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Home , Dynamics (mechanics), Rigid body, Rigid body dynamics

Classical

Dipan Kumar Ghosh UM-DAE Centre for Excellence in Basic Sciences Kalina, Mumbai 400098 November 14, 2016

1 Introduction

A rigid body is defined as a collection of particles such that the relative between any pair of particles in the body remains fixed. The of the rigid body gets fixed if we specify the positions of three non-colinear points in the body with respect to a fixed (external to the body) OXYZ. Such a -fixed coordinate system will also be called as the inertial coordinate system. Since three particles have a total of 9 degrees of freedom, the rigid body itself has because the distance between three possible pair of particles are fixed by three holonomic constraints. (Adding another particle would introduce another three degrees of freedom but it will bring in three additional constraints as well!). It is convenient to define a second set of axes which is fixed with the body, i.e. it moves or rotates with the body. We denote this set of axes by Cxyz where C is the origin of this “body-fixed” system of coordinates. Frequently (but not necessarily) C is chosen to be the centre of of the body. The six degrees of freedom of the rigid body is specified by the position vector R~ of the origin C of the body fixed system (with respect to the origin O of the inertial frame) and three independent angles that Cxyz makes with OXYZ (e.g. we can fix Cx with resoect to OX two angles - polar and azimuthal; Cy can then be chosen by one more angle; Cz gets fixed since it is perpendicular to both Cx and Cy).

1 c D. K. Ghosh, IIT Bombay 2

Z

z r’ y

C

r x R

O Y

X Let us consider the situation where Cxyz axes rotate about the OZ axis where the points O and C coincide. Our interest will be to find the relationship between dynamical quantities in the rotating with those in the fixed frame.(Note that what follows in the next section is true for a rotating frame which and is not just for the rigid body).

2 and of a particle as seen from a Rotating Frame

Let us consider what happens to a vector when it is rotated about a fixed axis by an angle δϕ with an Ω.~

r r’ r Under let the vector

~r → ~r0 = ~r + δr~

Clearly, δr~ is perpendicular to the plane containing Ω~ and ~r and its magnitude is δϕ | δr~ |= 2r sin( ) ≈ rδϕ 2 Thus, if we fix the direction of δr~ by the , we have,

δr~ = δϕ~ × ~r c D. K. Ghosh, IIT Bombay 3 where δϕ~ = δϕΩ.~ Note that while the unit vectors of the Cxyz system ˆi, ˆj, kˆ do not change in the body fixed frame but they change in the space fixed frame. For instance,

δˆi = δϕ~ × ˆi

Thus ! dˆi d~ϕ = × ˆi = Ω~ × ˆi dt dt fixed The subscript “fixed” on the l.h.s. is actually superfluous because in the rotating frame the unit vectors would not change. What we want to do is to relate the of a vector A~ in the fixed frame with the of the same quantity with respect to the rotating frame. Note that a vector is the same in both the frames though its representation may differ from frame to frame. Consider a vector A~ as seen from a rotating frame. Its derivative in the rotating frame can be wrtten as ! dA~ dA dA dA = xˆi + y ˆj + z kˆ dt dt dt dt rot The derivative, as seen from the fixed frame is the sum of this with a term which arises ˆ because the unit vectors themselves change with . Denoting ˆi, ˆj, k byx ˆi with i = 1, 2, 3, we have,

~ ! dA X dAi X dxˆi = xˆ + A dt dt i i dt fixed i i ! dA~ X = + A (Ω~ × xˆ ) dt i i rot i ! dA~ X = + Ω~ × A xˆ dt i i rot i ! dA~ = + Ω~ × A~ (1) dt rot

Note that Ω~ itself is frame independent.

2.1 and Acceleration in Rotating Frame Note that if O and C do not coincide, the position vectors in the two frames would be different ~r0 = R~ + ~r c D. K. Ghosh, IIT Bombay 4 where ~r0 is the position vector of a P from the origin of fixed frame while ~r is the position vector in the rotating frame. Thus ! dR~ d~r ~v = + f dt dt f f d~r = V~ + dt f d~r = V~ + + Ω~ × ~r dt rot ~ ~ = V + ~vr + Ω × ~r (2)

Here V~ is the velocity of C with respect to the fixed frame. A further differentiation will give relationship between the acceleration as seen from the two frames. ! ! dV~ d~v  dΩ~ d~r ~a = + r + × ~r + Ω~ × f dt dt dt dt f f f f ~ ~ ~˙ ~ ~ = A + [~ar + Ω × ~vr] + Ω × ~r + Ω × (~vr + Ω × ~r) ~ ~ ~˙ ~ ~ = A + ~ar + 2Ω × ~vr + Ω × ~r + Ω × (Ω × ~r) (3)

Thus the acceleration of a point P as observed in the rotating frame is related to acceleration of P as seen from the fixed frame by

~ ~ ~˙ ~ ~ ~ar = (~af − A) − 2Ω × ~vr − Ω × ~r − Ω × (Ω × ~r) (4)

We know that ’s laws are valid in inertial reference frame. In such a frame the cause of acceleration is a whose physical origin can always be traced to a real source.

