PHY6426/Fall 2007: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT ...maslov/classmech/h5_07sol.pdf ·...
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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,
Transcript of PHY6426/Fall 2007: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT ...maslov/classmech/h5_07sol.pdf ·...
PHY6426/Fall 2007: CLASSICAL MECHANICSHOMEWORK ASSIGNMENT #5:
SOLUTIONSdue by 9:35 a.m. Mon 10/01
Instructor: D. L. [email protected] 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.
