# Compound ballistic pendulum

Hello. I am doing a ballistic pendulum lab, and I have gotten stuck at a preparatory exercise. The problem is that the pendulum must be treated as a compound pendulum and not a simple pendulum.

## Homework Statement

We have a compound pendulum which is a metal rod of mass M suspended at some point O at a distance d from the center of mass. We fire a bullet of mass m and it hits the pendulum at a distance R from O. The bullet sticks to the pendulum and the pendulum gets an angular velocity. The pendulum has a maximum angle of $$\theta_{max}$$. The rod's moment of inertia is I.

Find an expression for the velocity of the bullet.

## Homework Equations

Angular momentum: $$L=I\omega_{p}$$

Max. kinetic energy of the rod with bullet: $$\frac{1}{2}(M+m)v^{2}_{p}$$

Max. potential energy of the rod with bullet: $$mgh=(M+m)gd(1-cos \theta_{max})$$

## The Attempt at a Solution

At the moment of impact angular momentum is conserved (right?): $$mv_{b}R = I\omega$$

After the bullet has stuck to the rod mechanical energy is conserved: $$\frac{1}{2}(M+m)v^{2}_{p}=(M+m)g(1-cos \theta_{max}) \Leftrightarrow v_{p} = \sqrt{2gd(1-cos\theta_{max})}$$

$$\omega_{p} = \frac{v_{p}}{d}$$

$$mv_{b}R = I\omega_{p} = I\sqrt{\frac{2g(1-cos\theta_{max})}{d}} \Rightarrow v_{b} = \frac{I}{mR}\sqrt{\frac{2g(1-cos\theta_{max})}{d}}$$

This, however, is not the correct answer which should be $$v_{b} = \frac{1}{mR}\sqrt{2IMgd(1-cos\theta_{max})}$$

What have I done wrong?

Thanks!
/I