Calculating Speed for Sliding and Rotating Sphere with Friction

[daudaudaudau]

Homework Statement

A sphere(mass m) is sliding on a frictionless surface until it meets a surface with friction(coefficient mu). This causes the sphere to start rotating, while it still slides. Calculate the speed of the sphere while it is both sliding and rotating on the surface with friction.

The Attempt at a Solution

My first attempt was just to consider a constant frictional force acting on the sphere, and hence the speed is

V=V0-mu*g*t

But then I heard that I also had to take into account that the sphere starts to rotate, and then the above is incorrect.

I'd say that the frictional force on the sphere is the same no matter how fast it is rotating. Can anyone help me?

 

[HallsofIvy]

Don't worry about the individual forces- look at the energy.

As long as the sphere is sliding, it's energy is its tranlational kinetic energy: 1/2 mv².

When the sphere starts rolling without slipping, it has both translational kinetic energy and rotational kinetic energy which must add to give the above. When "rolling without slipping, note that when the sphere has made one complete rotation, which takes t= 2\pi/\omega seconds, where \omega is the angular speed, in radians per second, the center of the sphere has moved a distance equal to the circumference of the sphere, 2\pi R. At speed v, that will take time t= 2 \pi R/v so that \omega= v/R. Write the rotational kinetic energy in terms of v, set the sum of the two kinetic energies equal to the initial kinetic energy and solve for v as a function of the original sliding speed.

You can't calculate both without some additional information.

 

[Doc Al]

A sphere(mass m) is sliding on a frictionless surface until it meets a surface with friction(coefficient mu). This causes the sphere to start rotating, while it still slides. Calculate the speed of the sphere while it is both sliding and rotating on the surface with friction.

For clarity, we can view the motion of the sphere in three phases:
(A) The sphere slides on the frictionless surface (no rotation)
(B) The sphere meets the frictioned surface and begins rotating as well as sliding
(C) The sphere reaches a speed such that it rolls without slipping

I assume that you only care about phase B? If so, the question is as easy as you think. (Usually one is asked to find the speed at which phase C--rolling without slipping--begins; that takes a little more effort.) Be sure you have described the problem exactly as given.

My first attempt was just to consider a constant frictional force acting on the sphere, and hence the speed is

V=V0-mu*g*t

Looks good to me.

But then I heard that I also had to take into account that the sphere starts to rotate, and then the above is incorrect.

The translational acceleration of the sphere is entirely due to the force of friction, so that equation holds during phase B.

I'd say that the frictional force on the sphere is the same no matter how fast it is rotating.

That's correct. As long as the sphere continues slipping against the frictioned surface, the friction force is the same.

Don't worry about the individual forces- look at the energy.

As long as the sphere is sliding, it's energy is its tranlational kinetic energy: 1/2 mv².

In phase A, the sphere's energy is pure translational KE. In phase B, the energy is both translational and rotational KE. (Note that the total KE decreases during phase B due to the work done by friction.) In phase C, the energy is both translational and rotational KE and remains constant.

The sphere loses total KE due to friction as it slips along the surface in phase B.

 

[daudaudaudau]

Okay. Actually there is a second part of the question where I have to determine how much heat is generated during phase B, when the sphere is sliding. In this question, do I have to take into account that the sphere is rotating? Otherwise I think the solution is the following:

The angular speed of the sphere is:
omega=5/(2*R)*g*mu*t

Which means that phase C starts when v=R*omega:
v0-mu*g*t=R*(5/(2*R)*g*mu*t) => t=2/7*v0/(mu*g)

Which means that(in phase B) the sphere slides a distance of

x = v0*t - 1/2*mu*g*t^2 = v0*(2/7*v0/(mu*g)) - 1/2*mu*g*(2/7*v0/(mu*g))^2 = 12/49*v0^2/(mu*g)

The total work due to friction is
Wf = Ff*x = m*g*mu*x = 12/49*v0^2*m

Or am I missing that the sphere is rotating? Because I guess if the sphere was spinning really fast, it would generate more heat.

 

[Doc Al]

Okay. Actually there is a second part of the question where I have to determine how much heat is generated during phase B, when the sphere is sliding. In this question, do I have to take into account that the sphere is rotating?

Absolutely! To find the work done by friction directly, you need the force times the distance that the contact area slipped. But that's the hard way.

Otherwise I think the solution is the following:

The angular speed of the sphere is:
omega=5/(2*R)*g*mu*t

Good.

Which means that phase C starts when v=R*omega:
v0-mu*g*t=R*(5/(2*R)*g*mu*t) => t=2/7*v0/(mu*g)

Good.

Which means that(in phase B) the sphere slides a distance of

x = v0*t - 1/2*mu*g*t^2 = v0*(2/7*v0/(mu*g)) - 1/2*mu*g*(2/7*v0/(mu*g))^2 = 12/49*v0^2/(mu*g)

This is the distance that the sphere's center travels. It's not the slipping distance over which the friction force is applied.

The total work due to friction is
Wf = Ff*x = m*g*mu*x = 12/49*v0^2*m

What you've calculated here can be called pseudo-work, which just tells you the change in translational KE. The real work is applied at the point of contact of sphere and surface--to calculate the work directly you need to consider how the surfaces slip.

Or am I missing that the sphere is rotating?

Yes, you are missing something in your attempted calculation of the work done by friction since the sphere rotates as well as translates.

But rather than try to calculate the work directly, why not just calculate the change in KE? The mechanical energy lost will equal the amount of thermal energy generated by friction. That's the easy way!

 

[daudaudaudau]

So the kinetic energy before phase B is

Kb = 1/2*m*v0^2

After phase C there is kinetic energy due to the translation and due to the rotation

Ke = 1/2*m*v^2+1/2*I*omega^2

where the moment of inertia is 2/5*m*R^2 for the sphere.

From my previous equations the speed v in phase C is

v=5/7*v0

and

omega = v/R = 5/(7R)*v0

hence

Ke = 1/2*m*(5/7*v0)^2+1/2*2/5*m*R^2*(5/7*v0/R)^2 = 5/14*m*v0^2

Which makes the work done by friction

W=1/7*m*v0^2
Right? Thank you so far, you are great !

 

[Doc Al]

Looks good to me.