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Conservation of angular momentum of a uniform thin rod

  1. Nov 13, 2005 #1
    A uniform thin rod of length 0.471 m and mass 4.0 kg can rotate freely on a frictionless horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet's velocity makes an angle of 60° with the rod.

    If the bullet lodges in the rod and the angular velocity of the rod is 6.99 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

    This is a hard one. I found the inertia of the rod to be .0740024 (4.003/12)*.471^2
    The angular velocity is 6.99.
    .0740024 * 6.99 is the momentum = .51727

    I also know that momentum is the cross product of the radius and the tangential momentum (bullet momentum). Since the radius and the momentum is perpendicular you just multiply. so (.471/2) * .003 *sin(60)*v = 6.18829e-4 * v. Then the .51727 / 6.18829e-4 = v. to find 851 m/s but this is not right, what did I do wrong?
  2. jcsd
  3. Nov 14, 2005 #2

    Doc Al

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    Staff: Mentor

    The only thing you neglected was the fact that the bullet lodges into the rod and thus adds to the rotational inertia of the system after the impact. Other than that, it looks OK.
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