Conservation of Energy/Momentum with Rotation

1. Nov 18, 2009

kelslee28

1. The problem statement, all variables and given/known data
Two balls of mass 2.26 kg are attached to the ends of a thin rod of negligible mass and length 72 cm. The rod is free to rotate without friction about a horizontal axis through its center. A putty wad of mass 145 g drops onto one of the balls, with a speed 2.7 m/s, and sticks to it. What is the angular speed of the system just after the putty wad hits?

2. Relevant equations
Conservation of momentum
p = mv
L= Iw
Lparticle = m rperpendicular v

Conservation of Energy
KE= 1/2mv2
Rotational KE = 1/2Iw2
I = 1/2mh2

3. The attempt at a solution
I tried this both with using conservation of momentum, conservation of energy, and torque. Torque seemed like the easiest but I have no way of getting angular velocity from that without a time.
For conservation of momentum I said that the initial momentum consisted of only the momentum of the putty. The final momentum I said was the momentum of ball 1, ball 2 and the putty put together. So I had m2v = m1(1/2L)w + m2(1/2L)v

I solved for w and got w = [m2v - m2(1/2L)v]/m1(1/2L)

For conservation of energy I said that the initial energy was only the kinetic energy of the putty KE = 1/2mv2
The final energy is the total rotational energy of ball 1, ball 2 and the putty.
1/2mv2 = 1/2(m1(1/2L)2) + 1/2((m1+m2)(1/2L)2) w2
I solved for w and got the square root of [m2v2]/ [m1L2/4 + (m1+m2)L2/4]

I am really just at a loss as to what I am doing wrong. None of my methods are working!

2. Nov 18, 2009

cepheid

Staff Emeritus
Well, angular momentum is conserved as well, in the absence of any net external torques.

Edit: I don't think that the momentum of either of the individual balls is conserved, because each one has a force acting on it (by virtue of the fact that it is attached to a rigid rod), a centripetal force that acts to keep the balls moving in a circular path. Hence, their momenta are continuously changing due to their velocities continuously changing direction.

Also, I don't think that momentum is conserved in the collision, because presumably some sort of rigid "stand" is being used to fix the whole rotating rod system to a surface (hence, the whole thing will not be pushed downward or tilt when the putty hits it). So, again, an external force of some kind is acting on the system.

I don't think that kinetic energy is necessarily conserved either, because the putty hits one of the balls and sticks to it (a perfectly inelastic collision). All of its kinetic energy is gone, going into deforming the putty, making sound, etc.

Angular momentum is conserved, however, because there are no net external torques on the system. Conservation of angular momentum should be sufficient to solve the problem.

Last edited: Nov 18, 2009
3. Nov 18, 2009

kelslee28

So this is what I did.
m1 = mass of ball1 and ball2
m2 = mass of putty

1/2Lm2V = Iw (of the ball on the left) + Iw (the ball on the right plus the mass of the putty)

1/2Lm2V = [m1((1/2)L)2 + (m1+m2)((1/2)L)2]w

solve for w

[1/2Lm2V]/[m1((1/2)L)2 + m1+m2((1/2)L)2]