A cylinder is kept at rest on a sufficiently rough horizontal surface. A horizontal constant force is applied at the centre of mass of cylinder.

Options

A. Acceleration of centre of cylinder is F/m
B. Acceleration of centre of mass of cylinder is 2/3m
C. Force of friction is F/3
D. Force of friction acts in the direction of motion

Answer

B, C

Explanation

Force acting on the cylinder are as shown in figure.

Let linear acceleration acquired by the cylinder is a then for pure rolling its angular acceleration must be alr. Force is applied at the centre hence cylinder tries to slide on the surface and friction acts opposite to its motion and provides necessary torque for rotational motion of cylinder to roll on the surface.
For linear motion along the surface we can write the following equation:
`F-F=ma " "...(i)`
For rotational motion about the centre we can write the following equation
`fr=(1)/(2) mr^(2)((a)/(r)) rArr f=(ma)/(2) " "...(ii)`
Substituting equation (ii) in equation (i) we get
`F-(ma)/(2)=ma rArr a=(2F)/(3m) " "...(iii)`
Substituting equation (ii) in equation (ii) we get
`f=(m)/(2)(2F)/(3m) rArr f=(F)/(3) " "...(iv)`
We can see that only options (b) and (c) are correct.