Does a Ball Rotate on a Frictionless Surface?

[anonymousphys]

Homework Statement

A ball is moving along a frictionless surface, after a force is applied. Does the surface need friction for the ball to actually rotate? If so why? Is the ball actually moving horizontally or is it just rotating?

Homework Equations

The force applied is that big arrow.
http://img709.imageshack.us/img709/2935/rollign.png
bottom line is the surface.

The Attempt at a Solution

I'm thinking that it doesn't need a a frictional force because the tangential force can already provide the rotation. What I'm thinking is that the ball is not moving unless there is a frictional force, but I'm not sure why.

 

[inutard]

You are quite right in saying that the ball needs a frictional force to move. However, this is only the case when you are applying a tangential force in the diagram shown. This is because the rotation of the ball creates an applied force backwards to the direction of motion, while the reactionary friction force pushes the ball forward. (I wish I could draw a diagram for you)

However, keep in mind that if the force applied is not tangential (ie. it has a horizontal component which acts to push the ball in translational motion as opposed to rotational motion), the ball will move and keep moving forever along a frictionless surface.

 

[anonymousphys]

You are quite right in saying that the ball needs a frictional force to move. However, this is only the case when you are applying a tangential force in the diagram shown. This is because the rotation of the ball creates an applied force backwards to the direction of motion, while the reactionary friction force pushes the ball forward. (I wish I could draw a diagram for you)

However, keep in mind that if the force applied is not tangential (ie. it has a horizontal component which acts to push the ball in translational motion as opposed to rotational motion), the ball will move and keep moving forever along a frictionless surface.

Thanks for the reply, but I'm still a bit confused.

So, I read in a book that for an object rolling down a hill:
http://img85.imageshack.us/img85/3273/rotation.png
So, since in my previous diagram, I did not have a force acting on the center of mass. The object is just rotating and not actually moving in the horizontal distance?
The ball mentioned in the first post is rotating right (assume no friction)?
It has the same force components as the one in my ball rolling down hill diagram (except for the center of mass force). Therefore, the ball is rotating but not actually moving horizontally?

 

[anonymousphys]

still need help would be glad for any replies.

 

[Doc Al]

I'm thinking that it doesn't need a a frictional force because the tangential force can already provide the rotation. What I'm thinking is that the ball is not moving unless there is a frictional force, but I'm not sure why.

Why would the ball not move? There's a net force acting on it. The applied force creates a translational acceleration and, since it's applied off center, also exerts a torque about the center which creates a rotational acceleration. The ball both rotates and translates.

 

[inutard]

But what portion of the force applied would go into rotational acceleration and translational acceleration? In my post I had said that the ball does not translate because such a force (from my understanding and intuition) would be negligibly small.

 

[Doc Al]

But what portion of the force applied would go into rotational acceleration and translational acceleration?

What do you mean 'what portion'? The applied force F provides both the translational acceleration and the rotational acceleration.

In my post I had said that the ball does not translate because such a force (from my understanding and intuition) would be negligibly small.

Again, I don't understand why you would say that. The applied force is not 'negligibly small'.

 

[inutard]

Hmm. So you are saying if i struck a ball tangentially with Force F, the translational acceleration would be F/m = a and the rotational acceleration of the ball would be T/i = a?

 

[Doc Al]

Hmm. So you are saying if i struck a ball tangentially with Force F, the translational acceleration would be F/m = a and the rotational acceleration of the ball would be T/i = a?

Exactly. Just to be clear, alpha = T/I = Fr/I, where I is about the center of mass.

 

[inutard]

Hang on then. Say you applied a certain amount of work to the ball. This would mean that it gains some translational kinetic energy and some rotational kinetic energy which would add up to the work done right?

 

[Doc Al]

Hang on then. Say you applied a certain amount of work to the ball. This would mean that it gains some translational kinetic energy and some rotational kinetic energy which would add up to the work done right?

Right.

 

[inutard]

OK. Say this work was a constant tangential force applied over a certain distance in the same direction.
That would mean the ball had a constant "(F/m) = a" acceleration for a distance d. Leading me to conclude that the ball would be accelerated to a certain vf and so
F*d = 0.5m(vf)^2. (1)
But the ball also has a constant rotational acceleration "T/I = alpha". Leading me to conclude that the ball would be accelerated to a certain wf so
F*d = 0.5I(wf)^2. (2)
But the total work applied is F*d = ma*d= 0.5Iw^2 + 0.5mv^2.
that means F*d doesn't just equal (1) or (2). So what went wrong in my thinking?

 

[Doc Al]

OK. Say this work was a constant tangential force applied over a certain distance in the same direction.
That would mean the ball had a constant "(F/m) = a" acceleration for a distance d. Leading me to conclude that the ball would be accelerated to a certain vf and so
F*d = 0.5m(vf)^2. (1)

That's true. This is a consequence of Newton's 2nd law. Note that 'd' is the distance traveled by the center of mass.

But the ball also has a constant rotational acceleration "T/I = alpha". Leading me to conclude that the ball would be accelerated to a certain wf so
F*d = 0.5I(wf)^2. (2)

It's certainly true that the ball will rotate, but the rotational KE does not equal F*d.

But the total work applied is F*d = ma*d= 0.5Iw^2 + 0.5mv^2.
that means F*d doesn't just equal (1) or (2). So what went wrong in my thinking?

Where you went wrong is in thinking that F*d, where d is the distance traveled by the center of mass, equals the work done on the ball. Realize that since the force is applied tangentially, the point of application moves a greater distance than the center of mass does, so the work done is greater than F*d.

 

[inutard]

Ah! yes I understand now. Thank you very much =]