Solve Rotational Motion Homework Problem Req. Acceleration 'a

[zorro]

Homework Statement

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?

Homework Equations

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

but the answer is 2gsin(theta).
Please help.

 

[Doc Al]

observing from the frame of the supporting surface, a pseudo force ma acts up the incline plane.

You have the wrong direction for the pseudo force.

 

[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?

 

[Doc Al]

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?

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

 

[zorro]

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

 

[Doc Al]

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

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.

 

[zorro]

There's a certain friction force acting up the plane that causes it to roll.

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

And applying your idea, what will be the equation?

 

[Doc Al]

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

Neither--it's static friction since there is no sliding.

And applying your idea, what will be the equation?

The equation you had is fine. Just correct the direction of the pseudo force.

 

[zorro]

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.

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)?

 

[Doc Al]

1 thing I wanted to make clear is - Which friction do we increase by pulling the surface upwards- is it rolling friction?

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.

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)?

The point of contact of the cylinder is momentarily at rest with respect to surface.

 

[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.

 

[Doc Al]

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.

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.

 

[zorro]

thanx a lot sir.