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Rot. + Trans. Motion: Conservation of E

  1. Mar 29, 2012 #1
    Suppose of uniform thin rod (mass M, length b) is attached to a pivot at the top. A piece of clay of mass m strikes the rod at distance x below the pivot at v0 perpendicular to the rod and sticks to it.

    I understand the E is not conserved b/c this is an inelastic collision (clay sticks to rod). But would the fraction of E loss be during the collision?

    I think:

    Fraction of E loss = E final / E initial where E initial = 1/2mv0^2 but what would E final be?

    Can someone please help with respect to the rod's rotational and translational E? What would it's moment of inertia be? Should I apply parallel axis theorem?
     
  2. jcsd
  3. Mar 29, 2012 #2
    So the problem is to find the final velocity of the system, and by that the final energy?
    [itex]Eini = 1/2mv_0^2\\
    Efinal = E_{rot} + E_{trans} = 1/2I\omega^2 [/itex]
    So no translation(pivot).
    The uniform rod has a moment of interia about its center of mass: [itex]I_{cm}=(M+m)b^2/12\\ I = I_{cm}+(M+m)r^2 = (M+m)(r^2+b^2\dfrac{1}{12})[/itex]
    Now use the conservation of angular momentum just before collision and after, cause the sum of external torque is zero.
     
  4. Mar 30, 2012 #3
    I don't think so...

    The axis of rotation is at the end of the rod (where the pivot is) so I = 1/3(M+m)L^2 + mx^2 (m is point particle at distance x from axis of rotation).
     
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