Rolling puck on a spinning disk

[Mr. T]

Homework Statement

A small puck is placed on a disk spinning with angular speed OMEGA around axis perpendicular to the disk and passing through its center. The coefficient of static friction between the disk and the puck is MU. At what distance(s) from the center of the disk will the puck be at rest?

Homework Equations

Rotational Inertia of Rotating Disk:

I=(1/2)*M*R^2

Force of Static Friction

F=MU*(NORMAL FORCE)

The Attempt at a Solution

By "at rest" I believe it means when the puck is rolling ON the spinning disk in a way that keeps it in the same place. But I have no idea how to determine the distance the puck needs to be at for that to happen.

 

[RoyalCat]

There are two possible and equivalent approaches to this problem. The first is just about applying Newton's Second Law.

Look at the puck on the spinning disk, what forces act on it? In order for it to be rotating with angular speed $$\omega$$ at a distance $$R$$ from the center of the disk, what must the force towards the center of the disk be?

The other approach involves fictitious forces. If we look at the puck from inside the rotating system, we see it at rest, but we also observe an additional force acting on it. It is "falling" outwards with a centrifugal acceleration.

The formulation of Newton's Second Law for both these systems is equivalent, find either, and apply the limitations on the static friction force, and you'll find the possible distance from the center.

 

[kerimek]

You are correct in your understanding of the problem. The goal is to determine the distance from the center of the disk where the puck will remain at rest while the disk is spinning. To solve this problem, we can use the concept of rotational equilibrium, where the sum of the torques acting on an object is equal to zero. In this case, the torque from the friction force must be equal and opposite to the torque from the angular velocity of the disk.

We can set up the equation as follows:

Torque from friction = Torque from rotation

MU*(NORMAL FORCE)*d = (1/2)*M*R^2*OMEGA

Where d is the distance from the center of the disk to the point where the puck is at rest. We can rearrange this equation to solve for d:

d = (1/2)*R^2*OMEGA/(MU*NORMAL FORCE)

To determine the normal force, we can use the fact that the sum of the forces in the vertical direction must be equal to zero. This means that the normal force is equal to the weight of the puck, which is equal to its mass multiplied by the acceleration due to gravity (g).

Therefore, our final equation becomes:

d = (1/2)*R^2*OMEGA/(MU*M*g)

This equation tells us that the distance at which the puck will remain at rest is dependent on the radius of the disk, the angular speed of the disk, the coefficient of static friction, the mass of the puck, and the acceleration due to gravity. We can use this equation to determine the distance for any given set of values.

I hope this helps to clarify the problem and provide a solution. Let me know if you have any further questions.