Thus in the inertial reference frame, the acceleration of the particle at P is ~af which must arise from a real force whose value is m~af . Likewise the acceleration of C must be traced to a real force mA~. Hence it is understandable that as seen from the rotating frame the acceleration of the particle is given by the first term of (4). The remaining terms on the right hand side of the above equation obviously have no force apparently associated with them and must be caused by the nature of the rotating frame. If we multiply these terms with the mass of the particle at P, we would get quantities having the dimensions of force. Such without apparent physical origin which seem to arise in rotating (in general in a non-accelerating frame) are called pseudo forces. Existence of such forces are postulated only to make Newton’s laws valid in the non-inertial frames as well. The last term in (4) −Ω~ × (Ω~ × ~r) is the familiar . The second term ~ −2Ω × ~vr is known as the which arises if there is a lateral velocity of the particle in the rotating frame. We will drop the third term by assuming that the angular velocity is constant. c D. K. Ghosh, IIT Bombay 5

2.2 Examples: Example 1: Effective on the earth’s surface

Consider a person standing on the surface of the earth at a latitude λ. The centrifugal force is −mΩ~ ×(Ω~ ×~r). Ω~ ×~r has a magnitude Rω cos λ and is directed tangentially along the latitude circle which has a radius of R cos λ.

centrifugal mg

The centrifugal term is directed away from the rotation axis of the earth and has a magnitude mRΩ2 cos λ. The force of gravity is directed towards the centre of the earth. Because of the centrifugal correction, the effective gravity is slightly towards south in the northern hemisphere. By taking the component of the centrifugal acceleration along the true gravitational force direction, it can be seen that the acceleration due to gravity gets reduced by an amount mω2R cos2 λ. At the equator the centrifugal term is directly opposite to the direction of m~g and it reduces the acceleration due to gravity by about 0.3%. Note that at any other latitude there is a horizontal component of the centrifugal force with the magnitude mω2R cos λ sin λ. As a result of this a plumb- at these places will not point towards true vertical but will deviate from it towards south in the northern hemisphere and towards north in the southern hemisphere. Example 2:

Fc

v

A person attempts to walk on the surface of a rotating turn table moving from its edge towards the centre. The turn table is rotating anticlockwise with an angular ~ ω when viewed from above. He will experience a Coriolis force Fc = −2m~ω × ~v, directed towards his right. In order to be able to walk straight, he will have to counter this with c D. K. Ghosh, IIT Bombay 6 a frictional force of equal magnitude to his left. How does this force arise? dL~ The acting on the person is . where L~ is the angular of the person dt having a magnitude mr2ω, ω being the angular velocity of the person as viewed from the laboratory frame which is equal to the angular speed of the turn table. Since the person d~r is moving towards the centre, = −~v. Thus dt dL dω = −2mrωv + mr2 dt dt = −2mrωv where we have assumed ω to be constant. Thus the torque acting on the person is r (−2mωv), the last expression being the tangential that is applied by the turn table on the man. In case the man chooses not to apply frictional force, in the rotating frame he will experience a tangential acceleration 2ωv because of the Coriolis force. In the lab. frame too, this acceleration will be there since the two frames are connected by angular velocity of the turn table which is in the vertical direction. The reason for the tangential acceleration in the lab frame is that as there is no horizontal force acting on the man, the is conserved. In order to keep the angular momentum conserved as the man tries to decrease his distance from the centre of the turn table, he will need to have a tangential acceleration which increases his tangential speed. Example 3: Eastward deflection of a falling body: An effect of the Coriolis force is that all bodies which are dropped from a height are deflected eastward (in both hemispheres). To understand this consider a body dropped from certain height at a latitude λ. We take the outward (true) vertical at the place to be the direction of the z-axis of the rotating coordinate system. The di- rection towards north is taken as the x axis and the westward direction is the y-axis. x z ~ω = ω cos λxˆ + ω sin λzˆ Since the deflection is small, we take the ve- locity ~v ≈ −gtzˆ. We then have,

−2~ω × ~v = −2ωgt cos λyˆ

Thus the Coriolis acceleration is towards east and is time dependent. It is straight- forward to integrate it and obtain eastward deflection d. The eastward speed varies with c D. K. Ghosh, IIT Bombay 7 time as follows: Z t v(t) = a(t)dt 0 Z t = 2ωg cos λ tdt = ωg cos λt2 0 The eastward deflection is given by √ Z t Z 2h/g d = v(t)dt = ωg cos λ t2dt 0 0 (2h/g)3/2 = ωg cos λ 3 The deflection calculated by this formula is small but is measurable being about 15 mm when dropped from a height of 100m. Example 4: Foucault’s In 1851, a French physicist, Foucault, demonstrated a very interesting result which gives a direct evidence of the earth’s rotation. The device known today as “Foucault’s Pendulum”, consisted of a pendulum suspended freely from the great dome of Pantheon in Paris by a wire nearly 70m long. The pendulum is free to move in any direction. A needle like projection from the metal bob of the pendulum touches a bed of sand kept on the ground so that the path of the bob is traced on the sand as the pendulum swings. It was found that the plane of of motion of the pendulum was rotating in a clockwise direction when viewed from above, returning back to the original plane in about 32 hours. The following diagram sketches the track on the sand. N

W E

S Let us look at some simple way of understanding this. Let us first consider the situation on the North pole. At this location, the gravity acts along the true vertical direction and the only other true force is the tension in the string. With respect to an inertial observer (outside the rotating earth), the motion of the pendulum takes place in the plane defined by these two forces while the earth continues to rotate below this plane c D. K. Ghosh, IIT Bombay 8 in a counterclockwise fashion. For a person on the earth, as he himself is rotating round the earth’s axis once in 24 hours, the plane would appear to rotate in a clockwise fashion completing the orbit in 24 hours. Let us look at what happens at a latitude λ. The coordinates are as was shown in the previous problem. We had shown that

~ω = ω cos λxˆ + ω sin λzˆ

In this case however, the velocity of the bob ~v is in the x-y plane. Resolving this velocity into its x and y component, we notice that so that we are primarily interested in the z component of the Coriolis force which will be responsible for a force acting perpendicular to the direction of the velocity. The magnitude of this force will be

F =| −2mω sin λzˆ × ~v |= (2mω sin λ) | v |

The situation seems to be the same as happens on the North pole with the angular velocity ω having been replaced by ω sin λ. This is responsible for the time period of the being increased by a factor 1/ sin λ.

