IIT JEE Advanced 2006 Paper 1

Here, mgsinθ drags the cylinder downwards. Thus, the friction acts upwards and hence opposes translation. Since it is the friction alone which creates a torque about CM (centre of mass) for rotation (anticlockwise), it supports rotation.

Therefore, aCM =

gsinθ

1+( k2 R2 )

=

gsinθ

1+( 1 2 )

= (

2

3

)gsinθ
Hence,
f =

1

2

mg sin θ
On decreasing the value of θ, the frictional force f also decreases.