Mechanics - marble rolling in dish

  • Thread starter micromass
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  • #26
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By the way if in a general situation we have a circular object rolling without slipping along some surface, and we define the angle for self rotation of the object with respect to a fixed vertical axis can we always claim the v_cm = b*dtheta / dt condition on top of whatever else may also crop up (e.g. in this case v_cm = (R - b)dphi / dt as well due to the cm sweeping out a circle of radius R - b)?
 
  • #27
ehild
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I think it holds for points on other curves, R meaning the radius of curvature at the specified point. But the angular acceleration has to take into account that R also changes as the sphere rolls on the surface.

ehild
 
  • #28
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I just meant does the v_cm = b*dtheta/dt part hold true in general for a fixed vertical axis and theta being measured relative to that at every instant, forget what I said about the R stuff
 
  • #29
ehild
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I think, considering rotation about the point of contact, the CM turns with the same angle as the point which was in contact with the ground turns about the CM. Am I right? :)

ehild
 
  • #30
TSny
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I just meant does the v_cm = b*dtheta/dt part hold true in general for a fixed vertical axis and theta being measured relative to that at every instant, forget what I said about the R stuff

I think, considering rotation about the point of contact, the CM turns with the same angle as the point which was in contact with the ground turns about the CM. Am I right? :)
ehild


That sounds right to me. In the inertial lab frame, the point of contact is the instantaneous center of rotation of the marble. So, the line connecting the point of contact with the CM would be rotating with an angular velocity ω = Vcm/b. In a nonrotating frame of reference moving with the CM, the CM would of course be at rest and the point of contact would be moving ("backward") at the same speed Vcm that the CM was moving in the lab frame. So, in this frame the line connecting the CM to the point of contact would be rotating with respect to the center of mass with the same angular speed ω. In the CM frame, I think it is clear that this angular speed is ##\dot\theta##.
 

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