3 of Tensor

Let us return to the discussion of rigid body. Once again we look at the body fixed and space fixed frames shown below (we have used a slightly different notation here in that the vectors with respect to the body fixed frames are written without the primes)

Z

P z r y

C r’

x R

O Y

X

We had seen that the velocity of a particle in the fixed frame ~vf is related to the velocity as observed from the body fixed frame is given by ~ ~ ~vf = V + Ω × ~r (5) c D. K. Ghosh, IIT Bombay 9

where we have used (2) with ~vr = 0 since the moving body in question is a rigid body with respect to which the particles do not have any . Here Ω~ is the angular velocity of the body about the instantaneous axis of rotation passing through the origin C of the body-fixed system, which itself moves with a velocity V~ with respect to the origin O of the space fixed system. Let us obtain an expression for the kinetic of the body with respect to the inertial frame.

Using (5) we have, if ~vi represents the velocity of the i−th particle with respect to the inertial frame, X 1 X 1 T = m v2 = m [V~ + Ω~ × ~r ]2 f 2 i i 2 i i i ~ where ~ri is the position of the i−th particle in the body-fixed frame. note that V = (Vx,Vy,Vz) which is the velocity of C with respect to O does not depend on the index i. ~ Further Ω = (Ωx, Ωy, Ωz) is also independent of i. Using (Ω~ × ~r)2 = Ω2r2 − (Ω~ · ~r)2, we get 1 1 X X T = MV 2 + m (Ω~ × ~r )2 + m V~ · (Ω~ × ~r ) f 2 2 i i i i i i 1 1 X X = MV 2 + m [Ω2r2 − (Ω~ · ~r )2] + m Ω~ · (~r × V~ ) (6) 2 2 i i i i i i i P where M = i mi. Note that ~ 2 2 (Ωi · ~ri) = (Ωxxi + Ωyyi + Ωzzi) 2 2 2 2 2 = Ωxxi + Ωyyi + Ωzzi + 2ΩxΩyxiyi + 2ΩyΩzyizi + 2ΩzΩxzixi 2 2 2 2 Further ri = xi + yi + zi . Substituting these in (6), we get 1 1 X T = MV 2 + m [Ω2(r2 − x2) + Ω2(r2 − y2) + Ω2(r2 − z2)] f 2 2 i x i i y i i z i i i X X ~ ~ − mi[ΩxΩyxiyi + ΩyΩzyizi + ΩzΩxzixi] + miΩ · (~ri × V ) (7) i i ~ P ~ Now the last term of (7) can be written as Ω · (( i mi~ri) × V ). If we make C coincide P with the centre of mass of the body, the term i mi~ri = 0, being M times the location of the position of the centre of mass with resoect to the centre of mass itself. The last term of (7) is therefore dropped. We define moments of inertia by X 2 2 X 2 2 Ixx = mi(ri − xi ) = mi(yi + zi ) i i X 2 2 X 2 2 Iyy = mi(ri − yi ) = mi(zi + xi ) i i X 2 2 X 2 2 Izz = mi(ri − zi ) = mi(xi + yi ) i i c D. K. Ghosh, IIT Bombay 10 and the products of inertia by X Ixy = − mixiyi = Iyx i X Iyz = − miyizi = Izy i X Izx = − mizixi = Ixz i More compactly, we call these as components of a second rank Intertia Tensor whose elements are given by

X 2 Iαβ = mi(ri δαβ − riαriβ) (8a) i Z d 2  = d rρ(~r) r δα,β − rαrβ (for a continuous body) (8b)

In the last formula ddr is a differential element in d dimensions. so that the matrix representation can be written as   Ixx Ixy Ixz I = Iyx Iyy Iyz Izx Izy Izz The representing a given coordinate direction is given by cos α nˆ = cos β cos γ with cos2 α + cos2 β + cos2 γ = 1. The about a given directionn ˆ can be written as T ←→ In = n I n where the double headed arrow over I has been used to represent a tensor, which is briefly introduced below.

A tensor is a generalization of a vector. Just as a vector consists of 3 quantities, which under rotation of coordinate system transform in a way that each new component is expressible as a linear combination of the three old components, under rotation, the components of a tensor get expressed as linear combination of 9 components. A tensor of n th rank has 3n components. A symmetric second rank tensor has only 6 indepen- dent components. A cross product of two vectors can be regarded as giving rise to an antisymmetric tensor of rank 2. One can form a vector by multiplying a tensor with a vector X bα = Tαβaβ β c D. K. Ghosh, IIT Bombay 11

By pre-multiplying with a row vector, we can get a scalar. Thus atT b is a scalar where a and b are column vectors. Since T is represented by a 3 × 3 matrix, the eigen vectors of T are expressible as a 3 × 1 column vector. Thus one can diagonalize T in a which defines three orthogonal spatial directions. Doing so would make the to be given by 1 X T = I Ω2 (9) rot 2 i i where Ii are called the Principal Moments of Inertia and the basis in which it is diagonalized are special directions within the rigid body known as the Principal Axes. The principal axes are a set of three directions in which the momentum of inertia tensor is diagonal. In a coordinate system whose axes are aligned with the principal axes, quantities such as the angular momentum, the kinetic energy etc. take particularly simple form. Letn ˆ be a unit vector about which the rigid body rotates. Let the direction cosines ofn ˆ relative to the principal axes be (cos α, cos β, cos γ). The moment of inertia of the rigid body about this axis is

