- #1

Izbitzer

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## 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 [tex]\theta_{max}[/tex]. The rod's moment of inertia is I.

Find an expression for the velocity of the bullet.

## Homework Equations

Angular momentum: [tex]L=I\omega_{p}[/tex]

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

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

## The Attempt at a Solution

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

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

[tex]\omega_{p} = \frac{v_{p}}{d}[/tex]

[tex]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}}[/tex]

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

What have I done wrong?

Thanks!

/I