Question

A uniform thin cylindrical disk 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 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 $\overrightarrow{V_0}=V_0\hat{i}$. The coefficient of friction is $\mu$.
The maximum value of $V_0$ for which the disk will roll without slipping is :

• A. $\mu g\sqrt{\frac{M}{k}}$
• B. $\mu g\sqrt{\frac{M}{2k}}$
• C. $\mu g\sqrt{\frac{3M}{k}}$
• D. $\mu g\sqrt{\frac{5M}{2k}}$

Answer 1

The correct option is C $\mu g\sqrt{\frac{3M}{k}}$

$Given:$ Mass$=M$ ; radius$=R$ ; velocity$=\overrightarrow{V_o}=V_o\hat{i}$ ; Coefficient of friction$=\mu$

$Solution:$ From (i) $\&$ (ii)

$\Rightarrow -2kx+f=-2f\Rightarrow f=\frac{2k}{3}\times x$

We see that the frictional force depends on $X$.

As $x$ increases, $f$ increases. Also, the frictional

force is maximum at $x=A$ where $A$ is the amplitude of S.H.M.

Therefore the maximum frictional force

$f_{max}=\frac{2k}{3}\times A$

The force should be utmost equal to the

limiting friction ($\mu Mg)$for rolling without

slipping. $\therefore \mu Mg=\frac{2k}{3}\times A$

For S.H.M. Velocity amplitude $=A\omega \therefore V_o=A\omega$

$\therefore V_0=\frac{3\mu Mg}{2k}\omega$ from (iv)

$\therefore V_o=\frac{3\mu Mg}{2k}\times \sqrt{\frac{4k}{3M}}$ from (iii)

$\therefore V_o=\mu g\sqrt{\frac{3M}{k}}$

$So,the$ $correct$ $option:C$