Rolling with slipping and linear velocity

[StatusX]

This was a question on a test that I couldn't figure out. You have a hoop, mass M, moment of inertia MR², and an initial angular velocity ω. It is dropped onto a plane with kinetic coefficient of friction μ. What will be its linear velocity when it stops slipping and rolls away smoothly if it doesn't bounce?

Conservation of energy can't be used since it must lose energy to friction. I'm lost on where to even start. Maybe work is extracted from the hoop as it is slipping in an amount of the friction force times the circumferential distance of the slip. This energy, which comes from the rotational kinetic energy of the hoop, must be lost to heat, but at the same time some will be translated into translational kinetic energy, and I don't know how much. Does anyone know how to do this?

 

[Doc Al]

The frictional force does two things; it produces (1) a translational acceleration of the center of mass, and (2) a rotational deacceleration about the center of mass. The translational & rotational acceleration will continue until the hoop rolls without slipping. Set up the kinematic equations and solve for the final speed. (What's the condition for rolling without slipping?)

 

[StatusX]

Yes, but how much of the force contributes to each?

The total force is Mgμ. Are you saying that a torque of MgμR and a horizontal force of Mgμ are both applied to the hoop? Or is it split somehow?

EDIT: I did the problem this way, with the torque and force written above, and I get that the final velocity will be 1/3 Rω. Is this right? It seems like it should depend on μ somehow.

 

[Doc Al]

Yes, but how much of the force contributes to each?

The total force is Mgμ. Are you saying that a torque of MgμR and a horizontal force of Mgμ are both applied to the hoop? Or is it split somehow?

There is only one force acting on the hoop! But it has no trouble creating both a rotational and translational acceleration! The force just acts. (I don't really know what you mean by a force splitting; perhaps you are thinking about the energy splitting between translational and rotational KE?)

EDIT: I did the problem this way, with the torque and force written above, and I get that the final velocity will be 1/3 Rω. Is this right? It seems like it should depend on μ somehow.

I get a different answer; try it again. No, it does not depend on μ.

 

[StatusX]

I guess the reason I'm confused is that the force is always perpendicular to the radius. When I asked how the force split, I was referring to how in most torque problems, there is a component perpendicular to the radius that rotates the body and one parellel that moves the center of mass. And you're right, I made a mistake, the answer should have been 1/2 Rω.

 

[StatusX]

I'm still not really clear on this. If you apply a force to an object at the center of mass over a certain distance, it acquires a certain speed. Now you're saying if you apply this force somewhere else on the body, it will acquire the same linear speed but also will gain angular momentum. But aren't you doing the same work each time?

 

[Doc Al]

If you apply a force to an object at the center of mass over a certain distance, it acquires a certain speed.

A net force applied anywhere on an object will cause an acceleration of its center of mass. If the center of mass moves a certain distance, it will have a corresponding speed.

Now you're saying if you apply this force somewhere else on the body, it will acquire the same linear speed but also will gain angular momentum.

In this case, that same force will impart both an acceleration of the center of mass and an angular acceleration about the center of mass. If the center of mass moves the same distance as before, it will acquire the same speed as before.

But aren't you doing the same work each time?

An excellent and subtle question! The work done equals the applied force times the displacement of the point of application. But the speed of the contact point depends on where the force is applied: The edge of the hoop moves faster than the center!

Imagine your hoop lying flat on a frictionless surface. Now apply the force F to the center. The acceleration is a = F/m. If you apply that force for a distance D, the work you do is W = F*D: All of that work goes into translational KE since there is no rotation.

Now apply the force F at the edge of the hoop. The acceleration of the center is the same (a = F/m), but the acceleration of the contact point is twice that. So to get the center of the hoop to travel the same distance D, you have to do twice as much work, since you are rotating the hoop as well as translating it.