T 2 2 2 I = n In = I1 cos α + I2 cos β + I3 cos γ where we have used the fact that the moment of inertia tensor is diagonal in the principal axes.We know that the angular velocity vector ~ω points in the same direction as the axes of rotation. Thus     I1 0 0 ω1 ~ ←→ L = I · ~ω =  0 I2 0  ω2 0 0 I3 ω3   I1ω1 = I2ω2 I3ω3

Ife ˆ1, eˆ2, eˆ3 be unit vectors along the principal axes, then ~ X L = eiIiωi (10) i Likewise, the kinetic energy of rotation about the axis can be written as 1 ←→ 1 T = ωT I ω = I ω2 + I ω2 + I ω2 (11) rot 2 2 1 1 2 2 3 3 c D. K. Ghosh, IIT Bombay 12

3.1 Angular Momentum of a Rigid Body The angular momentum of the rigid body about the inertial axis can be expressed as

~ X 0 0 Lf = mi~ri × ~vi i X ~ ~ = mi(R + ~ri) × (V + Ω × ~ri) i X ~ ~ ~ ~ X X ~ X = miR × V + R × (Ω × miri) + mi~ri × V + mi~ri × (Ω × ~ri) i i In the above, the primed quantities are with respect to the inertial frame and the unprimed quantities are with respect to the body-fixed frame. If we choose C to be the centre of P m ~r mass of the body, the second and the third terms drop out because i i i is the location M of the centre of mass of the body with respect to the centre of mass itself. We are then left with ~ X ~ ~ X Lf = miR × V + Mi~ri × (Ω × ~ri) i   ~ ~ X 2~ ~ = MRcm × V + mi ri Ω − ~ri(~ri · Ω) i ~ ~ ←→ ~ = MRcm × V + I · Ω ←→ The last relationship may be obtained by realizing that I being a tensor of rank two, P IαβΩβ is α component of a vector. Ignoring the first term which represents the angular momentum of the centre of mass about the fixed origin, we have, suppressing the suffix f for the fixed frame, " # X 2 X Lα = mi ri Ωα − ri,α ri,βΩβ i β X X 2  = mi ri δα,β − riαriβ Ωβ i β

The quantity in the parentheses is Iα,β, as per the definition (8a). The first term on the right is clearly the angular momentum of the centre of mass about the fixed origin while the second term is the angular momentum of the body about an axis passing through C. In matrix form     Ixx Ixy Ixz Ωx ←→ ~ I · Ω = Iyx Iyy Iyz Ωy Izx Izy Izz Ωz

This also shows that L~ is not necessarily parallel to the Ω axis. Example : A dumbbell c D. K. Ghosh, IIT Bombay 13

A dumbbell is rotated about an axis through C making an angle θ with the length of the dumbbell. Taking the axis and C on the page, the velocity of the two are ~v1 and ~v2 going in and coming out of the page. The magnitude of the velocities are ~ | ~v1 |=| Ω × ~r1 |= r1Ω sin θ

Since ~v1 is perpendicular to the page while ~r1 is on the page, ~r1 × ~v1 is on the page, perpendicular to the length. L~ is therefore as shown in the figure. Ω v1 m 1

θ L

C

m 2 v 2

It is easy to see that the contribution to the angular momentum due to both the masses are parallel and hence the magnitudes add up.

L = L1 + L2 2 2 = (m1r1 + m2r2)Ω sin θ

m2l m1 Taking the length of the dumbbell to be l, r1 = and r2 = , so that m1 + m2 m1 + m2 m m L = 1 2 l2 sin2 θΩ m1 + m2 4 Euler Equations

We have seen in equation (1) that for any arbitrary vector A~, ! ! dA~ dA~ = + (~ω × A~) dt dt f rot Note that the first term on the right is calculated in the body-fixed system and ω is the angular velocity of the rigid body about an axis in the body fixed system. Consider a c D. K. Ghosh, IIT Bombay 14 rigid body which is not subject to any external torque, we then have for the angular momentum vector L~ ! ! dL~ dL~ = + ~ω × L~ = 0 (12) dt dt f rot If there are external acting on the system, the right hand side of the above equation would be equal to the applied torque. In terms of the principal moments of inertia (ignoring the translational motion and taking C and O to coincide),

~ ←→ X L = I · ~ω = Iiωi i Equation (12) can therefore be written as follows. ! dL~ = −~ω × L~ dt rot Taking the components, we get

~ I1ω˙ 1 = −(~ω × L)1 = ω3L2 − ω2L3

= ω3I2ω2 − ω2I3ω3

= ω2ω3(I2 − I3)

The set of equations (for free particles) is then given by

I1ω˙ 1 = ω2ω3(I2 − I3) (13a)

I2ω˙ 2 = ω3ω1(I3 − I1) (13b)

I3ω˙ 3 = ω1ω2(I1 − I2) (13c)

The equations (13) are known as the Euler Equations for a free rigid body. If there are external torques, we would need to add to the right hand side of each of the equations above with the respective component of the torque acting on the body.

