Question

A uniform thin cylindrical disc of mass $M$ and radius $R$ is attached to two identical massless spring constant $k$ which are fixed to the wall as shown in 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 the horizontal plane. The unstretched length of each spring is $L$. The disc is initially at its equilibrium position with its centre of mass $\left(CM\right)$ at a distance $L$ from the wall. The disc rolls without slipping with velocity $V_0=V_0\hat{i}$. The coefficient of friction is $\mu$.
The maximum value of $V_0$ for which the disc 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 $\mu g\sqrt{\frac{3M}{k}}$

By Newton's 2nd law, we have

$2kx-f_{max}=ma$

$2kxr=I_p\alpha$ -(i.)

$f_{max}=\mu mg$ -(ii.)

From (i.) and (ii.), we have

$x=\frac{3}{2}\frac{\mu mg}{2k}$

$\frac{1}{2}\left(2k\right)x^2=\frac{1}{2}{I}_p{\omega }^2$

Hence, we have

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