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 Please help your instructor by doing your work neatly.
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 ·~vi dt 2 i dt For rotation, ~vi = ω ~ ×~ri so that dT 1 X d~vi mi = · (~ ω ×~ri ) . dt 2 i dt Using the cyclic property of the mixed product, µ ¶ d~vi d~vi · (~ ω ×~ri ) = ω ~ · ~ri × , dt dt and summing over the particles, we find dT ~, =ω ~ ·N dt where ~ = N
X i
µ ¶ X d~vi ~ri × mi = ~ri × 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~ ³ ´ Det Iˆ − λˆ1 = 0,
2 where ˆ1 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 & Lifshits, ”Mechanics”, problem 7, p. 104.
