# 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: 421 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: 421

## 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: 421 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: 421
 Quote by jason12345 I think you mean velocity where you state radius.
You're right. Change in radius vector per time (velocity) changes.
 P: 28 How does the radius vector not change? Doesn't it's magnitude stay the same, however the direction is changing?
 P: 421 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).

 Related Discussions Introductory Physics Homework 7 Introductory Physics Homework 1 Classical Physics 0 Introductory Physics Homework 1 Introductory Physics Homework 1