A long thin rod of mass M = 2:00 kg and length L = 75:0 cm is free to rotate about its

Question

center as shown. Two identical masses (each of mass m = .403 kg) slide
without friction along the rod. The two masses begin at the rod's point of rotation when
the rod is rotating at 10.0 rad/s.
(a) When they have moved halfway to the end of the rod, how fast (rad/s) is the rod
rotating?
(b) When the masses are halfway to the end of the rod, what is the ratio of the final kinetic energy to the initial kinetic energy (Kf=Ki)?
(c) When they reach the end, how fast is the rod rotating (rad/s)?

Answer 1

Use conservation of angular momentum.

I*w = constant

I is the moment of inertia, which is
(1/12)Mrod*L^2 + 2m*R^2

R is the distance of the masses from the center of the rod.

w = 10.0 when R = 0

For (a) and (b), R = L/4

For (c), R = L/2

Let's do (a)

Angular momentum with massesw m at R=0:
= (1/12)*2.00*(0.75)^2*10
= 9.38*10^-1 kg m^2/s
(This remains constant).
When R = L/4 = 0.1875 m,
I*w = (2/12)*(0.75)^2*w + 2*(0.403)*(0.1875)^2*w
= (9.38*10^-2 + 2.83*10^-2)w
= 1.221*10^-1*w = 9.38*10^-1
w = 7.68 m/s

For (b), compare initial and final values of (1/2) I w^2

For (c), repeat the process of (a), but use R = L/2