# Rotational Motion

1. Sep 5, 2010

### zorro

1. The problem statement, all variables and given/known data

The acceleration 'a' of the supporting surface required to keep the centre G of the circular cylinder in a fixed position during the motion, if there is no slipping between the cylinder and the support will be?

2. Relevant equations

3. The attempt at a solution

The torque acting about the point of contact of cylinder and surface is equal to the product of angular acceleration and moment of inertia about the point of contact.
observing from the frame of the supporting surface, a pseudo force ma acts up the incline plane.

(mgsin(theta) -ma) x radius=(3mr^2)/2 x a [ I about the point of contact is (3mr^2)/2 ]
on solving, a=2gsin(theta)/5

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2. Sep 6, 2010

### Staff: Mentor

You have the wrong direction for the pseudo force.

3. Sep 6, 2010

### zorro

Since the supporting surface is accelerating down the plane, if we observe from this reference frame, a pseudo force -ma acts on the sphere ...i.e. ma is directed up the plane.
Where am I wrong?

4. Sep 6, 2010

### Staff: Mentor

Why do you think the support surface is accelerating down the plane?

5. Sep 6, 2010

### zorro

It is understood. To avoid the sphere from rolling down, it must accelerate down the plane.

6. Sep 6, 2010

### Staff: Mentor

Think of it this way. If the surface didn't move, the cylinder would roll down the incline. There's a certain friction force acting up the plane that causes it to roll.

To make the center of gravity stationary, we need to increase that friction force. We do that by pulling the surface upwards, dragging the cylinder surface along. We're not trying to stop the cylinder from rotating, just from translating.

7. Sep 6, 2010

### zorro

Which frictional force are you talking about? Is it rolling friction or sliding friction?

And applying your idea, what will be the equation?

8. Sep 6, 2010

### Staff: Mentor

Neither--it's static friction since there is no sliding.
The equation you had is fine. Just correct the direction of the pseudo force.

9. Sep 6, 2010

### zorro

1 thing I wanted to make clear is - Which friction do we increase by pulling the surface upwards- is it rolling friction? If yes, then how can increasing this friction stop the cylinder from translating? And when we're trying to stop the cylinder from translating, does it rotate about the point of contact (fixed)?

10. Sep 6, 2010

### Staff: Mentor

No. Rolling friction has nothing to do with this problem. (We're ignoring it.) The friction that makes a ball roll down an incline, instead of just slide, is static friction.
The point of contact of the cylinder is momentarily at rest with respect to surface.

11. Sep 6, 2010

### zorro

This is what I understood. Tell me if I'm wrong anywhere.

The static frictional force acting up the plane produces a torque about the centre of mass - which causes the cylinder to roll down the plane. When we increase this frictional force by pulling the surface upwards, the tendency of the cylinder to roll down the surface increases but that of sliding down the surface decreases. In a nutshell, the static frictional force is responsible for both translational and rotational motion of the cylinder.

12. Sep 6, 2010

### Staff: Mentor

Good! Here's how I would put it. The static friction produces a torque about the center of mass, causing the cylinder to rotate. The net force on the cylinder, a combination of gravity down the incline and friction, produces the translational acceleration of the cylinder. By accelerating the surface upwards, we increase the friction up to the point where it just equals the gravitational force down the incline, making the net force equal to zero.

13. Sep 7, 2010

### zorro

thanx alot sir.