A uniform solid cylinder of mass `m` rests on two horizontal planks. A thread is wound on the cylinder. The hanging end of the thread is pulled vertically down with a constant force `F`.

Find the maximum magnitude of the force `F` which may be applied without bringing about any sliding of the cylinder, if the coefficient of friction between the plank and the cylinder is equal to `mu`. What is the maximum acceleration of the centre of mass over the planks?

Options

A. Since horizontal acceleration is provided by the friction acting on the cylinder, its translation kinetic energy `(1/2mv^(2))` is equal to work done by this friction,
B. Since moment about instantaneous axis of rotation is produced by force `F`. kinetic energy `(1/2 Iomega^(2))` is equal to work done by `F` where `I` is moment of inertia about the instantaneous axis of rotation and `omega` is the-angular velocity.
C. Since cylinder is moving, energy is lost against friction.
D. Work done by `F +` work done by friction on cylinder `="total" KE` of the cylinder.

Answer

B, D

Explanation

When force `F` acts vertically downwards on the thread, the thread unwinds and the cylinder rotates clockwise. Hence, its surface tries to slip over the planks in a backward direction. Hence, a friction comes into existence and acts on the cylinder in a forward direction (rightward direction). It is the only horizontal force acting on the cyliner. Due to the friction, cylinder experiences a rightward translational acceleration. But the point at which the friction acts remains always at rest, therefore no energy is lost against the friction.
since the work is done by `F` and no energy is lost against the friction therefore total kinetic enegy of the cylinder at any instant is equal to work done by `F`. Hence options a and c are incorrect and option d is correct.
If `I` is the moment of inertia of a body about its instantaneous axis of rotation, then the total `KE` of the rotating body is equal to `Iomega^(2)//2`. Hence option b is correct.