Consider a symmetric for which I1 = I2 = I and I3 = Is(6= I). In this case we have, Isω˙ 3 = 0 so that ω3 is constant. However, ω1 and ω2 are not constants and can be calculated from the Euler equations. Define a constant having the dimension of angular velocity I − I Ω = ω s (14) 3 I we have, from the first two of Euler equations (13),

(I2 − I3) ω˙ 1 = ω2ω3 = −Ωω2 (15) I3 and

ω˙ 2 = Ωω1 (16) c D. K. Ghosh, IIT Bombay 15

Equations (15) and (16) have the solution

ω1 = ω0 cos(Ωt)

ω2 = ω0 sin(Ωt) where ω0 is a constant. Since ω3 is constant, the total angular velocity ω = p 2 2 2 p 2 3 ω1 + ω2 + ω3 = ω0 + ω3 is constant. This shows that in the inertial frame, the ~ angular velocity vector ω describes a circle of radius ω0 about L with an angular velocity Ω. This is known as the angular velocity of .

5

We have seen the need to define a set of three body fixed axes (Cxyz) while describing the dynamics of rigid bodies. How does one define body fixed axes with respect to an inertial set (OXYZ)? A rigid body has 6 degrees of freedom out of which three are taken up in specifying the coordinates of the origin C of the body fixed system, leaving us with three to specify the of the three axes. One could, for instance, specify each of the three orthogonal body fixed axes by a set of three Miller indices. These 9 quantities are obviously not independent and one can derive 6 constraints which reduce the number from 9 to 3. In d dimensions, an R ∈ O has d(d − 1) independent parameters. This follows from the fact that the matrix is symmetric, since RtR = 1. A 1 d dimensional symmetric matrix has d(d + 1) independent entries ( along the diagonal 2 and along the upper triangle of the matrix). The orthogonality condition thus provides these many constraints for d2 elements of the matrix, giving us d(d − 1)/2 independent degrees of freedom. Thus in three dimensions, we have only three parameters. There is one additional requirement that out of all orthogonal matrices, we need only those which have a determinant equal to +1. This is because an overall inversion of coordinates is inconsistent with the orientation that a rigid body can assume. Thus 3 × 3 orthogonal matrix having only diagonal elements with each entry entry equal to −1 is not a permit- ted transformation. In d = 3, the permitted are specified by three parameters. Euler Angles provide one such convenient parameterisation. (The way one defines the sequence of operation in Euler angles is not standard and one can do it in more than one ways, but we will follow a more commonly used practice as used by Goldstein.)

1. Start with the Cxyz axes coinciding with the inertial frame S0, OXYZ. Rotate anticlockwise by an angle ϕ about the Z (ζ) axis. The new set of coordinates will

be denoted by Cξηζ. The new frame is shown as S1 in the figure. The old Z axis is same as the new ζ axis. The angle ϕ is called the angle of precession. Denoting the x position vector of a point P in the frame S0 as ~x = y, and that in the frame S1 z c D. K. Ghosh, IIT Bombay 16

ξ as ξ~ = η, we have ζ ξ~ = B~x where the B is given by

 cos ϕ sin ϕ 0 B = − sin ϕ cos ϕ 0 0 0 1

The old X-axis has now rotated to what we call as the new ξ axis. This line is denoted as the line of nodes.

2. The frame S1 with (ξ, η, ζ) as the axes, is rotated about the line of nodes (ξ axis )by 0 0 0 an angle θ, generating a new frame S2 with coordinate axes (ξ , η , ζ ). The old ξ axes is the same as the new ξ0 axis. θ is called the angle of . If the position ξ0 0 0 vector of P in frame S2 is denoted by ξ = η  then we have the corresponding ζ0 rotation matrix C such that Cξ~ = ξ~0, where

1 0 0  C = 0 cos θ sin θ 0 − sin θ cos θ

This rotation brings ζ0 to its final orientation ζ00. Clearly the line of nodes is per- pendicular to the plane determined by ζ and ζ0 = ζ00. The specification of the final ζ00 with respect to the original XYZ axes thus required specification of two angles ϕ and θ. A moment’s reflection convinces us that the angle θ corresponds to the familiar polar angle and ϕ to the azimuthal angle of the spherical polar coordinates. Clearly, these two are independent. A third rotation independent of these two would now complete our description. c D. K. Ghosh, IIT Bombay 17

0 0 0 0 3. The frame S2 with (ξ , η , ζ ) axes is rotated about the ζ axis through an angle ψ to 00 00 00 get the frame S3 with the coordinate axes (ξ , η , ζ ). S3 is the body fixed frame. The angle ψ is the angle of spin. If the position vector of P in frame S3 is denoted by ξ00 ξ00 = η00 then we have the corresponding rotation matrix D such that Dξ~0 = ξ~00, ζ00 where  cos ψ sin ψ 0 D = − sin ψ cos ψ 0 0 0 1

The three angles ϕ, θ and ψ are known as the Euler Angles. If a coordinate (ξ00, η00, ζ00) is obtained from (X,Y,Z) by an arbitrary rotation about an arbitrary axis passing through the origin, one can find a unique set of (ϕ, θ, ψ) such that the coordinate of an a arbitrary point in both frames are related by this transformation. It may be mentioned that the three rotations defined in terms of the Euler angles are independent because rotation about any of the axes cannot be obtained by combination of rotations about the other two and each of the rotations can be varied independently of the other two. These triad of Euler angles, therefore, is a very appropriate set of

5.1 Angular Velocity Unlike the translational velocity, the angular velocity, being a cross product, does not, in general, add up vectorially in a simple way. However, if different frames have a common origin, we can get the following relationship. Suppose a frame S2 has an angular velocity ω2 with respect to a frame S1, which, in turn, has an angular velocity ω1 with respect to a third frame S0. What is the angular velocity of the frame S2 with respect to S0? Let P be a fixed point in S2. Clearly, the velocity ~v2 of P in frame S2 is zero. The velocity of P in S1 is given by ~ ~ ~v1 = ~v2 + ~ω2 × OP = ~ω2 × OP

