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.