L. i=1

The Hamiltonian has units of energy. ∂L Notice that if ∂t = 0, the Hamiltonian is conserved. If L does not explicitly depend on t, then ∂L d = 0 =⇒ H = 0 =⇒ H = constant. ∂t dt

Example 6.6.1

Compute the Hamiltonian for a 1D particle. The kinetic energy is T = mx˙ 2/2, and the potential energy is some function U = U(x), so the Lagrangian is 1 L = T − U = mx˙ 2 − U(x). 2 98 Lagrangian Mechanics

Then the Hamiltonian is ∂L 1 1 H =x ˙ − L = mx˙ 2 − mx˙ 2 − U(x) = mx˙ 2 + U(x) = E. ∂x˙ 2 2

So the Hamiltonian of the particle is its total mechanical energy.

In the example of the 1D particle, we found that the Hamiltonian is the same as the total mechanical energy H = T + U = E.

In general, this happens when T is a quadratic function ofq ˙i. This occurs when there is no explicit time dependence of the positions of the particles. That is, when

~rn = ~rn(q1, . . . , qn).

Each qi may be a function of time, but as long as ~rn does not explicitly depend on time, then qi are called the natural generalized coordinates and then H = T + U = E.

Example 6.6.2

Compute the Hamiltonian for the bead on a spinning hoop (done earlier). We know that the Lagrangian is 1 L = mR2(θ˙2 + ω2 sin2 θ) − mgR(1 − cos θ). 2 Notice that L does not explicitly depend on t, but since the hoop is spinning, ~r must include t in order to specify the angle φ through which the hoop has rotated. So θ is not a natural generalized coordinate. The Hamiltonian is ∂L 1 1 H = θ˙ −L = θmR˙ 2θ˙−L = mR2θ˙2 − mR2ω2 sin2 θ+mgR(1−cos θ) 6= T +U. ∂θ˙ 2 2

∂L Notice that H is conserved since ∂t = 0, but it is not equal to the total mechanical energy. This is because the rotating hoop is performing work on the mass. We can write the Hamiltonian as 1 H = mR2θ˙2 + U (θ), 2 eff where the eﬀective potential energy is 1 U (θ) = − mR2ω2 sin2 θ + mgR(1 − cos θ). eff 2 So H is the “mechanical energy” with respect to θ in the hoop’s frame of reference. If you ﬁnd Ueff , you can graph it and easily see where there are equilibria and whether those equilibria are stable or unstable by looking at the maxima and minima.

6.7 Lagrange Multipliers

The beauty of Lagrangian mechanics is that we can churn out equations of motion while implicitly ignoring the constraint forces. But what if we want to know the constraint 6.7. Lagrange Multipliers 99 forces? Suppose we have a system with generalized coordinates qi and a constraint force f(q1, q2, . . . , qn) = constant. Normally, we reduce the number of coordinates we use by one for each constraint force. But if we want to calculate the constraint forces themselves, we include all the coordinates and use a modiﬁed form of the Euler-Lagrange equations

∂L d ∂L ∂f − + λ = 0, ∂qi dt ∂q˙i ∂qi where λ is the Lagrange multiplier. Then, it turns out that the qi component of the constraint force is ∂f Fcqi = λ . ∂qi 100 Lagrangian Mechanics

6.8 Summary: Lagrangian Mechanics

Skills to Master • Remember the expressions for position, velocity, and acceleration in Cartesian, polar, spherical, and cylindrical coordinates • Calculate the kinetic energy in Cartesian, polar, spherical, cylindrical, and generalized coordinates • Interpret derivatives of the Lagrangian as generalized forces and momenta • Calculate the Lagrangian for various constrained and unconstrained systems and obtain the equation(s) of motion by applying the Euler-Lagrange equations • Given an equation of motion (i.e. second order ODE in the generalized coordinate), determine if there is an equilibrium position of that coordinate, locate it, determine its stability, and calculate the frequency of small oscillations about that point • Given the Lagrangian of a system, compute its Hamiltonian • Use the method of Lagrange multipliers to calculate constraint forces

The path that a particle takes from one point to 2. Obtain the Euler-Lagrange equations another is that which minimizes the action. We can ∂L d ∂L easily compute this path by ﬁnding the Euler-Lagrange − = 0, equations of the Lagrangian ∂qi dt ∂q˙

L = T − U, where qi is a generalized coordinate. 3. Identify the equations of motion by solving the where T is the kinetic energy and U is the potential Euler-Lagrange equations. energy. 4. Identify any equilibria and their stability. Given 1. Be able to write the Lagrangian an equation of motion of the formx ¨ = f(x), L = T − U, you can identify equilibrium points x0 by letting x = x0 andx ˙ =x ¨ = 0. If the equation of motion in different coordinate systems. The hardest part reduces to the form is usually to write the kinetic energy T = kx2/2 of the object. We should be able to write the x¨ = −ω2x, kinetic energy in different coordinate systems, but we can always fall back to the Cartesian then the equilibrium is stable. If it does not re- representation of the kinetic energy, which is duce to this form, you have to look at small os- T = m(x ˙ 2 +y ˙2 +z ˙2)/2. Velocity in cylindrical cillations as described next. and spherical coordinates (respectively) is 5. Compute the frequency of small oscillations about the equilibria. There are several possible ~v =ρ ˙ρˆ + ρφ˙φˆ +z ˙zˆ ways of doing this. ~v =r ˙rˆ + rφ˙φˆ + r sin θ φ˙φˆ. • If the equation of motion for small oscillations reduces to the form Note, the Lagrangian may or may not depend on 2 time. When writing down v for an object, the x¨ = −ω2x, general procedure is to write down the coordinates of the object at time t, differentiate them then the equilibrium is stable, and the fre- with respect to time (you may have to apply the quency of small oscillations is ω. You may chain and product rules) to obtain ~v, and then have to make a small-angle approximation take the dot product with itself to get v2. to do this. 6.8. Summary: Lagrangian Mechanics 101

• Let The first one is obtained by binomial expansion and the x(t) = x0 + ε(t), other two are obtained by trig identities in conjunction where ε is small. Plug this into the equation with small angle approximations. of motion and simplify to get an equation of Suppose we have a system with generalized co- the form ordinates qi and a constraint force f(q1, q2, . . . , qn) = ε¨ = −ω2 ε. constant. Normally, we reduce the number of coordinates we use by one for each constraint force. But if You may have to use Taylor series to expand we want to calculate the constraint forces themselves, trig functions containing x + ε(t) and keep 0 we include all the coordinates and use a modified form only the first several terms since ε is small. of the Euler-Lagrange equations 6. Determine the generalized force. In the Euler- Lagrange equation, the generalized force is ∂L d ∂L ∂f − + λ = 0, ∂L ∂qi dt ∂q˙i ∂qi , ∂q where λ is the Lagrange multiplier. Then, it turns out and the generalized momentum is that the qi component of the constraint force is ∂L . ∂f ∂q˙ Fcqi = λ . ∂qi Some helpful approximations are (1 + ε)α ≈ 1 + αε

sin(x0 + ε) ≈ sin x0 + ε cos x0

cos(x0 + ε) ≈ cos x0 − ε sin x0. Chapter 7

Orbits

Here we will only consider two-body central force orbits. Consider a generic two-body problem such as the sun-Earth system, the Earth-moon system, or even an electron- proton system. As shown in Fig. (7.1), mass 1 is at ~r1 relative to the origin and mass 2 is at ~r2, so the vector from mass 2 to mass 1 is ~r = ~r1 − ~r2. m For spherically symmetric central forces, the force that either mass exerts on the 1 ~r= other is parallel to the vector between them. Hence, the potential energy for such a pair ~r1 − ~r2 of masses is a function only of the distance between them m2

U(~r1, ~r2) = U (|~r1 − ~r2|) . ~r1 There are many examples of such forces including gravitation, Coulomb’s law, and even ~r2 Hooke’s law for springs. We can immediately write down the Lagrangian for this system 1 1 O L = m (~r˙ )2 + m (~r˙ )2 − U|~r − ~r |. 2 1 1 2 2 2 1 2 Figure 7.1: Two-body problem. Notice that we are using six coordinates here since each vector has three components. Our goal is to reduce this number of coordinates required to describe the system. We start by making a change of coordinates. Instead of ~r1 and ~r2, we will change to m1 the relative coordinate as shown in Fig. (7.2) ~r m2 ~r = ~r1 − ~r2, and the center of mass coordinate R~ m ~r + m ~r R~ = 1 1 2 2 . m1 + m2

Solving this pair of equations for ~r1 and ~r2, we get the transformation equations that O take us back to the original vectors

Figure 7.2: Two-body problem m2 ~r1 = R~ + ~r in center-of-mass coordinates. m1 + m2 m1 ~r2 = R~ − ~r. m1 + m2 Next we rewrite the Lagrangian in terms of the two new vectors. We start by rewriting the kinetic energy as 2 2 1 ˙ m2 ˙ 1 ˙ m1 ˙ T = m1 R~ + ~r + m2 R~ − ~r 2 m1 + m2 2 m1 + m2 1 ˙ 2 1 m1m2 ˙ 2 1 ˙ 2 1 ˙ 2 = (m1 + m2)R~ + ~r = MR~ + µ~r , 2 2 m1 + m2 2 2 Orbits 103 where M = m1 + m2, µ is the total mass, and m1m2 µ = , ~r m1 + m2 ˙ is called the reduced mass. We can think of MR~ 2/2 as the kinetic energy of the center of mass and µ~r˙ 2/2 as the relative kinetic energy. We can now rewrite the Lagrangian for the two-body problem as O Figure 7.3: A 2-body problem 1 ˙ 1 L = MR~ 2 + µ~r˙ 2 − U(r). as an eﬀective 1-body problem. 2 2

If there is an external force ﬁeld present then U will also depend on the center of mass R~ of the system. Notice that all three components of R~ are cyclic. That is, they do not appear in the Lagrangian. This means the Euler-Lagrange equation for R~ reduces to

d ˙ MR~ = 0, dt which implies conservation of total momentum for the system. The total momentum is ˙ m1~r1 + m2~r2 ˙ ˙ MR~ = (m1 + m2) = m1~r1 + m2~r2 = P~ . m1 + m2 ˙ ˙ Since the Euler-Lagrange equation tells us that MR~ is constant, we know that R~ is constant. That is, the center of mass is moving at a constant velocity. We can then put ourselves in the same reference frame as the center of mass—a reference frame where ˙ R~ = 0. We can do this since there is no external force. The Lagrangian now simpliﬁes to 1 L = µ~r˙ 2 − U(r). 2 Notice that we’ve gone from six generalized coordinates to three—the three components of ~r. This Lagrangian is the same as that for a single mass µ at position ~r. We’ve eﬀectively turned a two-body problem into a one-body problem as shown in Fig. (7.3). The Euler-Lagrange equation gives us

µ~r¨ = −∇~ U(r).

In the center of mass frame with R~ = 0, the angular momentum of the system is ˙ ˙ L~ = ~r1 × m1~r1 + ~r2 × m2~r2 m2 m2 ˙ m1 m1 ˙ = ~r × m1 ~r + ~r × m2 ~r m1 + m2 m1 + m2 m1 + m2 m1 + m2 2 2 m2 ˙ m1 ˙ = m1 ~r × ~r + m2 ~r × ~r m1 + m2 m1 + m2 m m = 1 2 ~r × ~r˙ = ~r × µ~r˙ = ~r × ~p. m1 + m2 In other words, the angular momentum of our system, when we are in the center of mass frame, is the same as the angular momentum of a particle of mass µ. Since the angular momentum of the two-body system is constant in the center of mass frame, in particular, the direction of the angular momentum is constant, we know 104 Orbits

that both bodies stay in a single plane. We can choose our z-axis to be in the direction µ of the angular momentum, then we know that the two bodies stay in the xy-plane. This y means we can eliminate one more coordinate, so that we have only two coordinates r and φ as shown in Fig. (7.4). So far, we have reduced the two-body problem in three r dimensions to an eﬀective one-body problem in two dimensions. Now the position is given by ~r = rrˆ, and the velocity is given by ~r˙ =r ˙rˆ + rφ˙φˆ, so our Lagrangian becomes 1 φ x L = µ r˙2 + r2φ˙2 − U(r). 2 Notice that φ is cyclic which implies a conserved quantity. By looking at the Euler Lagrange equation for φ, we get Figure 7.4: A 2-body problem in ∂L 3D as an eﬀective 1-body prob- = µr2φ˙ = constant. lem in 2D. ∂φ˙

The conserved quantity is the angular momentum L~ = ` zˆ = µr2φ˙ zˆ. From the Euler- Tip Lagrange equation for r, we get Given the equation of mo- dU µrφ˙2 − = µr.¨ tion of a 1D system, we can dr obtain the energy by multiplying both sides by the Substituting φ˙ = `/(µr2) gives us speedr ˙. `2 dU µr¨ = − . µr3 dr

This is an equation in the form F~ = m~a for a 1D particle that feels a regular force −dU/dr and also a ﬁctitious (or eﬀective) centrifugal force

`2 F = . cf µr3 To ﬁnd the energy, there’s a standard trick we can employ that always works for the 1D case. We multiply both sides of the equation of motion byr ˙, which allows us to rewrite it as the energy

 `2 dU r˙ (µr¨) =r ˙ − µr3 dr d 1 d `2 µr˙2 = − − U(r) . dt 2 dt 2µr2

This implies that 1 `2 µr˙2 + + U(r) = constant = T + U. 2 2µr2 The actual kinetic energy for our “particle” is

1 1 `2 T = µ r˙2 + r2φ˙2 = µr˙2 + . 2 2 2µr2 This looks like the energy of a 1D particle with an eﬀective potential of

`2 U (r) = + U(r). eff 2µr2 The ﬁrst term on the right is the “potential energy” associated with the ﬁctitious centrifugal force. 7.1. The Kepler Problem 105

To summarize, our two-body central force problem has been reduced to an eﬀective Tip 1D problem in the center of mass frame with the center of mass at the origin and equation of motion If one body in a two body `2 µr¨ = + F (r). system is much more mas- µr3 sive than the other, then µ ' m, where m is the mass where of the smaller body. dU F (r) = − rˆ, dr is the force between the two bodies, µ is the reduced mass of the two bodies, and r is the distance between the two bodies. The energy of the system is

1 E = µr˙2 + U (r) = constant. 2 eff where the eﬀective potential is

`2 U (r) = + U(r). eff 2µr2

Note that this gives the relative motion of the two bodies. Neither of the bodies are actually at ~r. Rather, the center of mass of the two bodies is at the origin, and ~r points from one body to the other. If the large body is much more massive than the second body, then the large body is essentially at the center of mass of the system. In that case, we can approximate µ ' m and interpret these as the equations of motion for the smaller body. The angular momentum of the orbiting (i.e. smaller) body is

` = µr2φ.˙

7.1 The Kepler Problem

In this section we assume that energy is being conserved, so there will be no decaying orbits. Often we are interested in the shape of the orbit of the smaller body. The shape of the orbit is given by r(φ). To start, we change variables from t to φ. By the chain rule,

d dφ d d ` d = = φ˙ = . dt dt dφ dφ µr2 dφ

This allows us to write d d ` d ` d µr¨ = µ r = µ r . dt dt µr2 dφ µr2 dφ

Equating this with µr¨ = `2/(µr3) + F (r) gives us

1 d 1 d 1 µ r = + F (r). r2 dφ r2 dφ r3 `2

This ODE becomes a lot easier to solve if we make the substitution u = 1/r. Then

d d 1 1 d r = = − u, dφ dφ u u2 dφ 106 Orbits

and after simpliﬁcation, our ODE becomes r d2u µ = −u − F (u). dφ2 `2u2 The Kepler problem is the generic problem of an attractive inverse square force of the form γ F (r) = − 2 , t r rmin which gives us the potential γ U(r) = − . Figure 7.5: The orbital altitude r as a function of time. For gravitation, γ = GmM. Our ODE now becomes the linear ODE d2u µγ Tip + u = . dφ2 `2 At the point of closest ap- The homogeneous ODE is just d2u/dφ2 = −u and has the solution u(φ) = A0 cos φ + proach,r ˙ = 0 for an orbit. 0 B sin φ = A cos(φ − φ0). The full solution is just the homogeneous solution plus the constant, µγ u(φ) = A cos(φ − φ ) + . 0 `2 Substituting u = 1/r, we can write the solution as c = 1 + ε cos(φ − φ ), r 0 or as c r(φ) = . 1 + ε cos φ This is the equation of a curve in polar coordinates. It’s a conic section with one focus at the origin. The two constants are

