Ball starting at rest down a frictionless incline

[TyloBabe]

This is not a question from homework, but one about which I was curious. If a ball starts at rest at the top of frictionless plane, would it roll or slide down the incline?

The thing that confuses me is that it seems like it should just slide, since there is no friction and thus nothing to apply a torque. However, if you draw an instantaneous axis of rotation at the contact point between the incline and the ball, with the positive x-axis down the incline, and the y-axis in the direction of the normal force acting on the ball, then it would appear that the gravitational force acting on the ball's center of mass does apply a torque at the center of the ball (so a distance of radius=r away from the origin). Where am I wrong in this assumption?

 

[Driftwood1]

By definition for the motion to be considered perfectly frictionless, the contact surface area between the ball and inclined plane is zero (actually you can get pedantic and invoke some sort of Planck limit here if you wish)

Are you saying that objects cannot undergo a ratational motion from a rest position in a perfectly firctionless system?

 

[TyloBabe]

Not exactly. I'm currently just in the classical mechanics section of freshman physics for engineers, so when we're discussing torques we don't bring into account things like Planck yet. To be perfectly honest, I wouldn't know how Planck applied in this situation; I only have a basic understanding of classical mechanics.

I'm basically just asking is assuming an instantaneous axis of roation at the contact point, as described above, invalid?

Take, for example, a ball simply sitting at rest on a flat frictionless table. If I use an instantaneous axis of rotation at the contact point, an apply a horizontal force at the center of the ball, it would appear that there is a torque with force F acting at a distance r (the center of the ball), and so the ball should rotate. Is there something wrong with using this instantaneous axis of rotation? Just using common sense, it seems like the ball in this case should definitely not rotate.


edit:
I think I may have realized the problem. In the case of the flat surface, what it means to apply a torque about the instantaneous axis means that the ball rotates "about" an infitesimal point, and there really just translates. I think the answer to my original question is a ball would slide down a frictionless incline if it started at rest.

 

[quantum123]

It would slide.
For it to roll , you need to have static friction.

 

[rcgldr]

If the incline is frictionless, then both gravity and the normal force vectors go through the ball's center of mass, and no torque is generated.

 

[K^2]

I think he wants to consider point of contact as the pivot.

Yes, around point of contact there is torque. But all that means is that angular velocity around the point of contact is increasing, and it is so even if the ball is simply sliding.

This is why looking at torque around center of mass is far more useful in this problem. You know how the CM accelerates, since it's given by sum of all forces acting on an object. Knowing that, if you found a torque around CM, you know the object will rotate. Here, torque through CM is zero, so the ball simply slides down.

 

[Driftwood1]

I'm basically just asking is assuming an instantaneous axis of roation at the contact point, as described above, invalid?

Take, for example, a ball simply sitting at rest on a flat frictionless table.

Your problem is an excellent one.
What would be the nature of the contact surface (ie betweem ball and plane) in a perfectly frictionless situation?

You may as well have no plane and have the gravitational force act through the centre of gravity of the ball parallel to the direction of the plane.

In other words a perfect ball in a perfect vacuum that suddenly has a force placed on it at the same angle as the plane.

In this situation the ball will not rotate and move in the same direction as the force

 

[tiny-tim]

Hi TyloBabe!

I agree with rcgldr …

If the incline is frictionless, then both gravity and the normal force vectors go through the ball's center of mass, and no torque is generated.

If there is a friction force, that supplies the torque which rotates the ball.

But no friction, then no torque, no rotation.

You can only take moments (for τ = dL/dt) about either the centre of mass or the centre of rotation.

So, for the friction case, moments about either the centre of mass or the contact point will work,

but for non-friction, there's no reason to believe the contact point is the centre of rotation (and of course it isn't!).

 

[Driftwood1]

you may as well get rid of the incline - it is really irrelevant to the problem - it only dictates the direction.

Its really a ball in a perfect vacuum that has been subjected to a uniform force in a particular direction (and in this case parallel to the inclined plane).

The ball can only move in this direction without spining

 

[TyloBabe]

Thanks to everyone who replied. K^2 and Driftwood's answers were the ones for which I was looking. It's not that the using the contact point as the axis is invalid, it's just that, as K^2 said, the angular velocity about that point is increasing due to the torque, but in this case all that really means is that the ball is translating. Thanks again for the quick responses.

 

[rcgldr]

Yes, around point of contact there is torque.

This isn't taking into account that the ball is accelerating parallel to the plane. Since the plane is frictionless, there can't be any component of force parallel to the plane at the contact point. The ball is accelerating parallel to the plane in response to the component of gravity parallel to the plane, effectively it's in a type of free fall parallel to the plane since the plane is frictionless. The reaction force to this acceleration is equal and opposing to the force of gravity parallel to the plane, so there is no component of force parallel to the plane at the contact point. The net pair of Newton third law forces at the contact point are perpendicular to the plane, so there is no torque. It might be easier to visualize this if you had a horizontal frictionless plane, and had a string attached at the center of mass of the ball to accelerate the ball along the plane, the reaction force to acceleration would be equal and opposing to the tensionin the string, and again no net torque about the contact point.

 

[quantum123]

angular velocity around the point of contact is increasing

Not sure about that. Can you prove it?