# Angular Momentum question

1. Jan 11, 2010

### conana

1. The problem statement, all variables and given/known data

On a frictionless horizontal table, a stick of mass m and length l spins around a pivot at one of its ends with angular frequency $$\omega$$. It collides and sticks to an identical stick (which is presumably stationary), with an overlap length equal to x. Immediately before the collision, the pivot is removed. What should x be so that after the collision the double-stick system has translational but no rotational motion?

3. The attempt at a solution

From the diagram provided, it appears that the collision is inelastic so I am pretty sure I can not use conservation of energy. I guess my approach would be to use conservation of angular momentum and possibly linear momentum as well. I am pretty lost as to how to get started. I tried writing out an expression for my angular momentum but I am not sure what to use as my origin and how to find the angular momentum after the collision.

2. Jan 12, 2010

### ideasrule

Do the sticks stick together? (All puns intended.) I'll assume they do, because I don't think the problem is solvable otherwise.

You're right that you need both the conservation of linear momentum and the conservation of angular momentum. Since the sticks must move in the same direction, you know that their final momentum, 2mv, must be equal to the initial momentum of the moving stick.

For angular momentum, you can use any origin you want. I'd pick the middle of the overlap of the 2 sticks because that simplifies calculations quite a bit--final angular momentum ends up being 0 being the two-rod combination's center of mass moves directly away from the origin. Initial angular momentum is pretty easy to calculate.

3. Jan 13, 2010

### conana

For linear momentum I get

$$p_i=p_f$$

$$\Rightarrow mv_i=2mv_f$$.

$$\Rightarrow \dfrac{1}{2}ml\omega=2mv_f$$

$$\Rightarrow v_f=\dfrac{1}{4}l\omega$$.

Then for the angular momentum, using the middle of the overlap of the two sticks as my origin, I get

$$L_i=L_f$$

$$\Rightarrow mv_ir-I\omega=0$$

(Initially I typed this out as a sum of two Ls instead of a difference which gave me a nonsensical answer x>l. I now believe it to be a difference because the CM of the stick is rotating in one direction about the origin while the stick is rotating in the opposite direction about its center of mass. Is this logic correct?)

where the distance from the origin to the CM of the rotating stick $$r=\dfrac{l-x}{2}$$ and the moment of inertia of the stick about its CM $$I=\dfrac{1}{12}ml^2$$. Then

$$ml\omega\dfrac{l-x}{4}-\dfrac{1}{12}ml^2\omega=0$$

$$\Rightarrow x=\dfrac{2}{3}l$$.

Is this looking to be on the right track? Also, (assuming this is correct) it appears as though I didn't end up needing to use conservation of linear momentum, just my initial linear momentum. Thank you so much for your help.

Last edited: Jan 13, 2010