`2 `2A c = , ε = . µγ µγ

The constant c has units of length and the constant ε, which is called the eccentricity of the orbit, is dimensionless. When r = rmin we have that φ = φ0. Here we chose φ0 = 0 without loss of generality since it just speciﬁes the starting position of the orbit. The energy of a Kepler orbit is 1 `2 γ E = µr˙2 + − . 2 2µr2 r

As the orbiting body approaches rmin it is going toward the center of force, so its radial velocity isr ˙ < 0. When it’s going away from rmin, its radial distance is increasing so r˙ > 0. When r = rmin,r ˙ = 0 as shown in Fig. (7.5). Plugging this into the orbit equation gives us c/rmin = 1 + ε. Solving this for 1/rmin, plugging it into the energy equation, substituting c = `2/(µγ), and simplifying gives us µγ2 E = (ε2 − 1). 2`2 Solving this for ε gives us the eccentricity in terms of the energy of the orbiting body s 2E`2 ε = 1 + . µγ2 7.1. The Kepler Problem 107

Bounded Orbits U For bounded orbits, the total energy is negative E < 0 as shown in Fig. (7.6), the eff eccentricity is between 0 and 1, and the shape of the orbit is an ellipse. Plotting r(φ) in polar coordinates gives us an ellipse as shown in Fig. (7.7). Notice that when φ = 90◦, we get r = c. rmin rmax We can also plot it in Cartesian coordinates by starting with r = c − εr cos φ then r substituting r cos φ = x and r = px2 + y2. After simplifying and completing the square, E we get (x + d)2 y2 + = 1, a2 b2 Figure 7.6: The eﬀective poten- where d = cε/(1 − ε2) is the horizontal shift, a = c/(1 − ε2) is the semi-major axis, and √ tial Ueff for a 2-body problem. 2 b = c/ 1 − ε is the semi-minor axis of the ellipse. Notice that d = εa. This is just the The energy E in this case im- equation of an ellipse shifted via x + d such that one of the foci is at the origin of the plies a bounded orbit. coordinate system. We can recenter the ellipse at the origin by plotting in x0 = x + d. y

c r r φ rmin φ 2b x0 = x + d εa

Figure 7.7: For bounded orbits E < 0 and the shape of the orbit 2a is an ellipse.

At the left end of the ellipse, is the point of greatest distance for the orbiting mass from the force center. This point is called the aphelion or apogee. at the right end is the point of closest approach called the perihelion or perigee. Together, these are called apsidal points. b r Notice that a = εa+rmin, so rmin = a(1−ε). Similarly, rmax = a(1+ε). Substituting a = c/(1 − ε2) gives us εa c c r = , r = . min 1 + ε max 1 − ε

Solving both for c and equating them, gives us Figure 7.8

r − r ε = max min . rmax + rmin

It is a feature of the elliptical Kepler orbit that the same path keeps getting traced out. If there is a perturbation term, then there will be perihelion precession. If the object returns to the same spot after some time, the orbit is said to be closed. Example 7.1.1

Suppose a satellite is rmin from the center of Earth at its closest approach and rmax at its greatest distance. What is its distance above the surface of Earth when it has rotated 90 degrees about Earth from its closest approach? How far is it from Earth when it has moved halfway between perigee and apogee? For the ﬁrst question, note that it is asking for c, which is easily obtained by rearranging rmin = c/(1 + ε) or rmax = c/(1 − ε). 108 Orbits

For the second question, we want to calculate the distance r shown in U eff Fig. (7.8). One way to calculate it is by starting with the Pythagorean theo- 2 2 2 2 rem r = b + ε a , plugging in the value for ε found using the formula√ for ε in 2 2 2 r = ` terms of rmin and rmax, and then substituting a = c/(1 − ε ) and b = c/ 1 − ε µγ to get r 1 r = (r + r ) . 2 min max A more elegant way to ﬁnd r is to recall how an ellipse can be drawn by a E pencil pushing against a string of length L = 2r that has each end pinned to a diﬀerent focus of the ellipse. When your pencil reaches rmin or rmax you can see Figure 7.9: The energy and ra- that the length of the string is also L = rmin + rmax. dius of a circular orbit. A special case of elliptical orbit is the circular orbit which occurs when ε = 0. In this case, a = b, c = r = `2/(µr) as shown in Fig. (7.9), and the energy reduces to E = −µγ2/(2`2). The central force is now always centripetal, and F = −γ/r2 = µv2/r Ueff implies that the speed of the orbiting body is a constant v = pγ/(µr). The angular momentum of the orbiting body is ` = rp = rµv, and its radius is r = `2/(µγ). Its kinetic 2 E energy is T = mv /2 = µγ/(2µr) = γ/(2r), and its potential energy is U = −γ/r. By the virial theorem, T = −U/2. r rmin Example 7.1.2

Suppose that at perigee, a satellite is rmin above the Earth’s center and has speed v0. What is the eccentricity of its orbit, and what is rmax? At apogee and perigee, the velocity of the orbiting body is perpendicular to Figure 7.10: The energy of an the direction of the force and to ~r. Given r and v at perigee, we can calculate unbounded orbit. min 0 its angular momentum as ` = ~r × ~p = rminmv0. So since the mass of Earth is much greater than the satellite, we use µ = m so µγ = Gm2M and

y `2 `2 r2 m2v2 r2 v2 c = = = min 0 = min 0 . µγ Gm2M Gm2M GM

Once we have c, we can calculate ε using rmin = c/(1 + ε), and then we can calculate rmax using rmax = c/(1 − ε).

r φ Unbounded Orbits α αε For unbounded orbits, the total energy is positive E > 0 as shown in Fig. (7.10), the eccentricity is greater than 1, and the shape of the orbit is a hyperbola. The particle reaches some closest approach rmin then escapes to infinity with residual energy. For unbounded orbits, φ˙ → 0 and ~r˙ → constant as r → ∞. Converting r(φ) to Cartesian coordinates using the same process as for elliptical orbits gives us the equation of a hyperbola (x − δ)2 y2 − = 1, α2 β2 √ where α = c/(ε2 − 1), β = c/ ε2 − 1, and δ = εα. The asymptotes of the hyperbola are β p y = − x0 = − ε2 − 1 x0, α where x0 = x − δ. This hyperbolic orbit is illustrated in Fig. (7.11). A special case occurs when E = 0 and ε = 1. The particle barely escapes to infinity, that is, its velocity goes to zero as r → ∞. The resulting orbit is a parabola, which can 7.2. Kepler’s Laws 109 be verified by letting ε = 1 in r(φ) and converting to Cartesian coordinates as before. In Figure 7.11: Hyperbolic orbit. this case, we get The horizontal axis is x0 = x−δ. y2 = c2 − 2cx, where c = `2/(µr).

7.2 Kepler’s Laws

Kepler’s First Law Kepler’s ﬁrst law states that planetary orbits are ellipses with the center of force at one focus.

Kepler’s Second Law Kepler’s second law states that a line drawn from a planet to the sun sweeps out equal areas in equal times. This can be shown by constructing a triangle with angle dφ and the two adjacent sides being ~r(t) and ~r(t + dt). The area of this triangle is dA = (1/2)r2 dφ. Then

dA 1 dφ ` = r2 = = constant. dt 2 dt 2µ

Here, we used ` = µr2φ˙.

Kepler’s Third Law Kepler’s third law states that the square of the orbital period is proportional to the cube of the semi-major axis. We start by integrating dt = (2µ/`) dA to get t = 2µA/`, where A is the area of the ellipse, and t is the time it takes to make a complete orbit. We know that the area of an ellipse is πab, so the orbital period is 2µ τ = πab. ` √ Squaring both sides and plugging in b = c/ 1 − ε2 and c = `2/(µγ) and using the fact that a = c/(1 − ε2), we get 4π2µ τ 2 = a3. γ This holds for any inverse square law where U(r) = −γ/r. For Earth orbiting the sun, we have γ = GMeMs. Since Ms >> Me, µ reduces to Me and we get 4π2a3 τ 2 = . GMs

7.3 Transfer Orbits

Let us consider a trip to Mars. Some relevant data includes

11 RE = 1.5 × 10 m ← Earth’s orbital radius 11 RM = 2.3 × 10 m ← Mars’s orbital radius

We will assume both orbits are circular, so εE = εM = 0. 110 Orbits

A transfer orbit is an orbit that will take us from Earth’s orbit and put us in Mars’ orbit like the dotted line in Fig. (7.12). The point A is the location where both the Earth and the rocket can be found just prior to the rocket’s liftoff. The point B is the location where both Mars and the rocket can be found when the rocket is ready to land. ~r All three of these orbits are Kepler orbits, so they all have the general equation φ B A c `2 γ r(φ) = , c = ,U = − . 1 + ε cos φ µγ r The three orbits are 2 È RE = ← Rocket orbiting with Earth Figure 7.12: Transfer orbit. µγ `2 R = M ← Rocket orbiting with Mars M µγ 2 `t rt = ← Rocket in transfer orbit µγ(1 + εt cos φt) Since Earth and Mars occupy circular orbits, their orbital shapes r(φ) are constant. Note that these are the three different orbits of the rocket (not the planets), hence the reduced mass µ is the same in all three cases. This also means, for example, that È is not Earth’s angular momentum, but rather the angular momentum of the rocket when it is on Earth. For the rocket to leave from Earth and arrive at Mars, we know the transfer orbit needs to have rmin equal to the radius of Earth and rmax equal to the radius of Mars. This implies that the eccentricity of the transfer orbit is

rmax − rmin RM − RE εt = = . rmax + rmin RM + RE Suppose our rocket is currently sitting on Earth, so its orbit about the sun is the same as Earth’s orbit. To make it to Mars, we need to insert our rocket into the transfer orbit. To achieve this insertion, what ratio v λ = r , vE

of speeds does our rocket need to have before and after leaving Earth? The speed vE is the orbital speed of the rocket before it leaves Earth. Hence, it is equal to Earth’s orbital speed. The speed vr is the speed of the rocket immediately after it leaves Earth’s orbit and enters the transfer orbit. I denote this speed vr rather than vt because the transfer orbit is elliptical, and hence the speed vt(φ) of an object in the transfer orbit is not constant. However, vr is constant—it is just a number. Incidentally, the factor λ > 1 is called the thrust factor. We need to relate the thrust factor λ to the transfer orbit. What thrust factor will put is in the proper orbit with eccentricity εt? Notice that at the point A noted in Fig. (7.12) Earth’s orbit and the transfer orbit coincide. Furthermore, at this point, the velocities of both Earth and the transfer orbits are perpendicular to ~r. This makes the angular momentum of the rocket before and after liftoﬀ easy to calculate

`E = REµvE

`t = REµvr.

Here, È is the angular momentum of the rocket when it is sitting on Earth, and `t is the angular momentum of the rocket after it has lifted off and entered the transfer orbit. Dividing the second equation by the first, REµ cancels from both and we get v ` λ = r = t . vE È 7.3. Transfer Orbits 111

Immediately after liftoﬀ, the rocket has entered the transfer orbit, its distance from the sun is rt(φ) = RE and its angle is φ = 0, so its orbit equation becomes

2 2 `t `t rt(φ) = → RE = . µγ(1 + εt cos φt) µγ(1 + εt) Compare that with its orbit equation right before liftoﬀ `2 R = E . E µγ Equating the two and simplifying gives us

`t √ = 1 + εt. `E

Noting that λ = `t/`E and plugging in what we calculated for εt gives us the required tangential thrust factor to enter the transfer orbit from Earth

r 2R r6 λ = M ≈ . RM + RE 5 Once we get to Mars, we have to ﬁre the rocket again to exit the transfer orbit and enter Mars’ orbit. So we need a second thrust factor

0 vM λ = 0 , vr 0 where vM is the orbital speed of Mars and vr is the orbital speed of the rocket as it reaches Mars, but while it is still in the transfer orbit. Again, the orbital velocity of Mars and the orbital velocity of the rocket in the transfer orbit near Mars are both perpendicular to the radial vector ~r, and so the angular momenta are easy to calculate

`M = RM µvM 0 `t = RM µvr. Dividing the ﬁrst by the second gives us

0 vM `M λ = 0 = . vr `t

Since the angular momentum of each orbit is conserved, the angular momentum `t of the transfer orbit is the same at Earth and at Mars. In the orbit with Mars, the rocket has orbit equation `2 R = M . M µγ

Just prior to reaching Mars, the rocket is at a distance RM from the orbital center and its angle is φ = 180◦ so its orbit equation becomes

2 2 `t `t rt(φ) = → RM = . µγ(1 + εt cos φt) µγ(1 − εt) Equating the two gives us ` 1 M = √ = λ0. `t 1 − εt

Plugging in what we found earlier for εt gives us

rR + R r5 λ0 = M E ≈ . 2RE 4 112 Orbits

We know that the angular momentum of the rocket is conserved throughout its journey along the transfer orbit. So the angular momentum at the beginning, which was 0 `t = REµvr is the same as the angular momentum `t = RM µvr at the end of the journey. This tells us that the rocket’s speed as it approaches Mars is related to its speed as it left Earth as 0 RE vr = vr. RM The speed of the rocket in Mars’ orbit is

0 0 0 RE 0 RE vM = λ vr = λ vr = λ λvE. RM RM We know that r 2R rR + R rR λλ0 = M E M = M , RE + RM 2RE RE so r r 0 RE RE RM RE vM = λλ vE = vE = vE. RM RM RE RM The result v r R M = E , vE RM is consistent with the fact that for circular orbits, 1 v ∝ √ . r How long will this trip to Mars take? We know that it’s half of the period of the transfer orbit. In general, the period of a Kepler orbit is given by

4π2a3 τ 2 = , GMs

where a is the semi-major axis of the orbit and Ms is the mass of the sun. This is Kepler’s third law. In our case, we want s 2 3 4π 2 τt = at . GMs

We can see from Fig. (7.12) that R + R a = E M . t 2 We know that for Earth, s 2 3 4π 2 τE = RE = 1 year. GMs Dividing one by the other gives us

3 τ a 2 t = t . τE RE So 3 3 2 2 at RE + RM τt = τE = τE ≈ 1.4 years. RE 2RE The trip time is half of this or 0.7 year. 7.4. Summary: Orbits 113

7.4 Summary: Orbits

Skills to Master • Reduce a 2-body central force problem in 3D to a 1-body problem in 2D • Derive the solution for a Kepler orbit • Be able to switch back and forth between the absolute coordinates of a 2-body problem and the center-of-mass coordinates for the associated eﬀective 1-body problem • For a circular orbit, be able to relate the force, kinetic energy, potential energy, speed, angular momentum, radius, and period • For an elliptical (Kepler) orbit, be able to relate the apogee and perigee altitudes, orbital speed, angular momentum, energy, eccentricity, and period • Know the diﬀerence between circular, elliptical, parabolic, and hyperbolic orbits with regards to eccentricity and energy • Know Kepler’s laws • Be able to solve a transfer orbit problem

Two Body Central Force which gives us

The Lagrangian for the two-body with central force `2 problem is µr¨ = − F (r), µr3 1 ~˙ 2 1 ˙ 2 L = MR + µ~r − U(r) dU 2 2 where ` is the angular momentum, and F (r) = − dr is = Lcm + Lrel, the central force. The energy of the system is where 1 E = µr˙2 + U (r) = constant. 2 eff ~r = ~r1 − ~r2 (relative coordinate) m ~r + m ~r where the eﬀective potential is R~ = 1 1 2 2 (center of mass) m1 + m2 2 M = m + m (total mass) ` 1 2 Ueff (r) = + U(r). m m 2µr2 µ = 1 2 (reduced mass) m + m 1 2 The angular momentum of the orbiting (i.e. smaller) Note that ~r1 is the vector from our chosen origin to body is mass 1 and ~r2 is the vector to mass 2. ` = µr2φ.˙ To convert back to the original vectors, we can use Angular momentum is always conserved. You can ~ m2 ~r1 = R + ~r often use that in an orbits problem. For example, when m1 + m2 m1 the orbiting object is at r = rmin then its angular mo- ~r2 = R~ − ~r. m1 + m2 mentum is ` = rmv. Since angular momentum is conserved, it is always equal to this quantity, which implies In the center of mass frame, R~ = 0. In the center that as r increases, v must be decrease. of mass frame then, If the mass of the smaller body is negligible compared to the larger body, you can replace the reduced 1 2 L = µ~r˙ − U(r), mass µ with the mass of the smaller body m. 2 114 Orbits

