EP 222: Classical Mechanics Tutorial Sheet 1

This tutorial sheet contains problems on the Newton’s laws of motion and Lagrangian formalism.

1. Show that for a single particle with a constant mass the equation of motion implies the following diﬀerential equation for the kinetic energy: dT = F · v, dt while if the mass varies with time the corresponding equation is

d(mT ) = F · p. dt

1 2 1 Soln: (a) We know T = 2 mv = 2 mv · v. Thus, if m is constant dT dv = m · v = ma · v = F · v dt dt

(b) For a variable mass particle, let us consider

d(mT ) d 1 dm dm = m2v · v = m v · v + m2v · a = v + ma · (mv) = F · p, dt dt 2 dt dt

dp dm because for a variable mass particle dt = dt v + ma. 2. Prove that the magnitude R of the position vector for the center of mass from an arbitrary origin is given by the equation

X 1 X M 2R2 = M m r2 − m m r2 i i 2 i j ij i i,j

Soln: R is deﬁned as P m r R = i i i M X =⇒ MR = miri i 2 2 X X =⇒ M R = ( miri) · ( mjrj) i j 2 2 X X 2 2 X =⇒ M R = mimjri · rj = mi ri + mimjri · rj i,j i i6=j

1 Using

2 2 2 2 rij = (ri − rj) = ri + rj − 2ri · rj 1 =⇒ r · r = (r2 + r2 − r2 ), i j 2 i j ij we obtain X 1 X M 2R2 = m2r2 + m m (r2 + r2 − r2 ) i i 2 i j i j ij i i6=j X 1 X 1 X 1 X = m2r2 + m m r2 + m m r2 + − m m r2 i i 2 i j i 2 i j j 2 i j ij i i6=j i6=j i6=j X X 1 X = m m r2 + m2r2 − m m r2 i j i i i 2 i j ij i6=j i i6=j X X 1 X = ( m )( m r2) − m m r2 j i i 2 i j ij j i i6=j X 1 X = M( m r2) − m m r2 i i 2 i j ij i i6=j X 1 X = M( m r2) − m m r2 i i 2 i j ij i i,j

In the second sum on the RHS, we can perform unrestricted sum over i and j, because for i = j, rij = 0. 3. Show that the Lagrange equations

d ∂T ∂T − = Qj, dt ∂q˙j ∂qj can also be written as ∂T˙ ∂T − 2 = Qj. ∂q˙j ∂qj These are sometimes called the Nielsen form of Lagrange equations.

Soln: Assuming that T = T (qi,q˙i, t), we have dT X ∂T X ∂T ∂T T˙ = = q¨ + q˙ + dt ∂q˙ i ∂q i ∂t i i i i so that ∂T˙ X ∂2T X ∂2T ∂T ∂2T = q¨ + q˙ + + ∂q˙ ∂q˙ ∂q˙ i ∂q˙ ∂q i ∂q ∂q˙ ∂t j i j i i j i j j

2 leading to

∂T˙ ∂T X ∂2T X ∂2T ∂2T − = q¨ + q˙ + ∂q˙ ∂q ∂q˙ ∂q˙ i ∂q˙ ∂q i ∂q˙ ∂t j j i j i i j i j X ∂ ∂T ∂ ∂T ∂ ∂T = q¨ + q˙ + ∂q˙ ∂q˙ i ∂q ∂q˙ i ∂t ∂q˙ i i j i j j d ∂T = dt ∂q˙j Substituting this on the LHS of Lagrange equation, we obtain d ∂T ∂T − = Qj dt ∂q˙j ∂qj ∂T˙ ∂T ∂T =⇒ − − = Qj ∂q˙j ∂qj ∂qj ∂T˙ ∂T =⇒ − 2 = Qj ∂q˙j ∂qj

4. If L is a Lagrangian for a system of n degrees of freedom satisfying the Lagrange equations, show by direct substitution that

0 dF (q , . . . , q , t) L = L + 1 n dt also satisﬁes Lagrange’s equations where F is any arbitrary, but diﬀerentiable, function of its arguments.

