A long thin rod of mass M=2.00 kg and length L=75.0 cm is free to rotate about its center. Two identical masses (each of mass m = .45kg) 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.

Question

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

The moment of inertia of the rod rotating about its center is M*L^2/12

When the idential masses m are at radius R, the moment of intertia due to their masses is 2*m*R^2

The total angular momentum remains unchanged. Let w1 be the initial angular velocity.

A) When the masses move half way to the ends of the rod, their radius is R = L/4. The angular velocity becomes a new value w2. You can solve for it using the momentum conservation equation.

(ML^2/12)*w1 = [(ML^2/12)+2*m*(L/4)^2
]*w2

B) Use the formula for kinetic energy, (1/2)I*w^2, and compare the before and after values.
C) Repeat A), but with R = L/2 instead of L/4.

Answer 2

Thank you so much!

Answer 3

Hi Miki! I think we are taking the same physics class...
I'd like to possibly share insights about the material and possibly help each other out in a mutually beneficial way. If you are interested, shoot me an email at William dot Cordoba at yaho dot com

Thanks
William