Kepler Orbits Orbit Eccentricity Energy The Kepler problem is the generic problem of an attractive inverse square force Circle ε = 0 E < 0 γ γ F (r) = − ,U(r) = − . r2 r Ellipse 0 < ε < 1 E < 0 For gravitation, γ = GmM. For the Kepler problem, the shape of the orbit is given by Parabola ε = 1 E = 0 c r(φ) = . 1 + ε cos φ Hyperbola ε > 1 E > 0 which is the equation for a conic section. The eccentricity is ε, and Transfer Orbits `2 c = . µγ A transfer orbit is used by a rocket to move from a Note that perigee and apogee are given by lower orbit to a higher orbit or vice versa. Suppos- c ing that the higher and lower orbits are circular, then rmin = the transfer orbit will be elliptical and needs to have 1 + ε c rmin equal to the radius of the lower orbit and rmax r = . max 1 − ε equal to the radius of the higher orbit. This makes it straightforward to calculate the eccentricity of the These can be readily deduced from the equation that transfer orbit. gives the shape of the orbit. Notice that at perigee All three of these orbits are Kepler orbits, so they and apogee, the velocity of the orbiting body is per- all have the general equation pendicular to the direction of the force. These imply that r − r c `2 ε = max min . r(φ) = , c = . rmax + rmin 1 + ε cos φ µγ y If v1 is the velocity of the rocket before it enters the transfer orbit and v2 is the velocity of the rocket c immediately after it enters the transfer orbit, then the r rmax necessary tangential thrust factor is φ 2b x0 = x + d v2 εa rmin λ = . v1

Given the standard problem setup, the rocket enters the transfer orbit at φ = 0 and leaves the transfer 2a orbit at φ = 180◦. This makes the angular momentum of the rocket easy to calculate before and after it The energy of a Kepler orbit is related to its ec- enters the transfer orbit at φ = 0. The two resulting centricity by equations can be equated and solved for v2/v1 to get µγ2 E = (ε2 − 1). λ. Similarly, the angular momentum of the rocket is 2`2 easy to calculate before and after it leaves the trans- The period of a Kepler orbit is related to its semi-major fer orbit at φ = 180◦. The two resulting equations axis as can be equated to obtain the “braking” thrust factor 4π2µ τ 2 = a3. necessary to leave the transfer orbit. γ Chapter 8

Noninertial Frames

Newton’s laws are only valid in inertial frames. Often we need to solve problems in noninertial frames. For example, we are in a rotating frame right now, since the Earth is spinning on its axis.

8.1 Frame with Linear Acceleration

Suppose you have a stationary frame S0 with coordinate axes (x0, y0, z0) and a frame S with coordinate axes (x, y, z) that is at R~ s with respect to frame S0 and is accelerating as ¨ R~ s = A~ with respect to the stationary frame S0. A mass m has position ~r0 with respect to the stationary frame S0 and position ~r with respect to the accelerating frame S. z

z0 A~

y R~ s y 0 x ~r ~r0 x0 m

Since frame S0 is an inertial frame, we know that Newton’s law holds and ¨ m~r0 = F~. In the accelerating frame S, we can write

~r0 = ~r + R~ s, and this implies that ¨ ¨ ¨ ¨ ~r0 = ~r + R~ s = ~r + A~. ¨ ¨ So we have that m~r0 = m(~r + A~) = F~. This gives us Newton’s second law modiﬁed for a mass m in a reference frame accelerating with linear acceleration A~

m~r¨ = F~ − mA~.

We can think of −mA~ as a ﬁctitious inertial force. It is negative because the eﬀective force due to the acceleration of the frame is in the direction opposite the acceleration of 116 Noninertial Frames

the frame. We can think of F~ − mA~ as the overall effective force on the mass in the accelerating frame. y0 y 1 2 Example 8.1.1: Accelerating Pendulum 2 At x, x0 Consider a pendulum whose pivot point is accelerating to the right with ac- φ celeration A as shown in Fig. (8.1). In the stationary frame S , the net force on the bob is the force of tension ` 0 plus the weight force. Newton’s law gives us ¨ m~r0 = T~ + m~g. m In the accelerating frame S, Newton’s law is modified as Figure 8.1: A pendulum accel- ¨ ~ ~ ~ erating to the right. m~r = T + m~g − mA = T + m~geff, ~ where the effective gravitational force on the pendulum is ~geff = ~g − A. Since ~g z0 points down and −A~ points left, we can construct a triangle such that g = z eff p 2 2 g + A is the hypotenuse and we can see that ~geff makes an angle φ0 = − tan−1(A/g) with the vertical. nˆ y P At equilibrium in the accelerating frame, the pendulum won’t hang straight ~r down. The forward acceleration makes the pendulum pull to the rear, making y0 an angle φ0 with the vertical. That is, it’s equilibrium position is determined by the effective gravitational force acting on it. The angular frequency of small oscillations about this equilibrium is x0 x s rg pg2 + A2 ω = eff = . Figure 8.2: Rotating frames. ` `

8.2 Rotating Frames

Consider a coordinate system (or reference frame) called So with fixed axes xo, yo, and zo. Next, imagine a rotating reference frame S with axes x, y, and z with the same origin as the fixed reference frame as shown in Fig. (8.2). The noninertial frame S is rotating about an axis in the nˆ direction with angular velocity Ω~ = Ω nˆ. P is a point at ~r that is moving with the noninertial rotating frame. That is, P is at rest in the rotating frame. A simple example is you on the surface of Earth. Think of the non-rotating frame S0 as being the space outside of Earth, and think of the Earth as being the rotating frame S. Earth is a noninertial rotating reference frame since it is spinning about its axis. Now think of your position on the surface of Earth as being the point P . How can an observer in the stationary frame of outer space describe your motion in the rotating frame? We can give the components of ~r in either frame S or So. It is just a vector. It is the same ~r, just with different components. To determine ~r, we have to specify which frame we’re in. The vector ~r is fixed in S, so ~r˙ = 0 in frame S. However, as seen from frame ˙ So, ~r is spinning about nˆ, so ~r 6= 0 in frame So. When looking at the time dependence of ~r, we get different answers depending on which frame we’re referencing. In the noninertial frame S, the velocity is

d~r = 0. dt S 8.2. Rotating Frames 117

In the ﬁxed frame So, the velocity is

d~r = Ω~ × ~r. dt So If the point P is moving with respect to the noninertial frame S, we have to add its velocity as d~r d~r = Ω~ × ~r + . dt dt So S On the left-hand side, we have the velocity of P as seen by an observer in the fixed frame. The first term on the right gives us the velocity of P due to the rotating frame, and the second term on the right gives us the additional velocity of P as measured by an observer in the rotating frame. In general, the velocity of any vector Q~ as seen in the fixed frame is related to the velocity of the vector as seen by someone at rest in the rotating frame by ! ! dQ~ dQ~ = Ω~ × Q~ + . dt dt So S What about acceleration in the inertial frame? To calculate this, we take the time derivative (with respect to frame S0) being careful to apply the product rule. ! d d~r ˙ d~r d d~r = Ω~ × ~r + Ω~ × + dt dt dt dt dt So So S So So ˙ d~r = Ω~ × ~r + Ω~ × Ω~ × ~r + dt S d~r d d~r + Ω~ × + . dt S dt dt S S Notice that on the left-hand side, everything is measured in the inertial frame, and on the right-hand side, everything is measured in the rotating frame. On the left, we have ~r¨ as measured in the inertial/fixed frame. We know that Newton’s second law is true at least in the inertial frame So, so

d2~r m = F~, dt2 So where F~ is the real force that is acting on the particle of mass m. We want to write an eﬀective Newton’s law for the rotating frame. To simplify things, we’ll denote ~r˙ and ~r¨ as the velocity and acceleration as measured from the noninertial/rotating frame. That is,

2 ˙ d~r ¨ d ~r ~r = , ~r = 2 . dt S dt S Then the acceleration as measured by an observer in the ﬁxed frame is ! d d~r ˙ = Ω~ × ~r + Ω~ × Ω~ × ~r + 2Ω~ × ~r˙ + ~r¨. dt dt So So Plugging this result into Newton’s law gives us

˙ m~r¨ = F~ − mΩ~ × ~r − mΩ~ × Ω~ × ~r − 2mΩ~ × ~r˙. 118 Noninertial Frames

~ This is the eﬀective force acting on the particle as seen from the noninertial/rotating Ω ˙ frame. So far, we have not assumed that Ω~ is constant. If we do so, then Ω~ = 0, and we get ρ m m~r¨ = F~ − mΩ~ × Ω~ × ~r − 2mΩ~ × ~r˙.

We can write this result as ~r m~r¨ = F~ + F~ + F~ , θ cf cor

O where F~ is the regular force on m in an inertial frame, and

Figure 8.3: Side view of particle ~ ~ ~ rotating about an axis. F cf = m Ω × ~r × Ω,

is the centrifugal force on the mass, and ˙ F~ cor = 2m~r × Ω~ , Ω~ × ~r is the Coriolis force acting on the mass. ρ Ω~ m 8.3 Centrifugal Force

When using Newton’s law in a rotating frame, we have to account for the Centrifugal force Figure 8.4: Top view of particle F~ = m Ω~ × ~r × Ω~ , rotating about an axis. cf

Example 8.3.1: Centrifugal Force on Orbiting Particle y0 Ω Consider a particle of mass m rotating about an axis with angular velocity Ω~ m as shown in Fig. (8.3). A top view of the same thing is shown in Fig. (8.4). Here, Ω~ is coming out of the page. x0 By the right-hand rule, we can tell the direction of Ω~ ×~r as shown in Fig. (8.4). The centrifugal force acting on the particle is F~ cf = m Ω~ × ~r × Ω~ .

Figure 8.5: Bead on a spinning Applying the right-hand rule again, we note that this force points away from the wire. center of rotation. In the moment captured in the image above, F~ cf points to the right. In this case, ρ = r sin θ, and

|Ω~ × ~r| = Ωr sin θ = Ωρ,

so the centrifugal force on the particle has magnitude

2 2 Fcf = |F~ cf | = mΩr sin θ Ω = mΩ r sin θ = mΩ ρ.

Consider a bucket half-filled with water. Ordinarily, the surface of the water is flat. However, if the bucket is rotating about its vertical axis, the surface of the water becomes a paraboloid due to the centrifugal force acting on the water. When you have a fluid that is in equilibrium, the surface of the fluid is an equipotential (i.e. the potential energy is constant along the surface). In a rotating frame, the effective potential will contain ~g and the centrifugal force, so the effective gravitational potential acting on the water now points outward as well as downward. The buoyant force acting on an object floating on the surface is always perpendicular to the water’s surface. 8.4. Coriolis Force 119

Example 8.3.2: Bead on a Spinning Wire y Consider a straight wire of arbitrary length that is rotating with angular velocity Ω~ = Ω zˆ. On the wire is a bead of mass m that is free to slide. What is the equation of motion for the bead in the rotating frame of reference. m In the ﬁxed reference frame, we have the picture given in Fig. (8.5). In the x reference frame that is rotating with the wire, the wire is not moving. If we let the wire lie along the x-axis as shown in Fig. (8.6), then the position of the bead is given by ~r = x xˆ. Newton’s law for a rotating reference frame is

 Figure 8.6: Bead on the wire in m~r¨ = F~ + F~ + F~ = F~ + m Ω~ × ~r × Ω~ + 2m~r˙ × Ω~ . cf cor an inertial frame.

There is no other force (e.g. gravity) acting on the bead in the rotating frame, ~ so F = 0. In this example, the Coriolis force is perpendicular to the wire, so it P doesn’t contribute. Our equation of motion, therefore, reduces to the centrifugal force on the bead m~r¨ = m Ω~ × ~r × Ω~ . ~v F~ cor ~v By the right-hand rule, we know that Ω~ × ~r points in the yˆ direction. Applying F~ cor the rule again tells us that (Ω~ × ~r) × Ω~ and thereby the centrifugal force, points in the xˆ direction. Expanding the two cross products either by exploiting the orthogonality of the vectors and applying the right hand rule or by using the determinant deﬁnition, gives us 2 Figure 8.7: Coriolis force on a F~ cf = mxΩ xˆ. merry-go-round. So our equation of motion is

m~a = mx¨ = mΩ2x xˆ =⇒ x¨ = Ω2x.

This ODE has the solution

x(t) = AeΩt + Be−Ωt.

8.4 Coriolis Force

When using Newton’s law in a rotating frame, we have to account for the Coriolis force

˙ F~ cor = 2m~r × Ω~ = 2m~v × Ω~ .

Here, the velocity ~r˙ = ~v is the velocity of the object with respect to the rotating frame. Consider a large merry-go-round rotating with angular velocity Ω. Suppose you’re at the center of this merry-go-round and you throw a ball toward the point P with velocity ~v. Where does the ball ﬂy from your perspective? Initially, the ball’s velocity is in the direction toward P . We know that the angular velocity vector of the merry-go-round is out of the page, so using the right hand rule on the Coriolis force F~ cor = 2m~v × Ω~ , we know that the Coriolis force points to the right, perpendicular to the ball’s velocity. This force causes the ball to curve toward the right. At some later instant, the ball velocity now points to the right of P , and the Coriolis force is again perpendicular to it, causing the ball to curve even further to the right. From your perspective, at the center of the merry-go-round, the ball curves to the right and doesn’t come close to passing over P . This is illustrated in as shown in Fig. (8.7). 120 Noninertial Frames

Ω From the perspective of a person on the ground, the ball does indeed travel in a straight line, but the point P rotates away from where it was before the ball reaches yˆ zˆ it. Incidentally, since the centrifugal force on the ball is always outward, it does not contribute to the ball’s deﬂection.

θ Projectile Motion The equations we normally use for projectile motion do not include the Coriolis force. For high velocity projectiles, the Coriolis force can cause a signiﬁcant deﬂection. Suppose we are on Earth at some co-latitude θ. Co-latitude is simply the angle between the vector pointing to your position on the surface and the vector pointing to the north pole. Alternatively, you can think of it as the complement of your regular latitude. Figure 8.8: Earth with its an- We choose our coordinate system at our location on the surface such that yˆ points gular velocity vector Ω and a north and zˆ points upward (i.e. outward) as shown in Fig. (8.8). That way, the gravita- Cartesian coordinate system at tional force ~g is parallel to our z-axis. The xˆ direction now points to the east from our a point on its surface. perspective on the Earth’s surface. Including the Coriolis force, the equation of motion for a projectile of mass m is

Tip mr¨ = m~g + 2m~r˙ × Ω~ ,

When dealing with the where Earth’s angular velocity vector is Coriolis force on Earth’s ~ surface, remember that the Ω = h0, Ω sin θ, Ω cos θi. angular velocity Ω~ points Expanding the cross product in the Coriolis force, we get from the the center of

Earth toward the north xˆ yˆ zˆ pole. If you’re in the north- ~r˙ × Ω~ = x˙ y˙ z˙ = (y ˙Ω cos θ − z˙Ω sin θ) xˆ − x˙Ω cos θ yˆ +x ˙Ω sin θ zˆ. ern hemisphere, you have ~ a component of Ω com- 0 Ω sin θ Ω cos θ ing up out of the ground, and if you’re in the south- So our equation of motion can be broken into Cartesian coordinates as ern hemisphere, you have a component of Ω~ pointing x¨ = 2Ω (y ˙ cos θ − z˙ sin θ) into the ground. y¨ = −2Ωx ˙ cos θ z¨ = −g + 2Ωx ˙ sin θ.

These are the equations of motion at a given co-latitude for a projectile, and they include the Coriolis force. Example 8.4.1

Obtain a ﬁrst-order approximation for the motion of an object of mass m released at rest from a height h above Earth, taking the Coriolis force into account. The angular velocity of Earth Ω~ is very small, and as a zero order approximation, we can set it equal to zero, which gives us

x¨ = 0, y¨ = 0, z¨ = −g.

