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Angular Momentum Scenario

  1. Nov 27, 2007 #1
    A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the weights is to double the rotational inertia of the system. As he is rotating, the man opens his hands and drops the two weights. They fall outside the turntable. What happens to the man's angular velocity?

    The answer is that his angular velocity should stay relatively constant. I am confused to why this occurs. This is my reasoning for this:

    As the weights are in free fall, they still are roughly same distance from the axis of rotation. Thus I is constant, and hence the nothing in the system changes. Angular velocity is still constant.

    However, once the weights fall on the ground, the external friction force exerts a net torque on the weights, stopping them. However, since the system is not closed, the angular momentum need not be conserved, so the man's angular velocity remains unchanged.
     
  2. jcsd
  3. Nov 27, 2007 #2

    Doc Al

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    Staff: Mentor

    Do you have a question? A cleaner way to see the effect of dropping the weights is to ask: What torque is exerted by the weights on the man when he drops them?
     
  4. Nov 27, 2007 #3
    Oh I see. There is no net torque acting on the man when he drops them.
     
  5. Nov 27, 2007 #4
    Oh, and on a tangential question about angular momentum...

    I was reading about precession of a gyroscope in my physics text and it seems to take the angular momentum with respect to the axis, but the torque with respect to a point on the axis. I know that for torque and angular momentum to be coherent, we should always take both with respect to the same origin. So I am assuming it is reasonable to take torque with respect to any point on the axis of rotation for which the angular momentum of a rigid body in simple rotation is taken in respect to?

    This seems to make sense, as in my text, the proof of the relation, [tex]||L_z|| = I\omega[/tex], with respect to an arbitrary point on the axis of rotation seems to reduce to the situation of treating the situation as finding L for each particle making up the rigid body, with the point on the axis perpendicular to the particle. Since in simple rotation, the velocity and radius are on the same plane as any differential cross section of the rigid body which the point particle lies on, each individual L must be along the axis, perpendicular to the cross section. That would mean L is independent of any specific point on the axis, but rather only the axis itself. Therefore, as long as torque is taken with respect to a point on the axis, the two quantities are compatible. Is my reasoning correct?
     
  6. Nov 29, 2007 #5
    Doc Al where are youuuuu
     
  7. Nov 30, 2007 #6

    Doc Al

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    I'm here, I'm here. (Sorry for the delayed response.)

    If you're saying what I think you're saying, then I agree. Let me rephrase part of what you said. For a body rotating about an axis of symmetry, the total angular momentum of the body is the same about any point on the axis. So, you can analyze torque about any point on that axis without redefining the angular momentum. (I think that's what you meant about the quantities being "compatible".)

    So far, so good. But I will pick a nit with this paragraph:
    Just to be clear: It is not true that the angular momentum of each individual particle is along the axis of rotation. (Consider [itex]\vec{L} = \vec{r} \times \vec{p}[/itex] applied to a particle of the body. That angular momentum vector is at an angle with respect to the axis of rotation.) It's only the sum of the angular momenta that's parallel to the axis. That's why [itex]L = I \omega[/itex] is only true for symmetric bodies. (But [itex]L_z = I \omega[/itex] always holds.)
     
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