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MP350 Classical Mechanics

Solutions Problem set 2

1. (a) We use the general expression for the kinetic energy in spherical coordinates,

T =1

2

mr2 + r22 + r2 sin2 2 (1.1)In the case of a pendulum, the radial coordinate is fixed, r = , so we have

L = T V =1

2m2(2 + sin2 2) + mg cos . (1.2)

The canonical momenta are

p =L

= m2 (1.3)

p =L

= m2 sin2 (1.4)

(b) Since the lagrangian does not depend on the coordinate , the correspondingcanonical momentum p is conserved,

dp

dt=

d

dt

L

=

L

= 0 . (1.5)

(c) The two EulerLagrange equations for and are

d

dt

L

= m2 =L

= m2 sin cos 2 mg sin (1.6)

= = 2 sin cos g

sin . (1.7)

d

dt

L

= m2 sin2 + 2m2 sin cos =

L

= 0 (1.8)

= + 2cos

sin = 0 . (1.9)

2. (a) A uniform distribution in the plane x = 0 can not generate any dependenceon the second and third coordinates y, z. Therefore the momenta py, pz in they z directions are conserved. The distribution is also rotationally symmetric

about the x-axis, so the angular momentum Lx = pyzpzy is also conserved.(b) A uniform distribution in the half-plane x = 0, y 0 can not generate any

dependence on the third coordinate z. Therefore the momentum pz in thez-direction is conserved. There are no other manifest symmetries in theproblem, so this is the only conserved quantity.

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(c) This system is rotationally symmetric about the z-axis, so Lz (the z-componentof the angular momentum) is conserved. There is no translational symmetryin any direction, so no linear momentum is conserved, nor are the two otherangular momentum components.

(d) This system has no translation symmetry in the x, y or z direction, but itis translationally symmetric in the (1, 1, 0) direction. Therefore, the linearmomentum in this direction, p (1, 1, 0) = px + py is conserved.

The system is also rotationally symmetric about the (1, 1, 0) axis, so the

angular momentum about this axis is conserved. This angular momentum isproportional to Lx + Ly, so Lx + Ly is conserved.

3. (a) First we need to express r1 andr2 in terms of the new coordinates

R ,r .

We have

r = r1 r2

r1 =r +r2 (3.1)

R =

m1r1 + m2

r2

m1 + m2=

m1(r +r2 ) + m2

r2

m1 + m2= r2 +

m1

m1 + m2r , (3.2)

which gives

r2 = R m1m1 + m2

r (3.3)

r1 =r +r2 =

R +

m2

m1 + m2r (3.4)

We now plug (3.3), (3.4) into the expression for the kinetic energy,

T =1

2m1r1

2

+1

2m2r2

2

=1

2m1

R +

m2

m1 + m2r2

+1

2m2

R

m1

m1 + m2r2

=

1

2 m1

R

2

+

2m2m1 + m2

R

r +

m22

(m1 + m2)2

r

2+

1

2m2

R

2

2m1

m1 + m2

R r +

m21

(m1 + m2)2r

2

=1

2(m1 + m2)

R

2

+1

2

m1m

2

2

(m1 + m2)2+

m21

m2

(m1 + m2)2

r

2

=1

2MR

2

+1

2r

2

.

(3.5)

The total lagrangian is therefore

L = T V =

1

2 M

R

2

+

1

2

r

2

V(

R ,

r ) . (3.6)

(b) The canonical momentum conjugate toR is

Pi =L

Ri= MRi =

P = M

R . (3.7)

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This is just the total momentum of the system:

P = M

R = (m1 + m2)

m1r1 + m2

r2

m1 + m2= m1

r1 + m2r2 . (3.8)

For an isolated system, there are no external forces, which means that thepotential energy only depends on the relative position on the particles, not

the absolute (centre-of-mass) position. This means thatR does not appear

in the potential energy, and therefore not in the lagrangian, ie it is a cyclic

coordinate. Then the momentum conjugate to R , ie the total momentumP, is conserved.

(c) We consider only one-dimensional movement; three dimensions just adds anextra index without changing the argument. The centre-of-mass coordinateX is

X =

Ni=1

mixiNi=1 mi

1

M

Ni=1

mixi (3.9)

We now rewrite the coordinate of each particle in terms of X and N 1relative coordinates. We do not actually care what the relative coordinates

are, so we do not write them explicitly:

xi = XX + xi = X1

M

mjxj +

xi

M

mj = X +

1

M

j

mj(xi xj)

(3.10)The kinetic energy is

T =1

2

Ni=1

mix2

i =1

2

Ni=1

mi

X +

1

M

Nj=1

mj(xi xj)2

=

1

2

N

i=1

miX2

+

X

M

N

i,j=1

mimj(xi xj) + (. . .) ,

(3.11)

where (. . .) denotes terms that only depend on the relative coordinates. Thesecond term on the last line of (3.11) vanishes since every xi xj is cancelledby the equal and opposite xjxi term. So the kinetic energy is simply

1

2MX2

plus terms which only depend on the relative coordinates.

The canonical momentum conjugate to X is (as above, assuming V does notdepend on velocities)

P =

T

X = M

X = M

1

M

N

i=1 m

ixi =

N

i=1 p

i , (3.12)

which is just the total momentum of the system. Conservation of total mo-mentum for an isolated system follows as before.