# Angular momentum ball problem

1. Nov 15, 2008

### boredaxel

1. The problem statement, all variables and given/known data
A 5.00-kg ball is dropped from a height of 12.0 m above
one end of a uniform bar that pivots at its center. The bar has mass
8.00 kg and is 4.00 m in length. At the other end of the bar sits
another 5.00-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
L= r X p
L = I $$\omega$$
mgh =1/2 mv^2

3. The attempt at a solution
The solution seems to require the conservation of angular momentum. But i am not sure how angular momentum could be conserved? Doesnt the weight of the falling ball provide a net external torque to the system?

2. Nov 15, 2008

### tiny-tim

Welcome to PF!

Hi boredaxel! Welcome to PF!

Yes, it does provide a torque, but that doesn't matter, because it isn't an external torque …

at least, if you consider the whole system together, it isn't external!

Angular momentum, like momentum, is always conserved.

Hint: treat this exactly the same way as you would if the ball was hitting a block with another ball directly the other side …

you'd use conservation of (linear) momentum, and v1f = v2f, wouldn't you?

Well, do the same, except … "angularly"!