Is there an instantaneous angular acceleration for a conical pendulm?

jason12345

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?

olivermsun

Can you define what your angle and center refer to?

jason12345

The angle is between the string and the axis of symmetry the pendulum rotates around.

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."

jason12345

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

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.

As far as I can tell, the angular velocity is constant if defined around the axis of symmetry.

jason12345

I think you mean velocity where you state radius.

I agree that angular velocity is constant.

olivermsun

You're right. Change in radius vector per time (velocity) changes.

greentlc

How does the radius vector not change? Doesn't it's magnitude stay the same, however the direction is changing?

olivermsun

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).