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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?

Attempt-
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.
?
```
7 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										Dear student,
You have the wrong direction for the pseudo force.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.

We are all  IITians and here to help you in your IIT  JEE preparation.  All the best.

Sagar Singh
B.Tech IIT Delhi
sagarsingh24.iitd@gmail.com

```
7 years ago
sanki kurli
18 Points
```										the cylinder is rolling over the surface without slipping and the centre 'G' does not move ....
that means the surface should also move in opp. direction of the cylinder and with the same acc. as the cylinder.
now the eqn.
torque equaton         .....     (mg X r)=I*alpha               { ...alpha=ang. acceleration}
=> mg*r*sin(theta)=m(r^2)/2*alpha     { you should take I of the center & not from the point of contact}

=>alpha=2mg*sin(theta)/r
its given pure rolling...so..   a=alpha*r
=>a=2mg*sin(theta)

```
7 years ago
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