These are the familiar results that we use when we ignore Earth’s rotation entirely. For an item dropped from rest at a height h, for example, we have that

x˙ = 0, y˙ = 0, z˙ = −gt.

Integrating these and taking into account the initial conditions, gives us the familiar x(t) = 0, y(t) = 0, and z(t) = h − gt2/2. However, we want a better 8.4. Coriolis Force 121

z approximation. To get the first-order approximation, we plug ourx ˙,y ˙ andz ˙ back into the equations of motion that account for the Coriolis force. When we do that, we get x¨ = 2Ωgt sin θ, y¨ = 0, z¨ = −g. If we wanted an even better approximation, we could plug these results back into the equations. We could keep repeating this process of successive approximations until we get a good enough approximation. However, since we’re only looking for y the first correction, we integrate these, and accounting for the initial conditions, we get T~ 1 1 x Tx x(t) = Ωgt3 sin θ, y(t) = 0, z(t) = h − gt2. 3 2 m Ty Notice that after accounting for the Coriolis force, our x(t) is no longer zero. That is, the Coriolis force deflects our dropped object toward the East. We know that Figure 8.9: Foucault pendulum. the time it takes for the object to hit the ground is t = p2h/g. Plugging this into x(t) gives us the total eastward deflection

3 1 2h 2 x = Ωg sin θ. 3 g

It might seem surprising that the object is deflected eastward. Since the Earth is rotating toward the East, a naive conclusion might be that the deflection is westward since the Earth ‘rotates out from underneath the object’. The eastward deflection, however, is caused by the conservation of angular momentum. When the object is being held at rest at some height h, it has some constant angular momentum. When it is dropped, conservation of angular momentum requires that its angular velocity increases as its distance from Earth’s center decreases. In other words, the dropped object out-rotates Earth for the brief period before it hits the ground.

Example 8.4.2: Foucault Pendulum

For the Foucault pendulum, we have a pendulum of length L and mass m hanging from a great height as shown in Fig. (8.9). Because L is so large, we can ignore the displacement in z and treat the pendulum bob as moving only in the xy-plane. We are always in the small angle approximation here. What is the eﬀect of the Coriolis force on such a pendulum? We can write the tension on the bob in component form as

Tz ≈ T = mg x x T = − T = − mg x L L y y T = − T = − mg. y L L Our equation of motion for the bob of the pendulum at the surface of Earth is

m~r¨ = m~g + 2m~r˙ × Ω~ + T~,

where the angular velocity of Earth is

Ω~ = h0, Ω sin θ, Ω cos θi. 122 Noninertial Frames

Tip Expanding the cross product using a matrix determinant and givenz ˙ = 0, Ωz = p Ω cos θ, and ω0 = g/L, we get the equations of motion The Coriolis force does no 2 work since it is always per- x¨ − 2Ωzy˙ + ω0x = 0 pendicular to the direction y¨ + 2Ω x˙ + ω2y = 0. of motion. It is therefore, z 0 not a damping force. It So we have reduced the problem to a pair of coupled ODEs. To solve them, we cannot remove energy from multiply the second ODE by i and subtract the result from the ﬁrst ODE. Then, a system. if we let η = x + iy then iη = ix − y, and we get the linear second order ODE

2 η¨ + 2iΩzη˙ + ω0η = 0.

If we seek solutions of the form η = eλt, we can plug that and its derivatives into the ODE to get 2 2 λ + 2iΩzλ + ω0 = 0. Applying the quadratic formula gives us q 2 2 λ = −i Ωz ± Ωz + ω0 .

We can assume Ωz << ω0, then

λ ≈ −i (Ωz ± ω0) .

2 We can ignore the Ωz under the radical since Ωz is small and Ωz is even smaller. We can show this by performing a Taylor expansion. s q 2 2 2 2 Ωz 1 Ωz Ωz + ω0 = ω0 1 + 2 = ω0 1 + 2 + ··· . ω0 2 ω0

The solution for our pendulum is then

−i(Ωz −ω0)t −i(Ωz +ω0)t η = C1e + C2e .

If we start the pendulum swinging along the x axis, that is if we start it from rest at t = 0, we have x = A and y = 0, then we get C1 + C2 = A. Diﬀerentiating it and using the initial condition v(0) = 0, gives us

A AΩz C1 = − . 2 2ω0

Since Ωz << ω0, we can neglect the second term and assume C1 = C2 = A/2, then our solution becomes

−iΩz t η(t) = x(t) + i y(t) = Ae cos(ω0t).

Since Ωz << ω0, the cosine part of this is the rapid oscillation of the pendulum, and the exponential part is the slow oscillation due to the Coriolis force. Initially, the pendulum is swinging along the x-axis. At some later time, the angle the swinging pendulum makes with the x-axis is Ωzt. 8.5. Summary: Noninertial Frames 123

8.5 Summary: Noninertial Frames

Skills to Master • Write down the modiﬁed form of Newton’s second law in a linearly accelerating or a rotating frame • Apply Newton’s law in a linearly accelerating frame by using an eﬀective force • Calculate the centrifugal and Coriolis forces in a rotating frame • Solve projectile motion problems with Coriolis force included

Linearly Accelerating Frames where F~ is the regular force on m, and In a frame that is accelerating with constant acceler- F~ = m Ω~ × ~r × Ω~ ation A~, we can still use Newton’s law for a mass m cf provided that we add the fictitious force is the centrifugal force on the mass, and F~ = −mA~. f ˙ F~ cor = 2m~r × Ω~ For example, if we have a mass in an accelerating frame that is also under the influence of gravity, we have is the Coriolis force acting on the mass. In cylindrical coordinates, the centrifugal force on ¨ ~ m~r = m~g − mA = m~geff , a mass m a distance ρ from the z-axis and rotating ~ about zˆ with angular speed Ω is where m~geff = m(~g − A) is the effective gravitational force acting on the mass. This effective gravitational 2 F~ cf = mΩ ρ ρˆ. force allows us to calculate, for example, the equilibrium angle for a simple pendulum. You can calculate the potential acting on a continuous mass as

Rotating Frames ~r U = − F~ eff · d~`, To obtain the equation of motion for a moving body in Ô a reference frame that is rotating with constant angular ~ velocity Ω: where F~ eff is the effective force which includes the 1. Identify the rotating frame. This is simply the fictitious forces. frame in which whatever is rotating with angular ~ velocity Ω is stationary. Choose convenient axes Projectile Motion such that the motion of the object of interest in this frame is easy to calculate. For projectile motion near Earth’s surface taking into 2. Apply Newton’s modified law for rotating frames account the Coriolis force, our coordinate system is at Earth’s surface at some colatitude θ. The z-axis points ¨ ~ ~ ~ m~r = F + F cf + F cor. outward from Earth so it is parallel to ~g and the y-axis points toward the north pole. Then the equation of Note that all quantities used here except Ω~ are motion for a projectile of mass m is calculated with respect to the rotating frame. That is, you have become an observer that is sta- mr¨ = m~g + 2m~r˙ × Ω~ , tionary in the rotating frame. Newton’s modified law for rotating reference where Earth’s angular velocity vector is frames is ¨ m~r = F~ + F~ cf + F~ cor, Ω~ = h0, Ω sin θ, Ω cos θi. 124 Noninertial Frames

After expanding the cross product, we can break the into the equations of motion and solving them for the equation of motion into components as given initial conditions. When dealing with the Coriolis force on Earth’s x¨ = 2Ω (y ˙ cos θ − z˙ sin θ) surface, remember that the angular velocity Ω~ points y¨ = −2Ωx ˙ cos θ from the the center of Earth toward the north pole. If z¨ = −g + 2Ωx ˙ sin θ. you’re in the northern hemisphere, you have a component of Ω~ coming up out of the ground, and if you’re Note the ﬁrst order approximation of the solution in the southern hemisphere, you have a component of ~ to these equations is obtained by plugging Ω pointing into the ground. A useful relation to remember is that the tangen- x˙ = 0 tial velocity of a rotating object is the cross product of y˙ = 0 its angular velocity and its position vector z˙ = −gt, ~v = ~ω × ~r. Chapter 9

Rigid Rotations

A rigid body is a collection of particles with their relative distances ﬁxed. For a collection of discrete masses m, we can represent their positions by ~rα. Recall that the center of mass of the collection is 1 X R~ = m ~r , cm M α α α where M is the total mass. If it is a continuous mass, then we use an integral instead of a sum 1 R~ = ~r dm. cm M ˆ The total momentum is the momentum of the center of mass

X ˙ ˙ P~ = mα~rα = MR~ cm. α The equation of motion of the rigid body is just the equation of motion of the center of mass ¨ MR~ cm = F~ ext. Suppose you have some point R~ within the rigid body. This point could be, but doesn’t have to be, the center of mass of the rigid body. Then given some origin O, the position of the nth particle can be written as

0 ~rn = R~ + ~rn ,

0 where ~rn is the vector from R~ to that particle. The kinetic energy of the rigid body can then be written as

2 2 1 X 2 1 X ˙ 1 X ˙ ˙ T = m ~r˙ = m R~ + ~r˙ 0 = m R~ + 2R~ · ~r˙ 0 + ~r˙ 02 2 n n 2 n n 2 n n n n n n 1 ˙ 2 ˙ X 1 X = MR~ + R~ · m ~r˙ 0 + m ~r˙ 02. 2 n n 2 n n n n

There are two interesting cases, that we want to look at. If the point of interest R~ ˙ is at rest (even only instantaneously), then R~ = 0, and we get

1 X T = m ~r˙ 02. 2 n n n

This is the rotational kinetic energy of the rigid body about R~ . Remember, this holds true even if R~ is not the center of mass of the rigid body. Consider for example, a wheel that is 126 Rigid Rotations

rolling without slipping. The point on the wheel touching the ground is instantaneously ˙ at rest. If we let R~ be that point, then R~ = 0, and the total kinetic energy of the wheel is the rotational kinetic energy of the wheel about that point. The second interesting case occurs when R~ = R~ cm is the center of mass of the rigid ~ P 0 body. If R is the center of mass then n mn~rn is the center of mass relative to the P ˙0 center of mass. That is, it is zero. Its time derivative is also zero, so n mn~rn = 0. So when R~ = R~ cm, the kinetic energy reduces to

1 ˙ 2 1 X T = MR~ + m ~r˙ 02. 2 cm 2 n n n

We can write this as T = Tcm + Trot. That is, the total kinetic energy of a rigid body is the kinetic energy of the center of mass plus the rotational kinetic energy of the body about the center of mass. ˙ In general, our problems will either involve a stationary point (i.e. R~ = 0) or the

center of mass (i.e. R~ = R~ cm). Angular momentum is L~ = ~r × ~p. The total angular momentum of our rigid body about the origin O for the case R~ = R~ cm is

X ˙ ˙ X 0 ˙ L~ = ~rn × mn~rn = R~ cm × MR~ cm + ~rn × mn~rn . n n

0 Here we used ~rn = R~ + ~rn . The ﬁrst term on the right is the angular momentum associated with the center of mass about O, and the second term is the angular momentum of the rigid body about the center of mass. We can write this as

L~ = L~ cm + L~ rot.

We can also think of this in terms of the orbital and spin angular momenta of the rigid body L~ = L~ orbital + L~ spin.

9.1 Rotation About a Fixed Axis

Suppose we have an arbitrary rigid body rotating about a ﬁxed axis with angular velocity ~ω = h0, 0, ωi. Here the z-axis is the axis of rotation. We again denote the position of the nth particle in the rigid body by ~rn. Then the time rate of change of the position of the nth particle is ˙ ~rn = ~ω × ~rn.

Expanding the cross product on the right using the determinant method gives us ˙ ~rn = ~ω × ~rn = −ωyn xˆ + ωxn yˆ.

Then the kinetic energy of the rotating body is

1 X 1 X 1 T = m r˙2 = m x2 + y2 ω2 = Iω2, 2 n n 2 n n n 2 n n where X 2 2 I = mn xn + yn , n 9.1. Rotation About a Fixed Axis 127 is the moment of inertia that we recognize from introductory physics. The angular momentum of the body is

X ˙ L~ = ~rn × mn~rn. n ˙ Plugging in ~rn = ~ω × ~rn then expanding the cross product gives us

˙ 2 2 ~rn × mn~rn = −xnznω xˆ − ynznω yˆ + (xn + yn)ω zˆ. So the components of the angular momentum are

X 2 2 Lz = mn(xn + yn)ω = Izzω n X Lx = − mnxnznω = Ixzω n X Ly = − mnynznω = Iyzω. n

For the moment of inertia Izz, the ﬁrst z indicates that this is for the z-component of the angular momentum, and the second z indicates the direction of ~ω. That is, it tells us the rotation is about the z-axis. The quantities Ixz and Iyz are called products of inertia. We have nonzero oﬀ-axis components of L~ because the angular momentum does not necessarily point in the direction of ~ω. Example 9.1.1

Consider a pair of particles of mass m orbiting about the z-axis with angular velocity ~ω. They are opposite each other with respect to the z axis, and at one moment in time, one mass is at (0, y0, z0), and the other mass is at (0, −y0, z0). Find the moment and products of inertia for this system. The products of inertia are X Ixz = − mnxnzn = 0 + 0 = 0 n X Iyz = − mnynzn = −my0z0 + my0z0 = 0. n

The products of inertia are both zero. The product Ixz is clearly zero since x = 0 for both particles at this moment in time. The product Iyz is zero because the masses are symmetric about the y-axis. So the products of inertia are zero as a consequence of symmetry. The moment of inertia is

X 2 2 2 Ixx = mn(xn + yn) = 2my0. n

Since the products of inertia are zero, we have that Lx = Ly = 0. Since only Lz survives, we know that the angular momentum of the system is parallel to ~ω which we know points in the z direction.

Example 9.1.2

Consider now a modiﬁed version of the above problem. One mass is still at (0, y0, z0), but the second mass is now at (0, −y0, −z0) at a given moment in time. 128 Rigid Rotations

In this case, the two masses are symmetric about the origin, but not about y = 0 as in the previous example. Assume that ~ω = ω zˆ. That is, the z-axis is still the axis of rotation. The products of inertia are now X Ixz = − mnxnzn = 0 + 0 = 0 n X Iyz = − mnynzn = −2my0z0. n The moment of inertia is still

X 2 2 2 Ixx = mn(xn + yn) = 2my0. n The components of the angular momentum of the system are

2 Lx = 0,Ly = −2my0z0ω, Lz = 2my0ω.

Notice that at this moment in time at least, L~ is not parallel to ~ω since Ly 6= 0. In fact, L~ is not constant in this example. If we let the system rotate a little bit and recalculate the components of L~ , we will get diﬀerent results. It turns out that L~ rotates about the z-axis. Since L~ is changing in time, there must be a torque being applied to the system to keep it rotating about the z-axis.

9.2 The Inertia Tensor

Suppose we have a rigid body rotating about an arbitrary axis instead of the z-axis. Given arbitrary angular velocity ~ω, we want to calculate the angular momentum L~ of the rigid body about our origin O. X ˙ X L~ = ~rn × mn~rn = mn~rn × (~ω × ~rn) . n n ˙ We again made the replacement ~rn = ~ω × ~rn. Next, we apply the vector identity A~ × (B~ × C~ ) = B~ (A~ · C~ ) − C~ (A~ · B~ ) to get X L~ = mn (~ω(~rn · ~rn) − ~rn(~rn · ~ω)) . n Writing this in terms of components, we have   X 2 X Li = mn rnωi − rni rnjωj . n j P where i and j can be x, y, or z. Notice that rnjωj is just the dot product ~rn · ~ω. We P j can write ωi = j ωjδij, so

X X 2 X Li = mn rnδij − rnirnj ωj = Iijωj, j n j

where X 2 Iij = mn rnδij − rnirnj . n 9.2. The Inertia Tensor 129

The Iij are the elements of a 3 × 3 matrix made up of products and moments of inertia called the inertia tensor.   Ixx Ixy Ixz   I =  Iyx Iyy Iyz .   Izx Izy Izz

Note that we are denoting matrices with underlines, e.g. I. The moments of inertia are

X 2 2 Ixx = mn yn + zn n X 2 2 Iyy = mn xn + zn n X 2 2 Izz = mn xn + yn n

This is for a system of particles. For a continuous mass distribution, convert the sums to integrals by making the substitutions

X m → dm, and x , y , z → x, y, z. n ˆ n n n n For example, I = (x2 + y2) dm. xx ˆ The products of inertia are

X Ixy = Iyx = − mnxnyn n X Ixz = Izx = − mnxnzn n X Iyz = Izy = − mnynzn n

Notice that Ixy = Iyx. The same pattern holds for the other products of inertia, so I is a symmetric matrix. That is, IT = I. In introductory physics, we learned that L~ = I~ω. However, this is only valid for a ﬁxed axis of rotation, usually taken to be the z-axis. The generalization of that is

L~ = I~ω.

