# Conservation of Angular Momentum

1. Feb 18, 2004

### Lomion

A double-pendulum, initially in a vertical position, is attached to a pin. The distance from the mass m (3.2 kg) at the top of the pendulum to the pin is 0.2 m, the distance from the mass m (3.2 kg) at the bottom of the pendulum to the pin is 0.4 m. A bullet of mass 0.05 kg is fired into the bottom mass at 300 m/s, 20 degrees up from the horizontal. What is the maximum angle that the pendulum reaches before it swings back?

EDIT: I forgot this info. The bullet becomes embedded in the bottom mass after it hits. And the initial &omega;=6 rad/s (counter-clockwise)

I'm not too sure how to approach this problem. I used the principle of Conservation of Angular Momentum and figured out &omega;=-2.78 rad/s after the impact. (Hpendulum + Hbullet = Htotal)

I know that &omega;=0 rad/s when the pendulum reaches the maximum angle. However, I'm not sure how to calculate that angle, since if &omega; is 0, then H=0?

I think that maybe I should use energy methods, but I'm not sure how to do so exactly.

Any help would be appreciated!

Last edited: Feb 18, 2004
2. Feb 19, 2004

### gnome

Why not try using linear momentum. Using the angular momenta, you can calculate the linear momentum of each of the masses immediately after the collision. From that you can calculate their velocities. And that will give you the kinetic energy immediately after the collision. When the rotation stops, all of the kinetic energy will have been converted to potential, which will give you heights of the two masses, and that in turn will give you the angle. Remember that the initial potential energy is NOT zero -- the masses start out at different heights.

Also, I think you need to re-check your calculation of the angular momentum. I'm getting a different result.