Soln: We have dF X ∂F ∂F = q˙ + . dt ∂q i ∂t i i Therefore 0 dF (q1, . . . , qn, t) X ∂F ∂F L = L + = L + q˙ + , dt ∂q i ∂t i i so that ∂L0 ∂L ∂F = + , ∂q˙j ∂q˙j ∂qj leading to d ∂L0 d ∂L d ∂F = + dt ∂q˙j dt ∂q˙j dt ∂qj d ∂L X ∂2F ∂2F = + q˙ + dt ∂q˙ ∂q ∂q i ∂t∂q j i i j j d ∂L X ∂2F ∂2F = + q˙ + dt ∂q˙ ∂q ∂q i ∂q ∂t j i j i j

3 and 0 ∂L ∂L X ∂2F ∂2F = + q˙ + . ∂q ∂q ∂q ∂q i ∂q ∂t j j i j i j

Thus

0 0 d ∂L ∂L d ∂L X ∂2F ∂2F ∂L X ∂2F ∂2F − = + q˙ + − − q˙ − dt ∂q˙ ∂q dt ∂q˙ ∂q ∂q i ∂q ∂t ∂q ∂q ∂q i ∂q ∂t j j j i j i j j i j i j d ∂L0 ∂L0 d ∂L ∂L =⇒ − = − = 0 dt ∂q˙j ∂qj dt ∂q˙j ∂qj QED.

5. Obtain the Lagrange equations of motion for a spherical pendulum, i.e., a point mass suspended by a rigid weightless rod.

Soln: It is best to use spherical polar coordinates here. In the lectures we showed that the kinetic energy of a particle in spherical polar coordinates is 1 T = m r˙2 + r2θ˙2 + r2 sin2 θφ˙2 . 2

If length of the rod is l, then r = l, and r˙ = 0, we are left with two generalized coordinates (θ, φ), with 1 T = m l2θ˙2 + l2 sin2 θφ˙2 , 2 and using the point of suspension as the reference for potential energy, we have

V = −mgl cos θ.

4 Thus 1 L = T − V = T = m l2θ˙2 + l2 sin2 θφ˙2 + mgl cos θ. 2 Now the two Lagrange equations are

d ∂L ∂L − = 0 dt ∂θ˙ ∂θ 1 =⇒ ml2θ¨ − ml2 sin 2θφ˙2 + mgl sin θ = 0 2 and d ∂L ∂L − = 0 dt ∂φ˙ ∂φ d(ml2 sin2 θφ˙) =⇒ = 0. dt

6. Obtain the Lagrangian and equations of motion for a double pendulum, where the lengths of the pendula are l1 and l2 with corresponding masses m1 and m2, conﬁned to move in a plane.

Soln: As discussed in the lectures, this system has two generalized coordinates θ1 and θ2, the angles which the upper and the lower pendula make with respect to the vertical.

With the motion of the pendula conﬁned in a plane (say, xy plane), then the Cartesian coordinates of the two particles can be written as

x1 = l1 sin θ1

y1 = −l1 cos θ1

and

x2 = x1 + l2 sin θ2 = l1 sin θ1 + l2 sin θ2

y2 = y1 − l2 cos θ2 = −l1 cos θ1 − l2 cos θ2

5 So that 1 1 T = m (x ˙ 2 +y ˙2) + m (x ˙ 2 +y ˙2). 2 1 1 1 2 2 2 2 Now ˙ x˙ 1 = l1 cos θ1θ1 ˙ y˙1 = l1 sin θ1θ1 ˙ ˙ x˙ 2 = l1 cos θ1θ1 + l2 cos θ2θ2 ˙ ˙ y˙2 = l1 sin θ1θ1 + l2 sin θ2θ2

Easy to verify

2 2 2 ˙2 x˙ 1 +y ˙1 = l1θ1, 2 2 2 ˙2 2 ˙2 ˙ ˙ x˙ 2 +y ˙2 = l1θ1 + l2θ2 + 2l1l2θ1θ2 cos(θ1 − θ2).

With this 1 1 T = (m + m )l2θ˙2 + m l2θ˙2 + m l l θ˙ θ˙ cos(θ − θ ), 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 2 and

V = m1gy1 + m2gy2

= −(m1 + m2)gl1 cos θ1 − m2gl2 cos θ2, so that

L = T − V 1 1 = (m + m )l2θ˙2 + m l2θ˙2 + m l l θ˙ θ˙ cos(θ − θ ) 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 2 + (m1 + m2)gl1 cos θ1 + m2gl2 cos θ2.