Since this is matrix multiplication, L~ and ~ω are 3 × 1 column vectors. Given the angular momentum, L~ , what is the rotational kinetic energy?

X 1 X 1 X 1 T = m ~r˙ 2 = m ~r˙ · ~r˙ = m ~r˙ · (~ω × ~r ) . rot 2 n n 2 n n n 2 n n n n n n Applying the triple product identity

A~ · (B~ × C~ ) = B~ · (C~ × A~) = C~ · (A~ × B~ ), allows us to rewrite it as 1 X T = ~ω · ~r × m ~r˙ , rot 2 n n n n 130 Rigid Rotations

z or 1 T = ~ω · L~ . rot 2

2 This is the generalization of the kinetic energy Trot = Iω /2 that we learned in introductory physics. The angular velocity ~ω and angular momentum L~ are in general, not y pointing in the same direction. Since T is expressed as a dot product, we can also write it in summation notation as x 1 X 1 X X 1 X T = ω L = ω I ω = ω I ω , 2 i i 2 i ij j 2 i ij j Figure 9.1: A cube with one cor- i i j i,j ner at the origin. That is, we can write the kinetic energy as the matrix multiplication

1 T = ~ωT I~ω. 2

which in expanded form is    Ixx Ixy Ixz ωx 1h i   T = ωx ωy ωz  Iyx Iyy Iyz  ωy . 2    Izx Izy Izz ωz

To calculate the components of the inertia tensor, we have to define a coordinate system. The inertia tensor I is a property of the object and a point about which the object is rotating. That is, the inertia tensor is with respect to a specific point in the body. Choose a different point, and I will be different. Once it is calculated for one moment of time, it will be the same for that point in the rigid body regardless of how the motion proceeds. Example 9.2.1

Find the inertia tensor for a solid cube of side length a and mass M with one corner at the origin as shown in Fig. (9.1). Since this is a continuous mass, we have to integrate to calculate the moments and products of inertia.

a a a 2 2 2 2 2 5 2 2 Izz = (y + x )dm = (x + y )ρ dx dy dz = ρa = Ma . ˆ ˆ0 ˆ0 ˆ0 3 3 In the ﬁrst step, we used dm = ρ dV = ρ dx dy dz, and in the last step we used 3 ρ = M/a . By symmetry, we know that Iyy and Ixx will give us the same result. For the products of inertia, we get, for example

a a a 1 5 1 2 Ixy = − xy dm = − xyρ dx dy dz = − ρa = − Ma . ˆ ˆ0 ˆ0 ˆ0 4 4 Again by symmetry, all the products of inertia are the same. So the inertia tensor for the cube about the origin O is

 2 1 1  3 − 4 − 4 2  I = Ma  − 1 2 − 1 .  4 3 4  1 1 2 − 4 − 4 3 9.2. The Inertia Tensor 131

Example 9.2.2 z

What is the angular momentum of the previous cube if it is spinning about the z-axis with angular speed ω? What is the kinetic energy? Recall that L~ = I~ω. y We know that since it is spinning about the z-axis,   0   x ~ω = ω 0 .   1 Figure 9.2: Cube with center at the origin. So the angular momentum of the cube is

 2 1 1    1  3 − 4 − 4 0 − 4 2    2   L~ = Ma ω − 1 2 − 1  0  = Ma ω − 1 .  4 3 4    4  1 1 2 2 − 4 − 4 3 1 3

In other words, 1 1 2 L~ = Ma2ω − xˆ − yˆ + zˆ . 4 4 3 Recall that kinetic energy is 1 T = L~ · ~ω. 2 In our case, 1 T = Ma2ω2. 3 This is the same as the answer we would have gotten in introductory physics where 2 Trot = Izzω /2.

Example 9.2.3

Calculate the inertia tensor for the same cube but with the origin at the center of the cube as shown in Fig. (9.2). This time, the moment of inertia is

a a a 2 2 2 2 2 2 2 1 5 1 2 Izz = (y + x )dm = (x + y )ρ dx dy dz = ρa = Ma . ˆ ˆ a ˆ a ˆ a 6 6 − 2 − 2 − 2

Again, by symmetry, Iyy and Ixx are the same. For the products of inertia we get, for example,

a a a 2 2 2 Ixy = − xy dm = − xyρ dx dy dz = 0. ˆ ˆ a ˆ a ˆ a − 2 − 2 − 2 The result is zero since this is a separable integral and the integrand is odd in both x and y so both of those integrals will yield zero. By symmetry, the other products of inertia are also zero. 132 Rigid Rotations

Our inertia tensor for the cube about its center of mass is   1 0 0 CM 1 2  I = Ma  0 1 0 . 6   0 0 1

No matter what axis you spin this cube about, the angular momentum vector is always parallel to that axis.

P Suppose you want to compute the moment of inertia Ixx for a rigid body of mass M about a point P . Suppose further that the center of mass of the body is at (X,Y,Z) with respect to the coordinate system at which P is at the origin. Then

P CM 2 2 Ixx = Ixx + M(Y + Z ),

CM where Ixx is the moment of inertia of the body about its center of mass. For the products of inertia, we have

P CM Ixy = Ixy − MXY.

This is the generalized parallel axis theorem, and it allows you to compute I about any point P given I about the center of mass. This is essentially a coordinate change.

Example 9.2.4

Compute the inertia tensor for a ﬂat circular disk of radius R about its center of mass. We start by setting up our coordinate system. We place the center of the disk at the origin of our coordinate system and orient it so it is laying in the xy-plane. That way z = 0 for the disk. The moment of inertia for rotation about the z-axis is

2π R 2 2 3 1 4 1 2 Izz = (x + y )dm = σ r dr dθ = σπR = MR . ˆ ˆ0 ˆ0 2 2 For rotation about the x-axis, the moment of inertia is

2π R 2 2 2 3 2 1 4 1 2 Ixx = (y + z )dm = y dm = σ r sin θ dr dθ = σπ R = MR . ˆ ˆ ˆ0 ˆ0 4 4

By symmetry, for rotation about the y-axis, the moment of inertia Iyy is the same. For the products of inertia, we know that Ixz = Izx = Iyz = Izy = 0 since z = 0. For Ixy and Iyx, we also get zero because of symmetry. So our inertia tensor is   1 0 0 1 2  I = MR  0 1 0 . 4   0 0 2 9.3. Principal Axes 133

9.3 Principal Axes

Diagonal inertia tensors are nice because we can read oﬀ the eigenvalues (and moments of inertia) directly.   λ1 0 0   I =  0 λ2 0 .   0 0 λ3

We can also immediately write the angular momentum vector as

L~ = hλ1ωx, λ2ωy, λ3ωzi, since L~ = I~ω. So if the inertia tensor is diagonal, we can obtain the angular momentum vector simply by scaling each component of ~ω. The rotational kinetic energy is also easily calculated as 1 1 1 1 T = ~ωT I~ω = λ ω2 + λ ω2 + λ ω2. rot 2 2 1 x 2 2 y 2 3 z For any rigid body, we can always ﬁnd three orthogonal axes for which I is a diagonal matrix. These are known as the principal axes. The principal axes are rotation axes for which the angular momentum is parallel to the angular velocity, L~ k ~ω. That is, L~ = λ~ω. So for principal axes, we have the eigenvalue problem

L~ = I~ω = λ~ω.

Note that ~ω tells you the direction of the principal axes, and λ gives you the moment of inertia about the principal axes. To ﬁnd the principal axes, we ﬁnd the nontrivial solutions of the eigenvalue problem

(I − λ1) ~ω = 0.

Recall from linear algebra that the eigenvalues λ are found by solving the characteristic equation det(I − λ1) = 0.

After ﬁnding the eigenvalues, we can substitute them back into the eigenvalue equation one at a time to ﬁnd the eigenvectors or principal axes. Example 9.3.1

Find the principal axes for a cube of mass M rotating about one of its corners. Earlier, we found that the inertia tensor for a cube about one of its corners is

 2 1 1    3 − 4 − 4 8 −3 −3 2    I = Ma  − 1 2 − 1  = β −3 8 −3 ,  4 3 4    1 1 2 − 4 − 4 3 −3 −3 8

where β = Ma2/12. Here, the axes of rotation are along the edges of the cube. We know these are not principal axes because the inertia tensor is not diagonal. To ﬁnd the principal axes, we have to solve the eigenvalue problem

L~ = I~ω = λ~ω. 134 Rigid Rotations

We start by ﬁnding the eigenvalues by computing the characteristic equation

8β − λ −3β −3β 2 0 = −3β 8β − λ −3β = (2β − λ)(11β − λ) .

−3β −3β 8β − λ

So the eigenvalues are λ = 2β, 11β, 11β. For λ1 = 2β, we ﬁnd that the eigenvector is ~ω = ωxh1, 1, 1i. Normalizing it, we get the eigenvector and principal axis 1 eˆ1 = √ h1, 1, 1i. 3 Since our origin is at one corner of the cube, this tells us that a principal axis is through the opposite corner of the cube. The other two eigenvalues are degenerate, so any pair of orthogonal vectors that are also orthogonal to eˆ1 are principal axes.

Example 9.3.2

Find the principal axes for a ﬂat a by a square body with one corner at the origin and two sides along the positive x and y axes. Right from the start, since z = 0 for the rigid body, we know that Ixz = Izx = Iyz = Izy = 0. For rotation about the z-axis, we have a a 2 2 2 2 2 4 2 2 Izz = (x + y )dm = σ(x + y ) dx, dy = σa = Ma . ˆ ˆ0 ˆ0 3 3 For rotation about the x-axis, we have a a 2 2 1 4 1 2 Ixx = x dm = σ x dx, dy = σa = Ma . ˆ ˆ0 ˆ0 3 3 2 By symmetry, rotation about the y-axis also yields Iyy = Ma /3. The nonzero products of inertia are a a 1 4 1 2 Ixy = Iyx = − xy dm = −σ xy dx dy = − σa = − Ma . ˆ ˆ0 ˆ0 4 4 So our inertia tensor is   4 −3 0   I = β −3 4 0 ,   0 0 8

where β = Ma2/12. To ﬁnd the principal axes, we solve the eigenvalue problem (I − λ1)~ω = 0. To ﬁnd the eigenvalues, we compute the characteristic equation

4β − λ −3β 0 2 2 det(I − λ1) = −3β 4β − λ 0 = (8β − λ)[(4β − λ) − 9β ]

0 0 8β − λ = (8β − λ)(7β − λ)(β − λ), 9.3. Principal Axes 135

which implies that our eigenvalues are λ = 8β, 7β, β. For λ3 = 8β, we get the eigenvector

eˆ3 = h0, 0, 1i.

As with any ﬂat object in the xy-plane, the z-axis is always a principal axis, and the eigenvalue can be read directly from the inertia tensor. Since eigenvectors are orthogonal, we know that both remaining ones line in the xy-plane. We ﬁnd that for λ1 = 7β, we get 1 eˆ1 = √ h1, −1, 0i. 2

For λ2 = β, we get 1 eˆ2 = √ h1, 1, 0i. 2

There are two different ways we can think of the origin O of principal axes: 1. We can think of the origin O as the fixed point in space about which the rigid body is rotating. In this case, the coordinate system is an inertial (or space) frame. 2. Or as a tumbling body, with the origin at the center of mass and the principal axes attached to the rigid body. In this case, the coordinate system is the body frame. That is, the coordinate system is attached to the tumbling body. In the body frame, the inertia tensor is diagonal. The body frame can then be defined by the principal axes. If we’re in the body frame (i.e. coordinate system is attached to the tumbling body) and ~ω = ω1eˆ1 + ω2eˆ2 + ω3eˆ3, then L~ = λ1ω1eˆ1 + λ2ω2eˆ2 + λ3ω3eˆ3. The problem with the body frame is that it’s not an inertial frame. The laws of physics are nice in the space frame. In this frame, ! dL~ = ~Γ, dt space about O. That is, the time derivative of the angular momentum vector as measured in the space frame is equal to the torque about the origin. Note that ~Γ is the same whether measured in the space or the body frame. A vector is a vector, but the time derivative of a vector depends on the coordinate system. That is, we don’t have to specify the frame for a vector, but we do have to specify the frame when talking about the time dependence of a vector. Recall that if frame S rotates with angular velocity Ω~ with respect to frame So, then for any vector Q~, ! ! dQ~ dQ~ = Ω~ × Q~ + . dt dt So S So in the body frame, ! ! dL~ dL~ = ~ω × L~ + . dt dt space body ~˙ dL~ If we let L = dt , then body ! ˙ dL~ L~ + ~ω × L~ = = ~Γ. dt space 136 Rigid Rotations

We have that Tip ˙ L~ = λ1ω˙ 1eˆ1 + λ2ω˙ 2eˆ2 + λ3ω˙ 3eˆ3. Steady rotation occurs only Expanding the cross product on the left gives us Euler’s equations for rotation about principal axes. Stable rotation oc- λ ω˙ + (λ − λ )ω ω = Γ curs only for rotation about 1 1 3 2 2 3 1 the two principal axes asso- λ2ω˙ 2 + (λ1 − λ3)ω3ω1 = Γ2 ciated with the largest and λ3ω˙ 3 + (λ2 − λ1)ω1ω2 = Γ3. smallest moments of inertia. Notice that is is a nonlinear set of ODEs even if ~Γ = 0.

9.4 Torque-free Motion

If ~Γ = 0, then Γ1 = Γ2 = Γ3 = 0. If λ1 6= λ2 6= λ3 then Euler’s equations can be written as

λ1ω˙ 1 = (λ2 − λ3)ω2ω3

λ2ω˙ 2 = (λ3 − λ1)ω3ω1

λ3ω˙ 3 = (λ1 − λ2)ω1ω2.

For example, gravity does not exert a torque about the center of mass of an object, so the rotational motion of a ﬂying object is torque-free motion. Imagine spinning a rigid body about the principal axis eˆ3 with constant angular speed. Then ω1 = ω2 = 0 and ~ω = ω3 eˆ3. Since ~ω is constant,ω ˙1 =ω ˙2 =ω ˙3 = 0 then L~ = λ3ω3eˆ3 = constant. Is this rotation stable? Initially, we have ω1 = ω2 = 0. If we add a little perturbation or kick, ~ω will be diﬀerent after the perturbation. Assuming a small ω1 and ω2, then ω˙3 ≈ 0 because if ω1 and ω2 are assumed to be small then ω1ω2 is basically small squared. This gives us ω3 ≈ constant. Then we can write the remaining two torque-free Euler equations as λ2 − λ3 ω˙ 1 = ω3 ω2 λ1 λ3 − λ1 ω˙ 2 = ω3 ω1. λ2

Diﬀerentiating the ﬁrst with respect to time and replacing the resultingω ˙ 2 with the right side of the second equation, we get (λ3 − λ2)(λ3 − λ1) 2 ω¨1 = − ω3 ω1. λ1λ2

Notice that if the quantity in brackets is positive then the rotation is stable. Otherwise, it is unstable. The products of inertia can be negative, but the moments of inertia (i.e. the eigenvalues) are always positive. So we have that the rotation is stable if and only if λ3 is greater than both λ1 and λ2 or if λ3 is less than both λ1 and λ2. In other words, if λ3 is the smallest or largest moment of inertia, the rotation is stable. Conversely, if λ1 < λ3 < λ2 or λ2 < λ3 < λ1 then the rotation is unstable. This is essentially the intermediate axis theorem also called the tennis racket theorem or remote control theorem. For some interesting rotational dynamics, see the video in the footnote. 1

1https://www.youtube.com/watch?v=1n-HMSCDYtM 9.4. Torque-free Motion 137

Previously, we looked at torque free motion where all the eigenvalues were diﬀerent. If two of them are the same, e.g., λ1 = λ2, then the Euler equations become ~ω λ − λ 1 3 eˆ3 ω˙ 1 = ω3 ω2 λ1 λ1 − λ3 eˆ ω˙ 2 = − ω3 ω1 2 λ2 eˆ ω˙ 3 = 0. 1

Notice that the quantities in parentheses are now the same. Again, we have that ω3 = constant. If we let λ1 − λ3 Ω = ω3, Figure 9.3: A prolate ellipsoid λ1 in torque-free rotation. then the Euler equations become

ω˙ = Ωω 1 2 ~ω ω˙ 2 = Ωω1. eˆ3 This is a pair of coupled linear ODEs. We could solve this pair of coupled ODEs either by diﬀerentiating one and plugging eˆ2 it into the other or we could take them into the complex plane. By multiplying the second eˆ1 equation by i and adding it to the ﬁrst, we get Figure 9.4: An oblate ellipsoid d (ω + iω ) = Ω(ω − iω ) = −iΩ(ω + iω ). in torque-free rotation. dt 1 2 2 1 1 2 Next, we make the substitution η = ω1 + iω2, then the pair of ODEs can be written as

η˙ = −iΩη.

