A uniform thin cylindrical disk of mass M and radius R is attaached to two identical massless springs of spring constatn k which are fixed to the wall as shown in the figure. 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 disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity `vecV_0 = vacV_0hati.` The coefficinet of friction is `mu.`

The maximum value of `V_0` for whic the disk will roll without slipping is-

Options

A. `mug_(1)sqrt((M)/(k))`
B. `mugsqrt((M)/(2k))`
C. `mugsqrt((3M)/(k))`
D. `mugsqrt((5M)/(2k))`

Answer

C

Explanation

Using temperature conservation law
`1/2Mv_(0^(2)) (1+1/2) = 2 xx (1)/(2)kx_(1^(2))`
`2kx_(1) - f_("max") = Ma`&
`f_(max) R = ((MR^(2))/(2))alpha`
But `f_(max) = muMg`
`rArr x_(1) = (3muMg)/(2K)`
`rArr 3/4Mv_(0)^(2) = Kx_(1^(2)) = (1)/(K) ((9mu^(2)M^(2)g^(2))/(4))`
`rArr v_(0) = mugsqrt((3M)/(K))`