Rubber on a rotating disk (angular velocity, forces)

In summary, a rubber placed on the edge of a rotating disk will stay in place as long as the centrifugal force and frictional force between the disk and the rubber are balanced. When the angular velocity reaches a certain speed, the centrifugal force will exceed the frictional force and the rubber will fly off the disk on a tangential path. The maximum angular velocity at which the rubber can stay on the disk can be found by equating the maximum static friction force and the centripetal force, giving the square root of ((μ*g)/r) as the solution.
  • #1
sopetra
2
0

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


b303d997c132f7e635d4a0ecd24488f5.png

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

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.
 
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  • #2
sopetra said:
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.
 
  • #3
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
drvrm said:
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.
Physics-Tutor said:
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.
 
  • #5
haruspex said:
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.
 
  • #6
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
image431.gif
, 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
sopetra said:
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
image431.gif
, 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.
 

What is the relationship between rubber and angular velocity on a rotating disk?

The rubber on a rotating disk experiences a change in angular velocity as the disk spins. This is due to the friction between the rubber and the surface of the disk which causes the rubber to rotate at the same speed as the disk.

How does the angular velocity of the rubber affect the forces acting on it?

The angular velocity of the rubber on a rotating disk is directly related to the magnitude of the forces acting on it. As the angular velocity increases, the forces acting on the rubber also increase. This is because a higher angular velocity means a higher speed, which results in a greater centrifugal force acting on the rubber.

What is the significance of the coefficient of friction in the rubber-disk system?

The coefficient of friction between the rubber and the disk is an important factor in understanding the forces acting on the rubber. It determines the level of friction between the two surfaces, which in turn affects the angular velocity and forces acting on the rubber.

How does the size and shape of the rubber affect its performance on a rotating disk?

The size and shape of the rubber can significantly impact its performance on a rotating disk. A larger or thicker rubber will have a greater surface area in contact with the disk, resulting in a higher frictional force and potentially a higher angular velocity. Additionally, the shape of the rubber can affect its ability to grip the surface of the disk and maintain a consistent angular velocity.

What factors can cause the rubber to slip or lose grip on the rotating disk?

Several factors can cause the rubber to slip or lose grip on the rotating disk, including an insufficient coefficient of friction, a decrease in angular velocity, and external forces acting on the rubber. Additionally, the presence of debris or moisture on the disk surface can also affect the grip of the rubber.

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