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Compound ballistic pendulum

  1. Mar 4, 2008 #1
    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.

    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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]

    3. 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?

  2. jcsd
  3. Mar 4, 2008 #2
    Never mind, I had got the problem wrong. It is solved now. Thanks!
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