The solution to our new ODE is −iΩt η = η0e .

If η0 is a real constant ω0, then our solution becomes

η = ω0(cos Ωt − i sin Ωt).

Then the solution we’re interested in is

ω1(t) = Re(η) = ω0 cos Ωt

ω2(t) = Im(η) = −ω0 sin Ωt

ω3(t) = constant.

−iΩt Notice that the solution η = ω0e when plotted in the complex plane is a vector of length ω0 at angle −Ωt at time t. As time proceeds, the angle increases, so this vector is rotating clockwise. That is, the vector is precessing. Consider the prolate ellipsoid shown in Fig. (9.3). By its symmetry, we can easily determine that its principal axes lie along eˆ3, eˆ2 and eˆ1. That is, we know a principal axis lies along eˆ3, and we know that any other orthogonal pair of axes are also principal axes. We also know by symmetry that λ1 = λ2. Because more mass is further from the axis for rotation about eˆ1 and eˆ2, we know that λ1, λ2 > λ3. Therefore, we know that if the angular velocity is ~ω, that is, the ellipsoid is spinning about ~ω with magnitude ω, then Ω > 0, and over time, ~ω precesses along the circle shown in the clockwise direction when viewed from above. This is despite there being no torque on the system. 138 Rigid Rotations

Similarly, consider the oblate ellipsoid shown in Fig. (9.4). Again, we ﬁnd the prin-

z eˆ3 cipal axes the same way. This time, λ1 = λ2 < λ3, which gives us Ω < 0, so for angular velocity ~ω, the angular velocity vector precesses in the counterclockwise direction along the circle shown. Consider a symmetric top of mass M spinning about its principal axis eˆ3 with ~ω = y ωeˆ3 as shown in Fig. (9.5). The angular momentum of the top about the origin of the coordinate system is L~ = λ3~ω = λ3ω eˆ3 because eˆ3 is a principal axis. Gravity acts x on the center of mass of the top, applying a torque about the origin. If the center of mass is located at R~ cm = Rcmeˆ3, then the torque about the origin is ~Γ = R~ cm × M~g = Figure 9.5: A spinning and pre- −MgRcm eˆ3 × zˆ. The equation of motion of the top is cessing top. ˙ L~ = ~Γ.

So once we turn on gravity, L~ begins to change, and is no longer conserved. This implies that ~ω also begins to change. Now, ω1 and ω2 are no longer 0. However, if ω3 is large enough, we can assume they’re small. So assuming a rapidly spinning top (i.e. large ω3), we get L~ ≈ λ3ω eˆ3. (9.1)

In Fig. (9.5), L~ points along the top’s axis of rotation eˆ3 and ~Γ points into the page. ˙ Since L~ and ~Γ are perpendicular to each other and we know that L~ = ~Γ, we know that the magnitude of L~ does not change. A simple proof of this is

d ˙ L~ 2 = 2L~ · L~ = 2L~ · ~Γ = 0. dt

So the length of L~ is not changing. Its direction is changing, but its magnitude has no ˙ time dependence. Diﬀerentiating Eq. (9.1) gives us L~ ≈ λ3ω eˆ˙3. But we also know that ˙ L~ = ~Γ = −MgRcm eˆ3 × zˆ. Putting the two together gives us an ODE in terms of eˆ3,

eˆ˙3 = Ω~ × eˆ3,

where MgR Ω~ = cm zˆ. λ3ω So our spinning top precesses with positive Ω~ . Note that Ω~ << ~ω. In other words, the top is spinning rapidly about eˆ3 with angular velocity ~ω, but the axis eˆ3 itself is slowly spinning about zˆ with angular velocity Ω~ . 9.5. Summary: Rigid Rotations 139

9.5 Summary: Rigid Rotations

Skills to Master • Calculate the moment and products of inertia for rotation of a rigid system of particles or a rigid continuous mass distribution about a given axis • Calculate the inertia tensor for rotation of a rigid body about an arbitrary axis • Given the angular velocity vector for a rigid body, calculate the angular momentum vector and the rotational kinetic energy • Calculate the inertia tensor for rotation about a point by calculating the inertia tensor for the center of mass and then applying the generalized parallel axis theorem • Calculate the three principal axes for a rigid body rotating about a given point • Given the shape of a rigid body, guess its principal axes, the relative magnitudes of its moments of inertia, and the stability of rotation about the diﬀerent axes • Analyze the precession of a rigid body undergoing torque-free rotation

Given an origin O and some point R~ within or on system at that point when calculating the inertia ten- a rigid body, then we can write the position of the nth sor   particle in the body as Ixx Ixy Ixz   ~ 0 I =  Iyx Iyy Iyz . ~rn = R + ~rn ,   Izx Izy Izz where ~rn is the position of the particle in terms of the 0 For a rigid body composed of discrete particles mn, the coordinate system with origin O, and ~rn is the vector from the point R~ to that particle. If the point R~ is at moments and products of inertia are calculated as ˙ rest (even only instantaneously), then R~ = 0, and the X 2 2 Ixx = mn(yn + zn) kinetic energy of the rotating body can be written as n 2 X 1 X Ixy = − mnxnyn. T = m ~r˙ 0 . 2 n n n n If the rigid body is a continuous mass, then the sums That is, the kinetic energy reduces to the rotational become integrals kinetic energy of the body about the point R~ . In general, if R~ cm is the center of mass of a rigid I = (y2 + z2) dm body then the kinetic energy of the body is xx ˆ 1 ˙ 2 1 X T = MR~ + m ~r˙ 02 Ixy = − xy dm, 2 cm 2 n n ˆ n

= Tcm + Trot. where dm = σ da or dm = ρ dV depending on whether its a 2D or 3D object. That is, the kinetic energy of the body is the kinetic If the rigid body is a flat lamina lying in the xy- energy of the center of mass plus the rotational kinetic plane, then all the products of inertia containing z will energy of the body about the center of mass. be zero, and Ixx + Iyy = Izz. The moment of inertia of a rotating body is a prop- These integrals are either two or three dimensional erty of the body, and it is with respect to a specified integrals, so they can be very messy. Make use of sym- point in the body. Place the origin of your coordinate metry whenever you can. If necessary, divide the rigid 140 Rigid Rotations body into convenient pieces and calculate the moments and the kinetic energy is of inertia separately for each and then add them to get the final result. Even better, divide the object into 1 T 1 2 1 2 1 2 T = ~ω I~ω = λ1ωx + λ2ωy + λ3ωz . symmetric pieces, calculate the moment of inertia for 2 2 2 2 the most convenient piece and then multiply this by the To find the principal axes given a non-diagonal number of pieces. This will only work if the moments inertia tensor I, you have to diagonalize the inertia of the pieces are symmetric such that at any given dis- tensor. To do so, you find the eigenvalues (i.e. the mo- tance from the axis of rotation, each piece contributes ments of inertia) and the eigenvectors (i.e. principal the same amount of mass. axes) of I. This requires solving The angular momentum of a rigid body rotating with angular velocity ~ω is (I − λ1) ~ω = 0, L~ = I~ω. for nontrivial solutions. Start by finding the eigenval- Notice that this is the generalization of L = Iω from ues by solving the characteristic equation introductory physics. The kinetic energy of a rigid body rotating with det(I − λ1) = 0. angular velocity ~ω is 1 1 After finding the eigenvalues, substitute them back into T = ~ωT I~ω = L~ · ~ω. 2 2 the eigenvalue equation one at a time to find the eigenvectors or principal axes. Notice that these are the generalization of T = 1 Iω2 2 The rate of change of a rotating body’s angular from introductory physics. ~˙ If the vector from an arbitrary point P in the rigid momentum L measured in a frame attached to the body to its center of mass is hX,Y,Zi, then the mo- body satisfies Euler’s equation ments and products of inertia about P and about the ˙ center of mass CM are related as follows L~ + ~ω × L~ = ~Γ. P CM 2 2 Ixx = Ixx + M(Y + Z ) If our coordinate axes are the principal axes, and our P CM Ixy = Ixy − MXY. origin is at the center of mass of the rigid body, then the equations of motion for the rigid body are the com- This allows us to easily compute the inertia tensor ponents of Euler’s equation. about some point P given the inertia tensor about the In torque-free motion, ~Γ = 0, and the compo- center of mass. nents of Euler’s equation are For any rigid body, there are three orthogonal ~ principal axes such that L k ~ω if the body is ro- λ ω˙ = (λ − λ )ω ω tating about one of those axes. If the axes of your 1 1 2 3 2 3 coordinate system correspond to those principal axes, λ2ω˙ 2 = (λ3 − λ1)ω3ω1 then the inertia tensor is diagonal, and the eigenvalues λ3ω˙ 3 = (λ1 − λ2)ω1ω2. are the moments of inertia for each axis. The products of inertia are all zero in this case, Steady rotation can occur about principal axes.   Stable rotation occurs about principal axes for which λ1 0 0 the moment of inertia is either the largest or the small-   I =  0 λ2 0 . est of the three principal axes.   Other useful relations are 0 0 λ3 Then the angular momentum is simply ~Γ = ~r × F~ ˙ L~ = hλ1ωx, λ2ωy, λ3ωzi, = L~ . Chapter 10

Coupled Oscillators

10.1 Two Masses and Three Springs

Recall the familiar problem of a mass m attached to a spring of spring constant k. The force on the mass is related to its displacement x from its equilibrium position by Hooke’s law F = mx¨ = −kx. Consider now the more complicated problem of two masses and three springs as shown in the image below.

ka kb kc mA mB

xA xB

Looking at the ﬁrst mass, we have the spring force kaxA pulling to the left, and we have the force kb(xB − xA) pulling to the right. This gives us

mAx¨A = −kaxA + kb(xB − xA).

For the second mass, we have the force kcxB pushing to the left as well as the force kb(xB − xA). This gives us

mBx¨B = −kb(xB − xA) − kcxB.

This is a pair of coupled ODEs, which we can write in matrix form as

M~x¨ = −K~x, where " # " # " # m 0 k + k −k x (t) M = A ,K = a b b , ~x = A . 0 mB −kb kb + kc xB(t)

Notice that M and K are symmetric matrices, and they always will be. This is the form that we will use for coupled oscillators. To solve this matrix equation, we try the solution

~x(t) = Re (~z(t)) , 142 Coupled Oscillators

where Tip " # " # a eiωt a ~z = A = A eiωt = ~aeiωt. For coupled oscillators, iωt aBe aB there is one normal frequency and normal mode Notice that the components of ~z have the same frequency but diﬀerent amplitudes. The for each oscillating mass. numbers aA and aB are complex constants. The second derivative of ~z is

~z¨ = −ω2~aeiωt.

The matrix equation M~x¨ = −K~x then simpliﬁes to

K − ω2M~a = 0.

This is a generalized eigenvalue problem. It’s not a normal eigenvalue problem since M is not an identity matrix in general. We want to ﬁnd frequencies ω for which there are nontrivial solutions for the generalized eigenvalue problem. Nontrivial solutions occur when

det K − ω2M = 0.

The ω2 are generalized eigenvalues. The frequencies ω are the normal frequencies. The associated motion characterized by eigenvector ~a is the normal mode. To play around with coupled oscillators via a Java applet, visit the link in the footnote.1

Identical Masses and Springs

Suppose you have two masses and three springs where mA = mB = m and ka = kc = k. Then our matrices are " # " # m 0 k + k −k M = ,K = b b . 0 m −kb k + kb

Notice that M is now the identity matrix times m. To solve the eigenvalue problem, we compute the determinant

2 2 2 2 det K − ω M = (k + kb − mω ) − kb .

This gives us the eigenvalues (and normal frequencies) r r k k + 2k ω = , ω = b . 1 m 2 m

The eigenvector ~a1 associated with ω1 is obtained by plugging ω1 into the generalized eigenvalue problem

2 K − ω1M ~a1 = 0 k K − M ~a = 0 m 1 " #" # " # k −k a 0 b b 1A = . −kb kb a1B 0

1http://falstad.com/coupled/ 10.1. Two Masses and Three Springs 143

In a1A and a1B, the ‘1’ in the subscript means this is normal mode 1, and the A and B refer to the components of ~a1. This implies that a1A = a1B, but since these are complex −iδ1 numbers, we can write them as a1A = a1B = C1e , where C1 and δ1 are real numbers. So " # 1 iω1t i(ω1t−δ1) ~z(t) = ~a1e = C1 e . 1

Our solution is then " # " # xA 1 ~x(t) = = Re(~z) = C1 cos(ω1t − δ1). xB 1

The motion of mass A and mass B in the ﬁrst normal mode are then

xA(t) = C1 cos(ω1t − δ1)

xB(t) = C1 cos(ω1t − δ1).

In this mode, the motion is that the two masses oscillate back and forth as one (i.e. in p lockstep). We could have guessed this since ω1 = k/m does not depend on kb. p For the second normal mode, we have the normal frequency ω2 = (k + 2kb)/m. Plugging this into the eigenvalue problem gives us

 k + 2k K − b M ~a = 0 m 2 " #" # " # −k −k a 0 b b 2A = . −kb −kb a2B 0

−iδ2 This implies that a2A = −a2B = C2e . So we have that " # 1 i(ω2t−δ2) ~z(t) = C2 e . −1

Our solution is then " # " # xA(t) 1 ~x(t) = = Re(~z) = C2 cos(ω2t − δ2). xB(t) −1

The motion of mass A and mass B in this normal mode are then

xA(t) = C2 cos(ω2t − δ2)

xB(t) = −C2 cos(ω2t − δ2).

In this mode, the motion is that of two masses oscillating 180◦ out of phase like an accordion. For example, both masses move outward at the same time then inward at the same time. The general solution of our system of coupled oscillators can be written as the sum of the normal modes " # " # 1 1 ~x(t) = C1 cos(ω1t − δ1) + C2 cos(ω2t − δ2). 1 −1

Notice that there are four integration constants; C1, C2, δ1, and δ2, as there should be. 144 Coupled Oscillators

Normal coordinates are where one normal mode survives and all others vanish. In our case, if we write x + x η = A B 1 2 x − x η = A B . 2 2

Then for the ω1 (i.e. sliding back and forth) mode, we get

η1 = C cos(ω1t − δ1), η2 = 0.

For the ω2 (i.e. accordion) mode, we get

η1 = 0, η2 = C cos(ω2t − δ2).

Weak Coupling We now modify the problem above to look at the weak coupling case. In the weak coupling limit the masses are only weakly coupled to each other. That is, kb << k. We start by deﬁning the frequency r k + k ω ≡ b . 0 m

We can write both normal frequencies ω1 and ω2 in terms of ω0 as

r s r k 2 kb kb ω1 = = ω0 1 − ≈ ω0 1 − m k + kb k r s r k + 2kb 2 kb kb ω2 = = ω0 1 + ≈ ω0 1 + . m k + kb k

Here, we made the approximation kb ≈ 0 in the denominator since kb << k. If we write n ε = ω0kb/(2k), we can make the approximation (1 + ε) ≈ 1 + nε, and then

ω1 ≈ ω0 − ε

ω2 ≈ ω0 + ε.

