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Izbitzer
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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.
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.
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})
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
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