The velocity of P in S0 is given by ~ ~v0 = ~v1 + ~ω1 × OP ~ = (~ω1 + ~ω2) × OP = ~ω × OP~

Thus

~ω = ~ω1 + ~ω2 Using this, we can find the angular velocity ~ω of the rigid body. Clearly, the angular velocity of the body consists of turning about the Z-axis (or ζ axis) with a rateϕ ˙, super- imposed on the rotation of θ˙ about the ξ axis with a rate θ˙ followed by a rotation at the c D. K. Ghosh, IIT Bombay 18 rate ψ˙ about the ζ axis. Since all these rotations occur about about a common origin, we can add up the angular velocity vectors

~ω = ~ωϕ + ~ωθ + ~ωψ

Since we are expressing this in the body fixed frame, we need to express these in terms of the unit vectors of the body fixed frame. The relationships are easily obtained following our earlier discussion of the Euler angles. A general rotation with angular velocity ~ω can be considered as three successive and in- dependent rotations, first with an angular velocityϕ ˙ about the Z = ζ axis, followed by an angular velocity of θ˙ about the ξ0 = ξ00 axis and finally an angular velocity of ψ˙ about the ζ0 = ζ00 axis. We can use the matrices that we defined to express these in terms of the unit vectors of the body fixed frame S3.

0 ˙ 1. A velocity of ψ in frame S2 has components 0. To express it in terms of ψ˙ components of S3 we use the rotation matrix D to get

0  cos ψ sin ψ 0 0 0 D 0 = − sin ψ cos ψ 0 0 = 0 ψ˙ 0 0 1 ψ˙ ψ˙

˙ 2. The angular velocity θ is in frame S1. To express it in terms of the unit vectors of the body fixed frame, we first apply rotation given by matrix C followed by a rotation D, i.e.

θ˙  cos ψ sin ψ 0 1 0 0  θ˙ DC 0 = − sin ψ cos ψ 0 0 cos θ sin θ 0 0 0 0 1 0 − sin θ cos θ 0  cos ψ sin ψ cos θ sin ψ sin θ θ˙ = − sin ψ cos ψ cos θ cos ψ sin θ 0 0 − sin θ cos θ 0  cos ψθ˙  = − sin ψθ˙ (17) 0

3. Finally consider the angular velocityϕ ˙ given in frame S0 about the ζ axis. This can be expressed in terms of components in the body fixed frame by application of the c D. K. Ghosh, IIT Bombay 19

rotation matrix DCB. Using the expression for the matrix DC obtained in (17), and using the matrix B given earlier, we have the components given by

0  cos ψ sin ψ cos θ sin ψ sin θ  cos ϕ sin ϕ 0 0 DCB 0 = − sin ψ cos ψ cos θ cos ψ sin θ − sin ϕ cos ϕ 0 0 ϕ˙ 0 − sin θ cos θ 0 0 1 ϕ˙  cos ψ cos ϕ − sin ψ cos θ sin ϕ cos ψ sin ϕ + sin ψ cos θ cos ϕ sin ψ sin θ 0 = − sin ψ cos ϕ − cos ψ cos θ sin ϕ − sin ψ sin ϕ + cos ψ cos θ cos ϕ cos ψ sin θ 0 sin θ sin ϕ − sin θ cos ϕ cos θ ϕ˙ sin ψ sin θϕ˙  = cos ψ sin θϕ˙ (18) cos θϕ˙

The angular velocity can be written as

~ω =ϕ ˙[sin θ sin ψξˆ00 + sin θ cos ψηˆ00 + cos θζˆ00] + θ˙[cos ψξˆ00 − sin ψηˆ00] + ψζ˙ 00

Thus the components of the angular velocity vector in the body fixed frame (ξ00, η00, ζ00) are given by

˙ ω1 =ϕ ˙ sin θ sin ψ + θ cos ψ (19a) ˙ ω2 =ϕ ˙ sin θ cos ψ − θ sin ψ (19b) ˙ ω3 =ϕ ˙ cos θ + ψ (19c)

It may be noted that these are the components of ~ω in terms of the unit vectors of the body fixed system.

[In terms of the unit vectors ˆi, ˆj, kˆ of the inertial frame, the components of the angular velocity are

˙ ˙ ωx = cos ϕθ + sin ϕ sin θψ (20a) ˙ ˙ ωy = sin ϕθ − sin˙ θ cos ϕψ (20b) ˙ ωz =ϕ ˙ + cos θψ (20c)

These relations are easily seen by realizing that the first rotation ofϕ ˙ is about the z-axis of the fixed frame. The final ζ axis makes an angle θ with the initial z axis. Hence the spin rotation of ψ about the ζ axis has a component along the original z-axis as well. Similarly, the θ rotation is about a rotated x-axis and so on.] Henceforth we drop the redundant primes on the basis (00) and talk of the basis as (ξ, η, ζ). c D. K. Ghosh, IIT Bombay 20

5.2 Angular Momentum We will now obtain expressions for the components of the angular momentum vector in the body fixed axes in terms of the Euler angles. This is useful because the torque free motion of a rigid body is best analyzed using the principle of conservation of the angular momentum.For the body fixed principal axes (denoted by 1,2,3) the angular momentum L~ about the centre of mass as origin is given by

Lc = I1ω1 + I2ω2 + I3ω3 (21) where ω1, ω2 and ω3 are given by (19) above. Since the angular velocity is constant, let us choose its direction to coincide with the direction of the z axis of the space fixed coordinates (which is the same as the ζ axis of the stage 1 of Euler operation). Recall that at the first stage, we rotate the fixed Cartesian axes by an angle ϕ about the z(ζ) axis. Thus the angular momentum vector continues to be aligned with the ζ axis after this rotation as well and its components are (0, 0,L) in (ξ, η, ζ) axes. In the next stage we rotate the axes by an angle θ about the new ξ axis. This will take (η, ζ) by the usual orthogonal transformation