Then our general complex solution is " # " # " # " # ! 1 1 1 1 i(ω0−ε)t i(ω0+ε)t −iεt iεt iω0t ~z(t) = K1 e + K2 e = K1 e + K2 e e . 1 −1 1 −1

−iδ1 −iδ2 Note that K1 = C1e and K2 = C2e are complex numbers. Since ε << ω0, the quantity e±iεt varies much more slowly in time than eiω0t. To get an idea of the kind of behavior that can be produced by this kind of system, consider the case in which the complex coeﬃcients are real and equal to each other. That 0 0 is, suppose K1 = K2 = K /2, where K is a real number. Then our solution simpliﬁes to " # cos(εt) ~z(t) = K0 eiω0t. − sin(εt)

Taking the real part gives us

0 xA(t) = K cos(εt) cos(ω0t), 10.2. The General Problem 145 for mass A. The plot of this is shown in Fig. (10.1). Notice that the slowly varying “amplitude” cos(εt) acts as an envelope. For mass B, we get xA 0 xB(t) = K sin(εt) cos(ω0t).

The plot is shown in Fig. (10.2). In these plots, we see that the oscillations move from one oscillator to the other, t back and forth. When one oscillator is oscillating at maximum amplitude, the other is at a minimum. Then as the amplitude of the oscillations of the ﬁrst one decreases, the amplitude of the oscillations of the second one increases. This is a fundamental Figure 10.1: A plot of the dis- phenomenon of coupled oscillators in physics. One oscillator drives the other until the placement of mass A. other picks up enough energy to begin driving the ﬁrst, and so on.

10.2 The General Problem xB

Consider a conservative (i.e. no damping) system with generalized coordinates q1, . . . , qn and Lagrangian L = T − U. We will assume that the Lagrangian has no explicit time t dependence. Assume that there’s a stable equilibrium when each qi has some special value qi0 . At equilibrium,q ˙i =q ¨i = 0. Since L has no explicit time dependence, every term contains eitherq ˙i orq ¨i (which are zero at equilibrium), and so Figure 10.2: A plot of the displacement of mass B. d ∂L = 0. dt ∂q˙i 0 This will happen no matter how complicated L is when it has no explicit t dependence. The Euler-Lagrange equation, therefore, implies that ∂L ∂T ∂U = 0 = − . ∂qi 0 ∂qi 0 ∂qi 0

Note, means that the expression is evaluated at the equilibrium point. 0 For convenience, we can choose our coordinate system such that the coordinates are zero at the equilibrium point: qi0 = 0 for all i. That is, we choose the coordinate system such that the equilibrium is at the origin. Then we can do the usual Taylor expansion of U about the equilibrium. We get

X ∂U 1 X ∂2U U = U0 + qk + qjqk + ··· . ∂qk 0 2 ∂qj∂qk 0 k j,k

We can ignore U0 since it is just a constant, and we are dealing with potential energy. The ﬁrst sum is zero, so the leading term in this expansion of U is quadratic

1 X U ≈ K q q , 2 jk j k j,k where ∂2U Kjk = , ∂qj∂qk 0 are the elements of the matrix K. These elements are just numbers. We will assume that the coordinates of the particles expressed in terms of generalized coordinates have no explicit time dependence. That is, ~rn = ~rn(q1, . . . , qn). This implies that X ∂~rn ~r˙ = q˙ , n i ∂q i i 146 Coupled Oscillators

which implies that the kinetic energy is Tip X 1 1 X T = m ~r˙ 2 = A {q }q˙ q˙ , When ﬁlling the matrices 2 n n 2 jk i i j K and M, remember that n j,k they must be symmetric. where X ∂~rn ∂~rn A = m · . jk n ∂q ∂q n j k The notation Ajk {qi} means that Ajk could depend on all qi. If we Taylor expand, we get Ajk {qi} = Ajk {qi} + ··· 0 Keeping only the zeroth order term, we deﬁne

Mjk = Ajk {qi} . 0 These are the elements of the matrix M. The Lagrangian is then

1 X 1 X 1 1 L = T − U ≈ M q˙ q˙ − K q q = q˙T Mq˙ − qT Kq. 2 jk j k 2 jk j k 2 2 jk jk

The Euler-Lagrange equations are

d ∂L ∂L − = 0, dt ∂q˙i ∂qi where ∂L X = − Kikqk ∂qi k ∂L X = Mikq˙k. ∂q˙i So the Euler-Lagrange equation becomes X X Mikq¨k = − Kikqk, k which we can write as the generalized eigenvalue problem

M~q¨ = −K~q.

Note: Because of the way we have deﬁned K and M, these matrices must be symmetric. To summarize, we start by ﬁnding the kinetic energy of a system

1 X T = A q˙ q˙ , 2 jk i j j,k

from which we deduce the elements of the matrix A. Then we obtain the elements of the matrix M by evaluating A at the equilibrium value of the coordinates qi

Mjk = Ajk . 0 10.3. Double Pendulum 147

Next, we ﬁnd the potential energy

1 X U = K q q , 2 jk j k j,k φa La from which we deduce the elements of the matrix K. Once we have the matrices M and ma Lb K, we ﬁnd the eigenvalues and eigenvectors of the generalized eigenvalue problem φb mb M~q¨ = −K~q, Figure 10.3: Double pendulum and this gives us the normal frequencies and modes of the system.

10.3 Double Pendulum

Consider the double pendulum shown in Fig. (10.3). We can specify its state with the generalized coordinates φa and φb. The kinetic energy of the system is 1 1 T = m v2 + m v2. 2 a a 2 b b ˙ If the length of the ﬁrst pendulum is La then the linear speed of ma is va = Laφa. For the ˙ second pendulum, we have length Lb. The linear speed of mb relative to ma is vr = Lbφb. So we have that the velocity of ma is ~va and the velocity of mb relative to ma is ~vr. The velocity of mb relative to our origin at the pivot of ma is the sum ~vb = ~va + ~vr. Its velocity squared is then 2 2 2 2 2 ˙2 2 ˙2 vb = (~vr + ~va) = vr + 2~va · ~vr + va = Lb φb + 2~va · ~vr + Laφa.

But the angle between ~vr and ~va is φb − φa. You can convince yourself of this with some simple geometry and the fact that ~va must be perpendicular to the line La, and ~vr must be perpendicular to the line Lb. So we can expand the dot product and write 2 2 ˙2 ˙ ˙ 2 ˙2 vb = Lb φb + 2LaLbφaφb cos(φb − φa) + Laφa. The kinetic energy is then 1 1 T = (m + m )L2φ˙2 + m L L φ˙ φ˙ cos(φ − φ ) + m L2φ˙2. 2 a b a a b a b a b b a 2 b b b The potential energy is

U = magLa(1 − cos φa) + mbg (La[1 − cos φa] + Lb[1 − cos φb]) .

Suppose the two masses are the same, ma = mb = m, and the two lengths are the same, La = Lb = L. Then the kinetic energy reduces to 1 1 X T = mL2 2φ˙2 + 2φ˙ φ˙ cos(φ − φ ) + φ˙2 = A φ˙ φ˙ . 2 a a b b a b 2 ij i j i,j This implies that " # 2 cos(φ − φ ) A = mL2 a b . cos(φa − φb) 1 These values are obtained by reading them from the kinetic energy T . Since this matrix 2 ˙ ˙ is required to be symmetric, we split up the term 2mL φaφb cos(φb − φa) to distribute it equally in A12 and A21. That is, we wrote ˙ ˙ ˙ ˙ ˙ ˙ 2 cos(φa − φb)φaφb = A12φaφb + A21φaφb. 148 Coupled Oscillators

To compute M, we evaluate A at the equilibrium point where our generalized coordinates are zero: φa = φb = 0. " # 2 1 M = A = mL2 . 0 1 1 Our potential energy is

U = 2mgL(1 − cos φa) + mgL(1 − cos φb).

1 2 Applying the small-angle approximation cos φ ≈ 1 − 2 φ gives us 1 1 X U ≈ mgL(2φ2 + φ2) = K φ φ . 2 a b 2 ij i j ij

From this, we can read oﬀ the elements of the K matrix " # 2 0 K = mgL . 0 1

p By writing ω0 = g/L, we get " # 2ω2 0 K = mL2 0 . 2 0 ω0

The generalized eigenvalue problem is

M~q¨ = −K~q.

We start by ﬁnding the eigenvalues and eigenvectors. Since these are 2 × 2 matrices, we expect two eigenvalues ω1 and ω2 and two eigenvectors ~a1 and ~a2. To ﬁnd the eigenvalues, we compute

2 2 2 2 4 det K − ω M = 2(ω0 − ω ) − ω = 0.

This gives us the eigenvalues √ 2 2 ω1 = (2 − 2)ω0 √ 2 2 ω2 = (2 + 2)ω0.

Each eigenvalue speciﬁes a normal mode or eigenmode of the coupled oscillator. The number of normal modes is the same as the number of oscillating masses. For each eigenvalue, we obtain the corresponding eigenvector by plugging the eigenvalue back into K − ω2M~a = 0. √ 2 2 For the normal mode ω1 = (2 − 2)ω0, we get the eigenvector " # 1 ~a1 = a1 √ . 2

This tells us that " # " # 1 1 iω1t −iδ1 iω1t ~z(t) = a1 √ e = A1e √ e . 2 2 10.4. Triatomic Molecule 149

−iδ1 Notice that we replaced the arbitrary complex number a1 by A1e , where A1 and δ1 are real. Then " # m M m φ φ~(t) = a = Re(~z(t)). φb Figure 10.4: Modeling a tri- This gives us atomic molecule as a coupled oscillator.

φa(t) = A1 cos(ω1t − δ1) √ φb(t) = 2A1 cos(ω1t − δ1).

The motion of this normal mode is that of both pendulums moving in phase. That is, both swing to the right at the same time, then to the left at the same time, and so on. Naturally, the amplitude of the bottom one is larger since it swings further outward than the top pendulum. √ 2 2 For the normal mode ω2 = (2 + 2)ω0, we get the eigenvector " # 1 ~a2 = a2 √ . − 2

This tells us that " # 1 −iδ2 iω2t ~z(t) = A2e √ e . − 2 Then

φa(t) = A2 cos(ω2t − δ2) √ φb(t) = − 2A2 cos(ω2t − δ2).

The motion of this normal mode is that of the pendulums moving out of phase. That is, as one swings to the right, the other swings to the left. Try the double pendulum at the link in the footnote.2

10.4 Triatomic Molecule

Consider a simple model of a triatomic molecule such as carbon dioxide. We have two masses m and a third central mass M. They are connected by a pair of springs with spring constant k as shown in Fig. (10.4). We want to ﬁnd the normal modes and frequencies.

k k m M m x

xa xb xc

The potential energy is 1 1 1 U = k(x − x )2 + (x − x )2 = k(x2 − 2x x + 2x2 − 2x x + x2). 2 a b 2 b c 2 a a b b b c c Since 1 X U = K x x , 2 ij i j i,j

2http://myphysicslab.com/ 150 Coupled Oscillators

We can read oﬀ the matrix elements as   1 −1 0   K = k −1 2 −1 .   0 −1 1

When reading oﬀ the matrix elements, remember that K has to be symmetric so the the elements have to be shared equally among the relevant oﬀ-diagonal entries. The kinetic energy is

1 1 1 1 X T = mx˙ 2 + Mx˙ 2 + x˙ 2 = M x˙ x˙ . 2 a 2 b 2 c 2 ij i j i,j

Reading oﬀ the matrix elements, we get   m 0 0   M =  0 M 0 .   0 0 m

To ﬁnd the normal frequencies, we compute

det(K − ω2M) = ω2(k − ω2m) ω2mM − k(2m + M) = 0,

which gives us

2 ω1 = 0 k ω2 = 2 m 2 1 ω2 = k + . 3 M m

2 The ω1 = 0 mode occurs in any system with translation symmetry. It is the “oscillation” in which all three atoms move in the same direction and never come back. It is an oscillation with frequency 0 and an inﬁnite period. That is, the whole system just 2 translates. When we plug ω1 = 0 into

2 K − ω M ~a1 = 0,

we get the eigenvector   1   ~a1 = A1 1 ,   1 which corresponds to uniform translation of all three masses. 2 For the normal frequency ω2 = k/m, we get the eigenvector   1   ~a2 = A2 0 .   −1

This motion corresponds to the central atom remaining ﬁxed and the outside atoms oscillating out of phase with each other. As the left atom moves left, the right atom moves right and so on. It is an accordion-like motion. 10.5. Parallel and Series Springs 151

2 For the normal frequency ω3 = k(2/M + 1/m), we get the eigenvector   k1 1  2m  ~a3 = A3 − . m  M  1

k2 The motion of this mode is that of the outer masses moving in the same direction with the same magnitude and the central mass moving in the opposite direction with a magnitude Figure 10.5: Springs in parallel that depends on the relative masses.

10.5 Parallel and Series Springs k1 k2 m Consider a mass attached to a set of parallel springs as shown in Fig. (10.5). The equation of motion of the mass is

mx¨ = −k1x − k2x = −(k1 + k2)x, Figure 10.6: Springs in series so the eﬀective spring constant is k1 + k2. In general, for parallel springs, the eﬀective spring constant is just the sum of the spring constants

keff = k1 + k2 + ··· .

Now consider a mass attached to a pair of springs in series as shown in Fig. (10.6). Now the eﬀective spring constant is

1 1 1 = + + ··· . keff k1 k2 152 Coupled Oscillators

10.6 Summary: Coupled Oscillators

Skills to Master • Setup the generalized eigenvalue problem of a coupled oscillator using the Newtonian approach by first writing the equations of motion as a system of equations • Setup the generalized eigenvalue problem of a coupled oscillator using the Lagrangian approach by first writing the kinetic and potential energies of the system • Solve a generalized eigenvalue problem by obtaining the eigenvalues and eigenvectors • Interpret the physical significance of the eigenvalues and eigenvectors and identify the physical behavior of the coupled oscillator in its normal modes

To find the equations of motion for coupled oscil- and T , you will often have to distribute a lators, follow the following procedure: term into a pair of off-diagonal elements. 1. Obtain the total kinetic energy T and the total c) There is also a third way to obtain all of potential energy U for the coupled oscillators in the above and that is to simply obtain the terms of generalized coordinates. equations of motion of the oscillating masses 2. In order to use this method, T needs to be a by applying Newton’s laws. Once the cou- homogeneous quadratic function of the velocities pled equations of motion are obtained, you and U needs to be a homogeneous quadratic func- can write them in the form M~q¨ = −K~q be- tion of the coordinates. If they are not, you can fore proceeding with the rest of these steps. (for a wide class of coupled oscillators at least) Generally, the methods detailed above are use small-angle approximations to convert them easier than to obtain the equations of mo- to the proper form. tion by directly applying Newton’s laws. 3. Obtain the K and M matrices. This can be done 4. Solve the generalized eigenvalue problem in several different ways: 2 a) Compute the Lagrangian L = T − U, then K − ω M ~a = 0. obtain the Euler-Lagrange equations for the system. After you have the Euler-Lagrange To do so: equations, write them in matrix form as a) Obtain the normal frequencies ω (i.e. the eigenvalues) by solving det K − ω2M = 0. M~q¨ = −K~q. b) Obtain the normal modes (i.e. the eigenvectors), by plugging the eigenvalues (one at a b) Compare T and U with the following time) into K − ω2M~a = 0 and solving for 1 X ~a. T = M q˙ q˙ 2 jk j k 5. Once the normal frequencies ωi and their associ- j,k ated eigenvectors ~ai have been found, write the 1 X complex solution as the linear combination U = K q q . 2 jk j k j,k X −iδi iωit ~z(t) = Aie ~aie . Note, if T is a function not only ofq ˙jq˙k but i also of q and/or q , then simplify T by eval- j k 6. The actual solution is the real part of the com- uating each of these terms at the equilib- plex solution rium of the system. The two matrices K and M must be symmetric matrices, so when fill- X ~q(t) = Ai~ai cos (ωit − δi) , ing the elements from the terms found in U i 10.6. Summary: Coupled Oscillators 153

where Ai~ai cos (ωit − δi) gives the motion of the tains the masses m which come from the kinetic energy ith normal mode. T and K contains the spring constants k, which come 7. You may be asked for the speciﬁc equations for from the potential energy U. each mass given some initial conditions. In that In most of these cases, we assume small oscilla- case, it’s helpful to remember that you can write tions, so you are free to use the small-angle approximations Ai cos (ωit − δi) = Bi cos ωit + Ci sin ωit. 1 Remember, the number of normal modes for a sys- cos x ≈ 1 − x2 tem of coupled oscillators is the same as the number 2 of oscillating masses. So if you have two masses and sin x ≈ x three springs there will be two normal modes, and if tan x ≈ x. you have three masses connected by two springs, there will be three normal modes. Also, remember the binomial expansion approximation To remember whether M goes with T or U and whether K goes with T or U, remember that M con- (1 + x)n ' 1 + nx. Chapter 11

Collision and Scattering

Consider a ﬁxed target situated at the origin of a scattering axis z as shown below. The target produces a spherically-symmetric central force that repels incoming particles, so the incoming particle has a potential energy V (r) that is a function of the distance r between it and the target.

b θ z

Suppose your incoming particle is a distance b above the scattering axis when it is incoming. We call this distance the impact parameter. Asymptotically, the scattered particle leaves in a straight line at some scattering angle θ. Note that the scattering angle is a function of the impact parameter. That is, θ = θ(b). In the lab, we have a whole beam of incoming particles with some intensity I, where the intensity is the number of incident particles per unit time per unit area. number of incident particles I = . unit time × unit area

dΩ

θ z

What we measure is the number of particles dN detected by our detector per unit time. The experimentalist knows I and dΩ, and measures dN. The diﬀerential dΩ is the solid angle subtended by our detector. Recall that for a circle, the total angle is 2π radians, and for a circle of radius R, an angle dθ is related dL to the subtended arc length dL via dθ = R . The 3D generalization is a sphere of radius R. For a sphere, the total solid angle is 4π steradians. The relation between the solid angle dΩ and the subtended area on the surface of the sphere is dA dΩ = . R2 11.1. Hard Sphere Scattering 155

Recall that dA = R2 sin θ dθ dφ. The solid angle in spherical coordinates is

dΩ = sin θ dθ dφ.

