# Ball Torque problem

1. Oct 22, 2008

### irun4edmund

1. The problem statement, all variables and given/known data
A 5.80 kg ball is dropped from a height of 13.5 m above one end of a uniform bar that pivots at its center. The bar has mass 8.50 kg and is 5.40 m in length. At the other end of the bar sits another 4.00 kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?

2. Relevant equations
U = mgh
torque = Iα

3. The attempt at a solution
I set my equation up like this m1gh1 - Iα = m2gh2
where everything is known except for h2 (what im trying to find) and α (angular acceleration).

Im stuck because i have one equation with 2 unknowns, im not sure what im missing. Any help would be greatly appreciated.

Last edited: Oct 22, 2008
2. Oct 22, 2008

### LowlyPion

Is there a picture with this problem?

Is the bar horizontal, or elevated toward the end the mass starts on? Or is the second mass constrained to only go straight up?

3. Oct 22, 2008

### irun4edmund

there was no picture to go with the problem. I wrote word for word from my physics book. I assume the bar is horizontal and that ball travels strait up, but the problem does not state that.

4. Oct 22, 2008

### LowlyPion

Well let's assume that they mean for it to travel straight upward and that it will leave the bar at the same level as the first mass falls on it - that is with its ends level.

In this case then I would suggest that you look at the conservation of momentum.

You know that the initial mass m1 and V1 is all there is with everything else at rest.

After the process of attaching then the momentum should remain the same. So ...

m1V1 = I * w

where I is the combined moment of inertia and w is the resulting angular speed. And this speed w * r = V2 of the second mass

The combined I or Itotal = Im1 + Im2 + Ibar

Im1 = m1 * (L/2)2
Im2 = m2 * (L/2)2
Ibar = ... ? Figure the moment of inertia for the bar of length L.

Then you should be able to solve ... at least for the assumptions I would suggest.

Last edited: Oct 22, 2008
5. Oct 22, 2008

### irun4edmund

I calculated the velocity of the first ball to be v = (2gh)1/2 or 16.27

Then the moment of inertia for the system to be I = m1(L/2) + m2(L/2) + (1/12)mbarL2 or 57.12

Using these numbers i got w = 2.00 and v2 = 10.8 and found that the ball goes 1.48 m high..... but this isn't the right answer.

6. Oct 22, 2008

### LowlyPion

My mistake, the moments of the masses are given by.

I = m*r2 = m*L2/4

I just scribbled out the torque.

7. Oct 23, 2008

### irun4edmund

ok, i corrected the moment of inertia, but i still didn't get the right answer... I took another look at linear momentum and figured this thing out.

I*w = m1v1r not where r = L/2
I*w = m1v1 like previously stated.

Thanks for your help. problem solved.