# Rubber on a rotating disk (angular velocity, forces)

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1. Mar 25, 2016

### sopetra

1. The problem statement, all variables and given/known data
We place a rubber on the edge of a rotating disk. What forces act on the rubber? At what angular velocity, why and in what direction will the rubber fly off the disk?
2. Relevant equations

http://images.tutorvista.com/cms/formulaimages/83/angular-speed-formula111.PNG [Broken]
3. The attempt at a solution
I guess there is a centrifugal force that acts against the frictional force between the disk and the rubber. As long as they balance each other out, the rubber seems to be at rest. When the angular velocity reaches a given speed, the centrifugal force will exceed the frictional force and the rubber will leave the disk. I suppose the rubber leaves the disk on a tangential path, but I cannot explain the reasons behind it and have no idea about the exact angular velocity at which the object flies off the disk.
Thank you in advance.

Last edited by a moderator: May 7, 2017
2. Mar 25, 2016

### drvrm

for an observer who is inertial -only 'centripetal force' will be there so that rubber can move on the circular path. draw a free-body diagram.
centrifugal forces are experienced in an accelerated frame/non inertial frame.

3. Mar 26, 2016

### Physics-Tutor

Well you know that the maximum static friction force on an object before it starts moving is μsN (with N=mg). And it is also the centripetal force, an expression of which contains ω.

By equating these two, you should be able to find the maximum angular velocity at which the rubber can stay on the disc.

Try it and submit what you find. If you still have trouble I will add on this thread a link to a video I posted recently that explores this kind of problem.

Good luck

4. Mar 26, 2016

### haruspex

Centripetal force is not 'there'. It is not an applied force. It is that component of net force normal to the velocity required to achieve a specified curvature. In the set up described, there is only one applied force in the inertial frame, friction.
Although that is almost surely what is expected of the student here, it is not quite that simple in reality. If the angular velocity is being increased until the rubber flies off, that implies an angular acceleration. The overall acceleration is therefore not quite radial. The frictional force must be in the direction of acceleration, so is also not quite radial. Thus, sliding will commence a little sooner than predicted by ignoring the angular acceleration.

5. Mar 27, 2016

### drvrm

by 'there' i meant 'to be provided' as i was replying to a post talking about presence of centrifugal force.
thanks for your post , however i agree with your observations.

6. Mar 27, 2016

### sopetra

Thank all of you for your help. I think all the professor expects from us, is this "simplified, ideal" solution of the problem. I equated the equations , and I got the square root of ((μ*g)/r) as a solution for the angular speed above which the rubber will fly off the disk.

7. Mar 27, 2016

### haruspex

Yes, I'm sure that is the desired answer.