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Is there an instantaneous angular acceleration for a conical pendulm?

  1. May 3, 2012 #1
    For a conical pendulum, there is an instantaneous centripetal acceleration. Does this mean there is an instantaneous angular acceleration of the pendulum towards the center?
     
  2. jcsd
  3. May 3, 2012 #2

    olivermsun

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    Can you define what your angle and center refer to?
     
  4. May 3, 2012 #3
    The angle is between the string and the axis of symmetry the pendulum rotates around.
     
  5. May 3, 2012 #4

    olivermsun

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    I see, you're talking about a pendulum which swings about the center axis in a cone.

    Your angle, as defined, rotates with the pendulum string and remains constant, so I would say "no."
     
  6. May 3, 2012 #5
    Thanks for your reply, although I disagree with it :) I could also argue that the radius of the circular motion is constant and so there isn't an acceleration towards the centre - but there is: v^2/r
     
  7. May 3, 2012 #6

    olivermsun

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    There is an acceleration (which happens to be toward the center) because the radius vector is not constant. Only the radius magnitude is constant.

    As far as I can tell, the angular velocity is constant if defined around the axis of symmetry.
     
  8. May 3, 2012 #7
    I think you mean velocity where you state radius.

    I agree that angular velocity is constant.
     
  9. May 3, 2012 #8

    olivermsun

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    You're right. Change in radius vector per time (velocity) changes.
     
    Last edited: May 3, 2012
  10. May 3, 2012 #9
    How does the radius vector not change? Doesn't it's magnitude stay the same, however the direction is changing?
     
  11. May 4, 2012 #10

    olivermsun

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    The radius does change (dr/dt is nonzero), so that there is a velocity, but he was talking about whether or not there is an acceleration. There is, since d^2/dt^2 = dv/dt is nonzero. A changing radius vector isn't enough to imply an acceleration, although it is enough that the magnitude stays the same while the direction is changing (as you say).
     
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