θ1 equation of motion d ∂L ∂L − = 0, ˙ dt ∂θ1 ∂θ1 leads to d (m + m )l2θ˙ + m l l θ˙ cos(θ − θ ) + m l l θ˙ θ˙ sin(θ − θ ) + (m + m )gl sin θ = 0. dt 1 2 1 1 2 1 2 2 1 2 2 1 2 1 2 1 2 1 2 1 1 Upon taking the time derivative, we obtain the ﬁnal form

2 ¨ ¨ ˙2 (m1 + m2)l1θ1 + m2l1l2 cos(θ1 − θ2)θ2 + m2l1l2 sin(θ1 − θ2)θ2 + (m1 + m2)gl1 sin θ1 = 0

For θ2 d ∂L ∂L − = 0, ˙ dt ∂θ2 ∂θ2

6 which yields d m l2θ˙ + m l l θ˙ cos(θ − θ ) − m l l θ˙ θ˙ sin(θ − θ ) + m gl sin θ = 0. dt 2 2 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 2 2 Upon taking the time derivative, we obtain the ﬁnal form ¨ 2 ¨ ˙2 m2l1l2 cos(θ1 − θ2)θ1 + m2l2θ2 − m2l1l2 sin(θ1 − θ2)θ1 + m2gl2 sin θ2 = 0

7. If we want to obtain the equations of motion for a charged particle of mass m, moving in an electromagnetic ﬁeld (E, B), the potential in the Lagrangian has to be velocity dependent U = qφ − qA · v, where q is the charge of the particle, and φ, and A, respectively, are the scalar and vector potentials of the electromagnetic ﬁeld so that ∂A E = −∇φ − ∂t B = ∇ × A. Show that using this Lagrangian, we obtain the correct equations of motion for the particle.

Soln: Using Cartesian coordinates and the fact that v =x ˙ˆi +y ˙ˆj +z ˙kˆ, and A = ˆ Axˆi + Ayˆj + Azk, we have 1 L = m(x ˙ 2 +y ˙2 +z ˙2) − qφ + q(A x˙ + A y˙ + A z˙). 2 x y z Lagrange equation for x component d ∂L ∂L − = 0 dt ∂x˙ ∂x d ∂φ ∂A ∂A ∂A =⇒ (mx˙ + qA ) + q − qx˙ x − qy˙ y − qz˙ z = 0 dt x ∂x ∂x ∂x ∂x dA ∂φ ∂A ∂A ∂A =⇒ mx¨ = −q x − q + qx˙ x + qy˙ y + qz˙ z dt ∂x ∂x ∂x ∂x But dA ∂A ∂A ∂A ∂A x = x x˙ + x y˙ + x z˙ + x , dt ∂x ∂y ∂z ∂t so that ∂φ ∂A ∂A ∂A ∂A ∂A ∂A ∂A mx¨ = q − − x − q x x˙ − q x y˙ − q x z˙ + qx˙ x + qy˙ y + qz˙ z ∂x ∂t ∂x ∂y ∂z ∂x ∂x ∂x ∂φ ∂A ∂A ∂A ∂A ∂A = q − − x + qy˙ y + qz˙ z − q x y˙ − q x z˙ ∂x ∂t ∂x ∂x ∂y ∂z Using the fact that ∂A E = −∇φ − ∂t B = ∇ × A,

7 we obtain above mx¨ = qEx + q(v × B)x, which is the x component of the Lorentz force equation. Using the same procedure for y and z components, we obtain

m¨r = qE + q(v × B).

8. The electromagnetic ﬁeld is invariant under a gauge transformation of the scalar and vector potential given by

A → A + ∇ψ(r, t), ∂ψ φ → φ − , ∂t where ψ is arbitrary (but differentiable). What effect does this gauge transformation have on the Lagrangian of a moving particle in the electromagnetic field? Is the equation of motion affected?

Soln: On performing the gauge transformations, we have

A → A0 = A + ∇ψ(r, t),

0 ∂ψ φ → φ = φ − , ∂t 0 1 0 0 ∂ψ L → L = m(x ˙ 2 +y ˙2 +z ˙2) − qφ + qA · v = L − qv · ∇ψ − q 2 ∂t Or 0 ∂ψ ∂ψ ∂ψ ∂ψ L = L − q x˙ + y˙ + z˙ + ∂x ∂y ∂z ∂t dψ = L − q dt Because L0 diﬀers from L by the total time derivative of a diﬀerentiable function ψ = ψ(r, t), hence, from the result of Prob 4, it will lead to the same Lagrange equations as L.