The diﬀerential cross section is dσ 1 dN (θ) = . dΩ I dΩ The total cross section is dσ dσ σ = dΩ = sin θ dθ dφ. ˆ dΩ ¨ dΩ The total cross section σ has units of area and is the total number of particles scattering per unit time divided by the number of incident particles per unit area per unit time. The total cross-section is a measure of how big the target is as seen by the scattered particles. For a hard object (i.e. the particles bounce oﬀ of its surface), σ is the actual cross-section of the target. Once σ is found in terms of R, it is useful to plot dσ/dR versus θ. This illustrates the angles at which the scattering is greatest for particles with a given b. Consider a ring of incoming particles as illustrated below. The incoming ring has an inner radius b and a width db, so it has area 2πb db. If we take a cross-section of the scattered particles, we see that they also form a ring. Suppose all of the incoming particles are scattered onto this ring-shaped detector, then

dN = I2πb |db| 2π dΩ = sin θ dθ = 2π sin θ|dθ|. ˆ0 The absolute value bars are there to ensure that the area can never be negative.

dσ 1 dN Plugging these values into dΩ (θ) = I dΩ gives us

dσ b db = . dΩ sin θ dθ In general, we find θ as a function of b, then we invert it to get b as a function of θ. dσ From there, we plug b and db into the equation above to get dΩ . The first step, finding θ(b) is just a mechanics (orbit) problem.

11.1 Hard Sphere Scattering

Suppose we have particles scattering from a hard sphere of radius R. In this case, the potential acting on the incoming particles is zero everywhere except inside the ball. This is a central force with a very large ( ∼ ∞) potential inside the ball. 156 Collision and Scattering

α α

α α θ z

b Notice from the symmetry of the problem that θ = π − 2α. We can see that sin α = R so −1 b α = sin R , then b θ(b) = π − 2 sin−1 . R Inverting this gives us

π − θ θ b(θ) = R sin = R cos . 2 2

Taking the derivative gives us

db 1 θ = − R sin . dθ 2 2

Plugging these into the equation at the end of the previous section gives us the diﬀerential cross-section

θ 2 θ θ 2 dσ b db R cos 2 1 θ R 2 sin 2 cos 2 R = = − R sin = = . dΩ sin θ dθ sin θ 2 2 4 sin θ 4

Integrating the diﬀerential cross-section gives us the total cross-section

dσ σ = dΩ = πR2. ˆ dΩ

This is the actual cross-section of a sphere of radius R.

11.2 The General Case

In the general case, we have some potential energy U(r). Assume that U(r) → 0 as r → ∞. As illustrated below, there is some minimum approach rmin that the particle makes to the target. The angle between rmin and the scattering angle θ is ψ0.

ψ0 b rmin θ z

The position of the particle at time t is given by ~r(t). It is convenient to orient our polar coordinates so that the angle ψ is measured from ~rmin to ~r as illustrated below. 11.2. The General Case 157

~r ψ b rmin z

If the particle comes in from a great distance and scatters to a great distance then its initial speed and ﬁnal speed v0 are the same by conservation of energy and the fact that the potential energy is zero at a great distance. So the particle’s total energy is 2 E = mv0/2. Throughout the scattering process, the particle’s energy is 1 `2 E = mr˙2 + + U(r), 2 2mr2 where ` = mr2ψ˙ is its angular momentum about the origin (the origin is located at the target). The energy equation is obtained by treating this as an orbit problem. The angular momentum is calculated simply as ` = √Iω. We can also calculate the angular momentum of the particle using ` = bmv0 = 2mEb. The direction of the angular momentum is into the page. Both E and ` are conserved.

Ueff

E r

rmin

Notice that each value of r in (rmin, ∞) is visited twice by the particle—once while it is coming toward the target and once while it is scattering away from the target. The closest approach that the particle makes to the target is rmin. It is convenient to have this occur at t = 0, so that scattering begins at t = −∞ and ends at t = ∞. From the equivalent representations of the angular momentum, we have that r ` 2E b ψ˙ = = . mr2 m r2

Recall that ψ0 is the angle between rmin and the scattering angle θ. This is the total angle swept out by ~r from t = 0 to t = ∞, so

∞ ∞ r ˙ 2E b 1 ψ0 = ψ dt = 2 dr. ˆ0 ˆrmin m r r˙ Here, we used dt = dr/r˙. We can change variables to r since r is a monotonic function from t = 0 to t = ∞. From the equation for energy, we getr ˙ = p2/mpE − `2/(2mr2) − U(r). Next, we make the substitution b2 = `2/(2mE), so we haver ˙ = p2E/mp1 − b2/r2 − U(r)/E. Making this substitution in the integral above gives us

∞ b r2 ψ0 = dr. q 2 ˆrmin b U(r) 1 − r2 − E 158 Collision and Scattering

Recall in the hard sphere example, that θ = π − 2α by the symmetry of the problem. Likewise, in the general case, we have that θ = π − 2ψ0. This can be shown rigorously by integrating ﬁrst from t = −∞ to t = 0 and from t = 0 to t = ∞ and comparing the results. This means ∞ b θ = π − 2 r2 dr. q 2 ˆrmin b U(r) 1 − r2 − E

This is the general case for θ(b). After evaluating this integral, we would invert the result to get b(θ) which we then use to compute the diﬀerential cross-section

dσ b db = . dΩ sin θ dθ

Note that at r = rmin,r ˙ = 0. That is, the radial component of the incoming particle’s velocity is zero at the moment of closest approach. Fromr ˙ = p2E/mp1 − b2/r2 − U(r)/E, we get

2 b U(rmin) 1 − 2 − = 0. rmin E

Given U(r) and b, we can use this to compute rmin.

11.3 Rutherford Scattering

Suppose we have a gold nucleus with charge Q = +Ze at the origin. The incoming particle is an alpha particle with charge q = +2e. The potential energy between the pair of particles is kqQ γ U(r) = = , γ = kqQ. r r This gives us the angle ∞ b r2 ψ0 = dr. q 2 γ ˆrmin b 1 − r2 − rE

If we let u = b/r then du = −(b/r2)dr, then

0 1 ψ0 = − p γu du. ˆ b 1 − u2 − rmin Eb

Flipping the limits of integration and completing the square under the radical gives us

b rmin 1 ψ0 = q du. ˆ0 2 γ 2 γ2 1 − u + 2Eb + 4E2b2

If we let β = γ/(2E) and y = u + β/b.

q 2 1+ β b2 1 ψ0 = q dy. ˆ β β2 2 b 1 + b2 − y 11.3. Rutherford Scattering 159

Next, we let w = r/p1 + β2/b2, then   1 β 1 −1 −1 b ψ0 = β √ dw = sin (1) − sin q  ˆ b 2 β2 r 1 − w β2 1 + 2 1+ b b2 β π b sin − ψ0 = 2 q β2 1 + b2 β b cos ψ0 = q β2 1 + b2 b tan ψ = 0 β π − θ b = β tan . 2

Since tan((π − θ)/2) = cot(θ/2), we get

θ b(θ) = β cot . 2 Our diﬀerential cross-section is

θ 2 dσ b db β cot 2 β 1 β 1 = = − = . dΩ sin θ dθ sin θ 2 2 θ 4 4 θ sin 2 sin 2

Substituting β = γ/(2E) = kqQ/(2E) back in gives us the Rutherford scattering formula dσ k2q2Q2 1 = . dΩ 16E2 4 θ sin 2 Chapter 12

Math Reference

12.1 Complex Numbers

eiθ = cos θ + i sin θ

1 cos x = (eix + e−ix) 2 1 sin x = (eix + e−ix), 2i

12.2 Vectors

The BAC-CAB identity is

A~ × (B~ × C~ ) = B~ (A~ · C~ ) − C~ (A~ · B~ )

The triple product identity is

A~ · (B~ × C~ ) = B~ · (C~ × A~) = C~ · (A~ × B~ ).

12.3 Curvilinear Coordinates

Cartesian Coordinates

In Cartesian coordinates, the gradient of a function of three variables f(x, y, z) is

∂f ∂f ∂f ∇~ f = xˆ + yˆ + zˆ. ∂x ∂y ∂z

The quantity ∇~ × F~ is called the curl of F~.

  xˆ yˆ zˆ   ∇~ × F~ =  ∂ ∂ ∂ .  ∂x ∂y ∂z  Fx Fy Fz 12.3. Curvilinear Coordinates 161

Polar Coordinates

The transformation equations to go from polar to Cartesian coordinates or vice versa are

x = r cos φ y = r sin φ p r = x2 + y2 y φ = tan−1 . x

φˆ rˆ •

~r φ

The area element is

dA = r dr dφ.

The unit vectors are

rˆ = hcos φ, sin φ, 0i φˆ = h− sin φ, cos φ, 0i.

Cylindrical Coordinates

The relation between Cartesian and cylindrical coordinates is

(x, y, z) = (ρ cos φ, ρ sin φ, z) x = ρ cos φ y = ρ sin φ z = z p ρ = x2 + y2 y φ = tan−1 . x 162 Math Reference

• P

y ρ

φ •P 0 x

The unit vectors are

ρˆ = hcos φ, sin φ, 0i φˆ = h− sin φ, cos φ, 0i zˆ = h0, 0, 1i.

The volume element is

dV = ρ dρ dφ dz.

Consider the units when ﬁguring out if you have the right form of the volume element. If you remember that the coordinates are (ρ, φ, z) then you know that the volume element contains dρ dφ dz. But the units of this quantity are L2 and volume is L3. Hence the volume element needs the factor of ρ to make the units correct. You can use the same reasoning for the volume elements of other coordinate systems as well as for line and area elements.

Spherical Coordinates

x = r sin θ cos φ y = r sin θ sin φ z = r cos θ. 12.4. Diﬀerential Equations 163

• P r θ

The volume element is dV = r2 sin θ dθ dφ dr. The gradient in spherical coordinates is ∂f 1 ∂f 1 ∂f ∇~ f = rˆ + θˆ + φˆ. ∂r r ∂θ r sin θ ∂φ

12.4 Diﬀerential Equations

The common ODE x¨ = −ω2x, has sinusoidal solutions of the form x(t) = A cos(ωt + δ) = B cos ωt + C sin ωt. Another common ODE x˙ = −λx, has exponentially decaying solutions of the form x(t) = Ae−λt.

12.5 Taylor Series

The Taylor expansion of f(x) about a point a is ∞ X f (n)(a) f(x) = (x − a)n. n! n=0 Some Taylor series commonly applicable in mechanics include the following: x2 x3 x4 ex = 1 + x + + + + ··· for x ∈ R 2! 3! 4! x3 x5 x7 sin x = x − + − + ··· for x ∈ R 3! 5! 7! x2 x4 x6 cos x = 1 − + − + ··· for x ∈ R 2! 4! 6! x2 x3 x4 ln(1 + x) = x − + − + ··· for x ∈ (−1, 1] 2 3 4 164 Math Reference

12.6 Approximations

Some helpful approximations for small x are

cos x ' 1 (very small angle approx.) x2 cos x ' 1 − (small angle approx.) 2 sin x ' x (small angle approx.) (1 + x)n ' 1 + nx sin(a + x) ' sin a + x cos a cos(a + x) ' cos a − x sin a. Index

Action, 84 Euler’s equations, 136 Air resistance,8 Euler-Lagrange equation, 74 Angular momentum, 28–30 Apsidal points, 107 Fermat’s principle, 72 Archimedes’ principle, 12 Friction,8 Atwood machine, 43 Functional, 72

Bounded orbits, 107 Galilean invariance,2 Brachistochrone, 77 Generalized eigenvalue, 142 Buoyancy,8 Generalized eigenvalue problem, 142 Buoyant force, 12 Gravitational mass,1

Calculus of variations, 72 Hamilton’s principle, 84 Center of mass, 26 Hamiltonian, 96 Central force, 29, 44, 102 Hooke’s law, 49 Centrifugal force, 118, 123 Ignorable coordinate, 88 Characteristic time, 10 Impact parameter, 154 Co-latitude, 120 Inertia Tensor, 128 Collision, 154 Inertia tensor, 129 Conservation of angular momentum, 29 Inertial frame,2 Conservative forces, 36 Inertial mass,1 Coriolis force, 118, 119, 123 Integration by parts, 74 Coupled Oscillators, 141 Intermediate axis theorem, 136 Critical damping, 55 Curl, 160 Kepler orbits, 105 Cyclic coordinate, 88 Kepler’s first law, 109 Cycloid, 78 Kepler’s laws, 109 Cyclotron frequency, 17 Kepler’s second law, 29, 109 Cylindrical coordinates,7, 162 Kepler’s third law, 109, 112 Kinetic energy, 36, 84 Damped harmonic oscillator, 52 Decay time, 54 Lagrange equation, 84 Decrement of motion, 54 Lagrange multiplier, 98 Degree of freedom, 87 Lagrangian, 84 Differential cross section, 155 Lagrangian mechanics, 84 Double pendulum, 147 Linear air resistance,8 Drag,8 Driven oscillators, 57 Mechanical energy, 37 Moment of inertia, 31 Eccentricity, 106 Morse potential, 51 Effective potential, 104 Eigenmode, 148 Natural frequency, 53 Energy, 36 Natural generalized coordinates, 98 Equilibrium point, 41 Newton’s laws,1 166 Index

Noether’s theorem, 95 Virial theorem, 108 Noninertial frame, 115 Normal frequency, 142 Weak Coupling, 144 Normal mode, 142, 148 Work, 36 Work-energy theorem, 36 Orbits, 102 Oscillations, 49 Overdamping, 55

Parallel axis theorem, 132 Parallel springs, 151 Parseval’s theorem, 68 Perigee, 107 Perihelion, 107 Phase angle, 18 Potential energy, 84 Principal Axes, 133 Principle of equivalence,1 Principle of least action, 84 Product of inertia, 127

Quadratic air resistance,8, 13 Quality factor, 62, 71

Reduced mass, 103 Remote control theorem, 136 Resonance, 61 Rigid body, 125 Root mean square displacement, 68 Rutherford Scattering, 158 Rutherford scattering formula, 159

Scattering, 154 Scattering angle, 154 Series springs, 151 Solid angle, 154 Spherically symmetric, 44 Stable equilibrium, 41 Stationary, 72 Steradian, 154 Symmetry, 95

Tautochrone, 79 Tennis racket theorem, 136 Terminal velocity, 15 Torque, 28 Total cross section, 155 Transfer orbit, 109 Translational invariance, 45 Triatomic molecule, 149 Two-body problems, 102

Unbounded orbits, 108 Underdamping, 54 Unstable equilibrium, 41

This handbook grew out of the extensive notes that I took as a physics undergraduate student.