The correct options are A
For pure rolling motion of the cylinder, the direction of frictional force is forward and its magnitude is Ma3 B
The maximum value of a
, so that the cylinder performs pure rolling is 3μg C
The acceleration of the center of mass of the cylinder under pure rolling condition for the given a
From the frame of reference, attached to the horizontal board, there is a pseudo force Ma
on the cylinder in the backward direction.
So to prevent slipping, the frictional force acts in the forward direction.
Let this force be f.
So the linear acceleration of the cylinder is, A=(Ma−f)/M in the backward direction.
Torque due to this frictional force is fR and corresponding angular acceleration is given as, α=fR/I=fR/0.5MR2=2f/MR
For pure rolling,
Now the maximum value of frictional force is, μMg
So the maximum value of a is given by, μMg=Mamax/3⟹amax=3μg, for pure rolling to happen.
Now acceleration of the CoM is, A=(Ma−f)/M=2a/3