η0 = η cos θ + ζ sin θ ζ0 = −η sin θ + ζ cos θ ξ0 = ξ

In the present case, we have

ξ0 = 0 η0 = L sin θ ζ0 = L cos θ

In the final stage, we rotate by an angle ψ about the ζ0 axis, which would give

ξ00 = ξ0 cos ψ + η0 sin ψ = L sin θ sin ψ η00 = −ξ0 sin ψ + η0 cos ψ = L sin θ cos ψ ζ00 = ζ0 = L cos θ

Using (21), we then have

˙ L sin θ sin ψ = I1ω1 = I1(ϕ ˙ sin θ sin ψ + θ cos ψ) (22a) ˙ L sin θ cos ψ = I2ω2 = I2(ϕ ˙ sin θ cos ψ − θ sin ψ) (22b) ˙ L cos θ = I3ω3 = I3(ϕ ˙ cos θ + ψ) (22c) c D. K. Ghosh, IIT Bombay 21

We can obtain expressions for φ˙ and θ˙ using equations (22a) and (22b) and obtain ψ˙ by substituting these in (22c)

cos2 ψ sin2 ψ  ϕ˙ = L + (23a) I2 I1 L  1 1  θ˙ = − sin θ sin(2ψ) (23b) 2 I1 I2  1 cos2 ψ sin2 ψ  ψ˙ = L − − cos θ (23c) I3 I2 I1 6 Dynamics of a Symmetric Top

Let us consider the torque free motion of a symmetric top, which has an axial about the ζ axis. (This can be obtained by throwing such an object into the air. Gravity would make the centre of mass move in a parabolic trajectory though no torque is exerted on the body.)

Let us use, as before, I1 = I2 = I and I3 = Is. In the following, as in the previous discussion, we have assumed that the fixed point O and the centre of mass coincides. In case they do not, the equations need to be modified by replacing I by I0 = I +Ml2, where l is the distance between the points C and O. Is Z ζ

C

θ Y O I

I’

X

The angular velocities, given in eqns. (23) are then as follows: c D. K. Ghosh, IIT Bombay 22

L ϕ˙ = (24a) I θ˙ = 0 (24b)  1 1 ψ˙ = L − cos θ (24c) Is I Using the above, we have ˙ Is(ϕ ˙ cos θ + ψ) = L cos θ Since θ˙ = 0, the angle between the axis of the top and the direction of L remains constant. The rate of precession is φ˙ = L/I and the angular velocity with which the top rotates about its own axis, i.e. ω3 = (L/Is) cos θ The Lagrangian consists only of the kinetic energy, which can be simplified using equations (19) 1 L = T = I(ω2 + ω2) + I ω2 2 1 2 s 3 1   = I(sin2 θϕ˙ 2 + θ˙2) + I (cos θϕ˙ + ψ˙)2 (25) 2 s where we have used I1 = I2 = I and I3 = Is 6= I. The Lagrangian is cyclic in ϕ and ψ, so that the corresponding canonical momenta are conserved, ∂L p = = I sin2 θϕ˙ + I cos θ(cos θϕ˙ + ψ˙) = constant (26) ϕ ∂ϕ˙ s ∂L ˙ pψ = = Is(cos θϕ˙ + ψ) = constant (27) ∂ψ˙

Equation (19c) shows that the quantity in the parentheses of (27) is ω3, giving

pψ = Isω3 = constant (28)

We had already seen this result when we discussed the problem of symmetric top using the Euler equations. This leaves us with only one equation that the canonical momentum corresponding to θ must satisfy. The Euler Lagrange equation for the coordinate θ is given by

d ∂L ∂L = dt ∂θ˙ ∂θ ¨ 2 ˙ Iθ = I sin θ cos θϕ˙ + Is sin θ(cos θϕ˙ + ψ)

=ϕ ˙ sin θ(I cos θϕ˙ − Isω3) (29)

It may be noted that the canonical momenta are actually the components of angular momentum along the axes about which the Euler rotations were made, viz. z = ζ, ξ0 c D. K. Ghosh, IIT Bombay 23

0 and ζ . The quantities ω1, ω3 and ω3 given in (19a)-(19c) are the components of angular velocity along the body fixed axes (ξ00, η00, ζ00). To see this more clearly, we need to obtain expressions for projections of the angular momentum along the three axes of rotation. For this we simply need to reverse the steps we took in arriving at the body fixed axes starting from the space fixed axes through the rotation matrices B,C and D. 0 00 Note that the last rotation was about ζ which is identical to the body fixed ζ . Thus ωψ is the same as ω3, which can be verified by comparing (19c) and (27). Thus

L3 = Lψ = Isω3

Next consider the axis of rotation which coincides with the line of nodes, i.e. about the axis ξ. We had shown that

B~x = ξ~ Cξ~ = ξ~0 Dξ~0 = ξ~00

ξ where the vector ξ~ stands for η. Since the matrices are orthogonal, we can easily ζ determine their inverse. We then have,

(DC)−1ξ~00 = (DC)T ξ~00

We had already computed the matrix DC. Taking its transpose, we get

 cos ψ − sin ψ 0  ξ00 ξ sin ψ cos θ cos ψ cos θ − sin θ η00 = η sin ψ sin θ cos ψ sin θ cos θ ζ00 ζ

