Can you have angular acceleration without centripetal acceleration?

[21joanna12]

Homework Statement

My guess is no because if you have a ball on a string, for there to be angular acceleration, the angular velocity must increase so you need an increasing tangential speed, so your centripetal acceleration must increase (α=v²/r)... but I'm not sure.


One other question: is there such a thing as tangential acceleration? I was watching a video that said that:
tangential velocity= r(angular velocity)
tangential acceleration = r(angular acceleration)

Isn't tangential acceleration just the rate of change of tangential velocity, which is the centripetal acceleration? If so, how can you equate the tangential and angular accelerations? Is the tangential acceleration the cross product of displacement and angular acceleration?

Very confused! Thank you in advance for any replies and I apologise for the mass of questions. I looked through several websites and PF posts before posting this but couldn't find anything. If anyone finds a good link on this topic I would also be very grateful!

:)

 

[BvU]

Yes, angular acceleration can occur: anything solid that starts rotating experiences angular acceleration ! The cause is called a torque. Just like with F = m a you have T = I ##\alpha##.

So if you get on your bike and step on the pedal, there you go !

In the string / ball case you describe there is no way to exercise a torque with the string (no stiffness -- a string can only transfer force along the string), so to get that going your hand (or whatever is holding the end) has to make little circles to get the ball to describe big circles.

Questions on the video would be easier to comment on if you provide the link. The video probably didn't talk -- and even if it did it wouldn't talk opening bracket - closing bracket. It would say something like

tangential velocity is equal to r times angular velocity
tangential acceleration = r times angular acceleration

Isn't tangential acceleration just the rate of change of tangential velocity, ...

I would say yes. It is the component of the acceleration that is perpendicular to the radius vector.

..rate of change of tangential velocity, which is the centripetal acceleration? ...

Definitely not. Now I need your help to help: how much math are you comfortable with ? (time derivatives, cross products, etc).

Is the tangential acceleration the cross product of displacement and angular acceleration?

In the mean time: "displacement" is the wrong term here. "Position vector" would be a lot better.

 

[rcgldr]

If a disk was rotating clockwise and then experienced a constant counter clockwise torque, the disk would experience angular deceleration, then angular acceleration in the counter clockwise direction. During the instant in time that the disk is not rotating, then there is zero centripetal acceleration.

For tangental acceleration, a puck sliding on a frictionless table and connected to string is a better example. Tangental acceleration is acceleration in the same direction as the velocity of the puck, and centripetal acceleration is perpendicular to the velocity of the puck. If the puck was sliding in a circle, and the string was pulled through a hole in the middle of the frictionless table, during the time the string is being pulled, the ball experiences both tangental and centripetal acceleration. The ball is moving in a spiral path (in this case inwards), so the string is not perpendicular to the puck during the time the string is being pulled inwards or released outwards. Example image - the short lines are perpendicular to the pucks path, and you can see that the string is not perpendicular:

 

[21joanna12]

Yes, angular acceleration can occur: anything solid that starts rotating experiences angular acceleration ! The cause is called a torque. Just like with F = m a you have T = I ##\alpha##.

So if you get on your bike and step on the pedal, there you go !

In the string / ball case you describe there is no way to exercise a torque with the string (no stiffness -- a string can only transfer force along the string), so to get that going your hand (or whatever is holding the end) has to make little circles to get the ball to describe big circles.

Questions on the video would be easier to comment on if you provide the link. The video probably didn't talk -- and even if it did it wouldn't talk opening bracket - closing bracket. It would say something like

tangential velocity is equal to r times angular velocity
tangential acceleration = r times angular acceleration

I would say yes. It is the component of the acceleration that is perpendicular to the radius vector.

Definitely not. Now I need your help to help: how much math are you comfortable with ? (time derivatives, cross products, etc).

In the mean time: "displacement" is the wrong term here. "Position vector" would be a lot better.

Thank you for your reply! The video link is

https://www.youtube.com/watch?v=uJDyurbKpbs&feature=youtu.be

and the part about tangential velocity/acceleration begins 6 minutes in. Regarding math, I'm comfortable with time derivatives and general calculus; cross products I have only looked at in passing, but it doesn't seem too difficult a concept...

With regards to angular acceleration with no centripetal acceleration, could you say that this happens when a pendulum is momentarily stationary when it reaches its maximum height?

Thank you :)

 

[21joanna12]

Extending the last comment I made on my previous post, is a pendulum an example of centripetal acceleration and angular acceleration together, or centripetal acceleration and tangential acceleration together... or all three??

 

[21joanna12]

... I think I may get it now. Please correct me if I am wrong, but are tangential and angular acceleration analagous, although not the same thing, because angular acceleration deals with how fast you cover angles but tangential acceleration deals with how fast you actually travel, and hence the tangential acceleartion is also dependent on the radius by a=rα.

So if you have a non-zero tangential acceleration, you also must have a non-zero angular acceleration, although they are not the same thing.

In the case of a pendulum, you have a tangential force of mgsinθ (where θ is the angle between the string and the downwards vertical) which also creates a torque because it is always perpendicular to the string, and you also have a centripetal force. However when the pendulum passes through the equilibrium position, the tangential force is zero and the centripetal force is maximum, and at the highest points the centripetal force is zero and the tangential force is maximum. Therefore you have a tangential acceleration (and so angular accceleration equal to the tangential accceleration divided by the length of the string) at all points except when the bob is at equilibrium, and you have a centripetal acceleration of v²/r at all points except when the bob is at its highest points.

Sorry for the really long comment, but I think I may be getting it... hopefully.

 

[haruspex]

... I think I may get it now. Please correct me if I am wrong, but are tangential and angular acceleration analagous, although not the same thing, because angular acceleration deals with how fast you cover angles but tangential acceleration deals with how fast you actually travel, and hence the tangential acceleartion is also dependent on the radius by a=rα.

So if you have a non-zero tangential acceleration, you also must have a non-zero angular acceleration, although they are not the same thing.

In the case of a pendulum, you have a tangential force of mgsinθ (where θ is the angle between the string and the downwards vertical) which also creates a torque because it is always perpendicular to the string, and you also have a centripetal force. However when the pendulum passes through the equilibrium position, the tangential force is zero and the centripetal force is maximum, and at the highest points the centripetal force is zero and the tangential force is maximum. Therefore you have a tangential acceleration (and so angular accceleration equal to the tangential accceleration divided by the length of the string) at all points except when the bob is at equilibrium, and you have a centripetal acceleration of v²/r at all points except when the bob is at its highest points.

Sorry for the really long comment, but I think I may be getting it... hopefully.

I believe you've got it.

 

[BvU]

I think you are "getting it" quite well.
Watched the video. Have some sour comments because I'm too old, but wholeheartedly subscribe to mr. Palmer's flipping concepts.
Find the part around 6 minutes rather murky and hard to follow. He's constantly working with scalars (and doing so correctly, to be sure), and I dearly miss a vector treatment (which may well be in a subsequent lesson).

subscripts radial and tangential are just like subscripts x and y in a cartesian coordinate system: they denote two perpendicular directions along which you can decompose the velocity (a vector) and the acceleration (also a vector). The radial/tangential decomposition is hooked up to a position, the x,y is hooked up to a (generally fixed) origin.

You've seen the formulas come by (at least when Tom steps aside to show what he's been mumbling and writing ;-) and I think they are clear enough.

I agree with your pendulum analysis, except that I advise to stay away from terms like "centripetal acceleration". But what you write is correct. In fact you also have "centripetal acceleration of v²/r when the bob is at its highest points: v = 0 there.