# Angular momentum after a collision

## Homework Statement

A disk of radius ##r## and mass ##m## rolls down an inclined plan. It reaches the end of the plane with velocity ##v_{f}## and collides with a vertical rod of length ##L## and mass ##M## sticking with it. See figure. What is the angular momentum magnitude and direction after the collision?

## Homework Equations

$$L=I_{disk}\omega$$

## The Attempt at a Solution

The moment of inertia of the disk relative to the pivot is (using the parallel axis theorem)
$$I_{disk}=\frac{1}{2}mr^{2}+m(L+r)^{2}$$.
Angular momentum is conserved. The initial (and hence the final i.e. right after the collision) angular momentum is

$$L_{initial}=I_{disk}\omega=I_{disk}\frac{v_{f}\cos\theta}{L+r}=m\left[\frac{1}{2}mr^{2}+m(L+r)^{2}\right]\frac{v_{f}\cos\theta}{L+r}$$

Angular momentum points out of the plane since rotation is counterclockwise

However the solution is different . Where am I wrong?

#### Attachments

$$L=I_{disk}\omega$$ $$I_{disk}=\frac{1}{2}mr^{2}+m(L+r)^{2}$$.
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