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PHY6426/Fall 2007: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT #5: SOLUTIONS due by 9:35 a.m. Mon 10/01 Instructor: D. L. Maslov [email protected]ﬂ.edu 392-0513 Rm. 2114 Please help your instructor by doing your work neatly. 1. Goldstein, Problem 5.3 (30 points) T = 1 2 X i m i ~v 2 i , sum is over all particles in the system. dT dt = 1 2 X i m i d~v i dt · ~v i For rotation, ~v i = ~ ω×~ r i so that dT dt = 1 2 X i m i d~v i dt · (~ ω×~ r i ) . Using the cyclic property of the mixed product, d~v i dt · (~ ω×~ r i )= ~ ω · ~ r i × d~v i dt , and summing over the particles, we ﬁnd dT dt = ~ ω · ~ N, where ~ N = X i ~ r i × m i d~v i dt = X i ~ r i × ~ F i and ~ F = m i d~v i dt is the force on the i th particle. 2. Goldstein, Problem 5.16 (30 points) Tensor of inertia ˆ I =2ma 2 5 0 0 0 3 -2 0 -2 3 . The eigenvalue problem ˆ I~v = λ~v reduces to ﬁnding the roots of the equation Det ˆ I - λ ˆ 1 · =0,
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SOLUTIONSdue by 9:35 a.m. Mon 10/01

Instructor: D. L. [email protected] 392-0513 Rm. 2114

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.