A uniform thin cylindrical disc of mass `M` and radius `R` is attached to two identical massless springs of spring constant `k` which are fixed to the wall as shown in Fig. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. The unstretched length of each spring is `L`. The disc is initially at its equilibrium position with its centre of mass (`CM`) at a distance `L` from the wall. The disc rolls without slipping with velocity `V_(0)=V_(0)hati`. The coefficient of friction is `mu`.

The net external force acting on the disc when its centre of mass is at displacement `x` with respect to its equilibrium position is

Options

A. `-kx`
B. `-2kx`
C. `-(2x)/(3)`
D. `-(4kx)/(3)`

Answer

D

Explanation

Applying equation of torque about lowest point
`(2Kx) R = ((3)/(2)MR^(2))alpha`
`alphaR = (4Kx)/(3M)`
as there is no slipping
`a= alphaR = (4Kx)/(3M)`
Net force `= Ma = (4Kx)/(3)`
Which is directed opposite to displacement
`F_("net") = (-4Kx)/(3)`