# Angular momentum

1. Mar 13, 2005

### UrbanXrisis

The problem is here.

I got a different answer from what my book tells me and I want to know why. Angular momentum is:
$$L=I \omega$$
So... the moment of inertia for the block would be:
$$I=((M+m)l^2)$$
since the rod has negligible mass, it will not have a moment of inertia.
$$\omega=v/r=v/l$$
$$L=\frac{(M+m)l^2v}{l}$$
$$L=vlM+vlm$$

my text gives an answer of $$L=mvl$$
I dont understand what I misunderstood

Last edited: Mar 13, 2005
2. Mar 13, 2005

### SpaceTiger

Staff Emeritus
Try again without assuming that the block/bullet combination are moving at the same speed as the original bullet. You'll need to consider conservation of linear momentum.