# Is there an instantaneous angular acceleration for a conical pendulm?

1. May 3, 2012

### 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?

2. May 3, 2012

### olivermsun

Can you define what your angle and center refer to?

3. May 3, 2012

### jason12345

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

4. May 3, 2012

### 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."

5. May 3, 2012

### 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

6. May 3, 2012

### 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.

7. May 3, 2012

### jason12345

I think you mean velocity where you state radius.

I agree that angular velocity is constant.

8. May 3, 2012

### olivermsun

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

Last edited: May 3, 2012
9. May 3, 2012

### greentlc

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

10. May 4, 2012

### 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).