As our interest is only along the component along ξ, we have

cos ψξ00 − sin ψη00 = ξ

[This can be also seen geometrically as the axis ξ lies in the plane of ξ00andη00.] Hence we have

ωθ = cos ψω1 − sin ψω2 = cos ψ(ϕ ˙ sin ψ sin θ + θ˙ cos ψ) − sin ψ(ϕ ˙ cos ψ sin θ − θ˙ sin ψ) = θ˙ so that ˙ Lθ = Iθ T ~00 ~ Finally, for ωϕ, we have (DCB) ξ = ξ. As before, taking the conjugate, we have c D. K. Ghosh, IIT Bombay 24

cos ψ cos ϕ − sin ψ cos θ sin ϕ − sin ψ cos ϕ − cos ψ cos θ sin ϕ sin θ sin ϕ  ξ00 ξ cos ψ sin ϕ + sin ψ cos θ cos ϕ − sin ψ sin ϕ + cos ψ cos θ cos ϕ − sin θ cos ϕ η00 = η sin ψ sin θ cos ψ sin θ cos θ ζ00 ζ As we are here interested only in the last component (since the rotation was about ζ(= z) axis, we have

ωϕ = sin ψ sin θω1 + cos ψ sin θω2 + cos θω3

Substituting the expressions for ω1, ω2 and ω3 from (19a)-(19c), we get 2 ˙ ωϕ = sin θϕ˙ + cos θ(ϕ ˙ cos θ + ψ) so that the angular momentum component is given by 2 ˙ Lϕ = I sin θϕ˙ + Is cos θ(ϕ ˙ cos θ + ψ) which is identical to the expression (26).

7 Motion of a symmetric top including gravity

Z (ζ) ζ ’(ζ" ) η’(η) η"

θ

R Mg Y

ψ φ ξ"

ξ’( ) X ξ The Lagrangian is given by 1 1 L = [ϕ ˙ 2 sin2 theta + θ˙2] + I (ϕ ˙ cos θ + ψ˙)2 − Mgl cos θ (30) 2 2 s

Once again pϕ and pψ are conserved since the Lagrangian is cyclic in the corresponding coordinates. 2 ˙ pϕ = I sin θϕ˙ + Is(ϕ ˙ cos θ + ψ) cos θ (31) ˙ pψ = Is(ϕ ˙ cos θ + ψ) (32) c D. K. Ghosh, IIT Bombay 25

Using these, we have

pϕ − pψ cos θ ϕ˙ = 2 (33) Is sin θ ˙ pψ − pϕ cos θ ψ = 2 (34) Is sin θ As the Lagrangian has no explicit dependence on time, the Hamiltonian is conserved and is equal to total energy. We have, using a bit of algebra 2 p2 1 ˙2 1 (pϕ − pψ cos θ) 1 ψ E = Iθ + 2 + + Mgl cos θ (35) 2 2 Is sin θ 2 Is p2 Let us define E0 = E − Mgl − ψ and 2Is

2 (pϕ − pψ cos θ) Ueff (θ) = 2 − Mgl(1 − cos θ) (36) 2Is sin θ so that equation (35) becomes 1 E0 = Iθ˙2 + U (θ) 2 eff 0 Physical motion is possible only for E > Ueff (θ). We may rewrite equation (35) by defining z = cos θ, so thatz ˙ = sin θθ˙,

2 0 2 2 2 Isz˙ = 2IsE (1 − z ) − (pϕ − pψz) + 2MIsgl(1 − z)(1 − z ) ≡ f(z) where 0 2 2 2 f(z) = 2IsE (1 − z ) − (pϕ − pψz) + 2MIsgl(1 − z)(1 − z ) (37) We will look at the behaviour of the function f(z) as a function of z, while remembering that the physical region corresponds to | z |≤ 1. The following observations may be made

1. If pϕ = pψ, then (1 − z) is a factor of f(z) so that z = 1, i.e. cos θ = 0 is a solution. Thus the top remains vertical at some stage of its motion.

2. If the top is set into motion on a flat ground, θ cannot exceed π/2, so that the actual range for physical motion is 0 ≤ θπ/2, i.e. 0 ≤ z ≤ 1.

3. The cubic equation f(z) = 0 either has one real root or three real roots.

3 4. For large z, the dominant term is 2IsMglz . Thus f(z) is positive for large positive z and negative for large negative z.

2 5. At z = ±1, f(z) = −(pϕ −pψz) is always negative, except for the case where z = ±1 are the roots of f(z) = 0 (which corresponds to the vertical top, as mentioned above). Thus one of the roots must lie in the region z > +1. c D. K. Ghosh, IIT Bombay 26

6. Since in the physical region f(z) ≥ 0, f(z) must be a positive function with a

maximum between ±1 and zeros at z1 and z2

The function f(z) must then have shape like the following if physical motion is considered. As there are two roots in the physical region, there are two turning points, i.e. the natuation angle lies in the limit θ1 ≤ θ ≤ θ2.

f(z)

z z z=+1 z=−1 1 2 z

Thus, excepting for the case of vertical motion for which the nutation angle remains zero, the following motion is realized.

1. Steady motion while precessing : If the precession velocity given by (33)ϕ ˙ = pϕ − pψ cos θ 2 does not change sign, the axis precesses monotonically while oscil- Iz sin θ lating up and down. The trajectory is depicted in the figure below.

2. Ifϕ ˙ changes sign, say ifϕ ˙ > 0 at θ = θ2 and is negative for θ = θ1, the motion would be depicted as a loop-the loop motion between the two limits of nutation angles. c D. K. Ghosh, IIT Bombay 27

3. Finally ifϕ ˙ = 0 at θ2, the motion is shown as follows