PHY6426/Fall 2007: CLASSICAL MECHANICSHOMEWORK ASSIGNMENT #5:

SOLUTIONSdue by 9:35 a.m. Mon 10/01

Instructor: D. L. Maslovmaslov@phys.ufl.edu 392-0513 Rm. 2114

Please help your instructor by doing your work neatly.

1. Goldstein, Problem 5.3 (30 points)

T =12

∑

i

mi~v2i ,

sum is over all particles in the system.

dT

dt=

12

∑

i

mid~vi

dt·~vi

For rotation,

~vi = ~ω×~ri

so that

dT

dt=

12

∑

i

mid~vi

dt· (~ω×~ri) .

Using the cyclic property of the mixed product,

d~vi

dt· (~ω×~ri) = ~ω ·

(~ri × d~vi

dt

),

and summing over the particles, we find

dT

dt= ~ω · ~N,

where

~N =∑

i

~ri ×(

mid~vi

dt

)=

∑

i

~ri × ~Fi

and ~F = mid~vi

dt is the force on the ith particle.

2. Goldstein, Problem 5.16 (30 points)

Tensor of inertia

I = 2ma2

5 0 00 3 −20 −2 3

.

The eigenvalue problem I~v = λ~v reduces to finding the roots of the equation

Det(I − λ1

)= 0,

2

where 1 is the unity matrix. The roots are λ = 2ma2 (5, 5, 1) . The double degenerate eigenvalue λ1 = λ2 =10ma2 corresponds to the following equation for the components of the eigenvalue vector

0 0 00 −2 −20 −2 −2

v1

v2

v3

=

0−2(v2 + v3)−2(v2 + v3)

000

.

Hence, v1 can be arbitrary, whereas v2 and v3 must satisfy v2 + v3 = 0. One of the eigenvectors satisfying theseconditions is

~v(1) =

01−1

.

The other eigenvector, orthogonal to the first one can be chosen as

~v(2) =

100

The non-degenerate eigenvalue λ3 = 2ma2 corresponds to

4 0 00 2 −20 −2 2

v1

v2

v3

=

4v1

2(v2 − v3)−2(v2 − v3)

000

→

~v(3) =

011

.

The principal axes can be chosen as the x− axis, and lines z = y and z = −y.

3. Goldstein, Problem 5.17 (40 points)

See Ch. 32 in Landau & Lifshits, ”Mechanics”, problem 7, p. 104.