Is there an instantaneous angular acceleration for a conical pendulm?

by jason12345
Tags: acceleration, angular, conical, instantaneous, pendulm
 P: 111 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?
 P: 483 Can you define what your angle and center refer to?
P: 111
 Quote by olivermsun Can you define what your angle and center refer to?
The angle is between the string and the axis of symmetry the pendulum rotates around.

P: 483

Is there an instantaneous angular acceleration for a conical pendulm?

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."
P: 111
 Quote by olivermsun 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."
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
 P: 483 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.
P: 111
 Quote by olivermsun There is an acceleration (which happens to be toward the center) because the radius vector is not constant. Only the radius magnitude is constant.
I think you mean velocity where you state radius.

 As far as I can tell, the angular velocity is constant if defined around the axis of symmetry.
I agree that angular velocity is constant.
P: 483
 Quote by jason12345 I think you mean velocity where you state radius.
You're right. Change in radius vector per time (velocity) changes.
 P: 29 How does the radius vector not change? Doesn't it's magnitude stay the same, however the direction is changing?
 P: 483 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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