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Rod, Clay, and an Inelastic Collision

  1. Jun 26, 2004 #1
    Problem: A thin rod of mass M and length L rests on a frictionless table and is struck at a point L/4 from its center of mass by a clay ball of mass m moving at a speed v (the velocity vector is perpendicular to the rod). The ball sticks to the rod. Determine the translational and rotational motion of the rod after the collision.

    I can use conservation of angular momentum to determine the rotational velocity of the rod and the clay about the center of mass of the rod. Then, I'd figure I could use this angular speed to find the velocity of the clay and then use conservation of linear momentum to find the velocity of the center of mass of the rod. According to my calculations, the center of mass of the rod is moving in the same direction of the velocity vector v. Does this make sense?
     
  2. jcsd
  3. Jun 26, 2004 #2

    Gza

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    That seems like an awful lot of work to simply calculate the velocity of the center of mass after collision. Just use conservation of linear momentum throughout to calculate translational velocity. Use conservation of angular momentum to find the angular velocity about its new center of mass after collision.
     
  4. Jun 26, 2004 #3
    Hmm...I thought the rotation was about the center of mass of the rod not about the center of mass of the system after collision. How do you figure?
     
  5. Jun 26, 2004 #4

    Doc Al

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    Staff: Mentor

    The motion of a rigid object (such as the composite object "rod + putty") can be described as a translational motion of its center of mass plus a rotation about its center of mass.

    You know how to find the speed of the center of mass after the collision: linear momentum is conserved.

    To find the rotational speed about the cm, do this. First find where the cm is just at the instant of collision. Then find the angular momentum about the cm prior to the collision. Since it's conserved, that's also the angular momentum about the cm after the collision. Find the rotational inertia of the "rod + putty" about the cm, then use it to find [itex]\omega[/itex].
     
  6. Jun 26, 2004 #5
    I think you meant "can be BEST describe...", since motion is relative to a reference frame and the point in the frame I'm describing the motion from. When you use the phrase about its center of mass, I imagine the object is rotating about an axis going through the center of mass, but how do I know, in this problem for example, where the rotation axis is (I know it's not fixed in some location in space, but that is all I know).
    Let me get this clear: I should do all of my calculations (pre- and post-collion) taking the center of mass of the rod + putty as my reference point.
     
  7. Jun 27, 2004 #6

    Doc Al

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    Staff: Mentor

    translation plus rotation

    As long as you realize that the motion of a rigid object is a combination of the motion of its cm plus its rotation about its cm.
    I'm not saying that the object is in pure rotation about its center of mass. It's also translating.
    I would, but I'm lazy. (All what caculations? It's just angular momentum.) As long as you realize that the total angular momentum of the object about any axis is the sum of its angular momentum about its center of mass plus the angular momentum of its mass (assumed concentrated at its center of mass), then you can use any axis. So, if you find the initial angular momentum about the cm, then that will equal the final angular momentum about the cm, which would equal [itex]I_{cm}\omega[/itex].
     
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