# Homework 5: Solutions ```PHY6426/Fall 2007: CLASSICAL MECHANICS
HOMEWORK ASSIGNMENT #5:
SOLUTIONS
due by 9:35 a.m. Mon 10/01
Instructor: D. L. Maslov
[email protected] 392-0513 Rm. 2114
1. Goldstein, Problem 5.3 (30 points)
T =
1X
mi~vi2 ,
2 i
sum is over all particles in the system.
dT
1 X d~vi
=
mi
&middot;~vi
dt
2 i
dt
For rotation,
~vi = ω
~ &times;~ri
so that
dT
1 X d~vi
mi
=
&middot; (~
ω &times;~ri ) .
dt
2 i
dt
Using the cyclic property of the mixed product,
&micro;
&para;
d~vi
d~vi
&middot; (~
ω &times;~ri ) = ω
~ &middot; ~ri &times;
,
dt
dt
and summing over the particles, we find
dT
~,
=ω
~ &middot;N
dt
where
~ =
N
X
i
&micro;
&para; X
d~vi
~ri &times; mi
=
~ri &times; F~i
dt
i
vi
th
and F~ = mi d~
particle.
dt is the force on the i
2. Goldstein, Problem 5.16 (30 points)
Tensor of inertia


5 0 0
Iˆ = 2ma2  0 3 −2  .
0 −2 3
ˆv = λ~v reduces to finding the roots of the equation
The eigenvalue problem I~
&sup3;
&acute;
Det Iˆ − λ1̂ = 0,
2
where 1̂ 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 0
v1
0
0
 0 −2 −2   v2  =  −2(v2 + v3 )   0  .
0 −2 −2
v3
−2(v2 + v3 )
0
Hence, v1 can be arbitrary, whereas v2 and v3 must satisfy v2 + v3 = 0. One of the eigenvectors satisfying these
conditions is


0
~v (1) =  1  .
−1
The other eigenvector, orthogonal to the first one can be chosen as
 
1
~v (2) =  0 
0
The non-degenerate eigenvalue λ3 = 2ma2 corresponds to


 
 
v1
4v1
0
4 0 0
 0 2 −2   v2  =  2(v2 − v3 )   0  →
0 −2 2
v3
−2(v2 − v3 )
0

~v (3)

0
=  1 .
1
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 &amp; Lifshits, ”Mechanics”, problem 7, p. 104.
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