Rubber on a rotating disk (angular velocity, forces)

[sopetra]

Homework Statement

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?

Homework Equations

http://images.tutorvista.com/cms/formulaimages/83/angular-speed-formula111.PNG

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.

 

[drvrm]

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.

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.

 

[Physics-Tutor]

Well you know that the maximum static friction force on an object before it starts moving is μ_sN (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

 

[haruspex]

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.

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.

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

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.

 

[drvrm]

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.

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.

 

[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.

 

[haruspex]

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.

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