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

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.