Maximum Velocity for Pure Rolling of Spring Attached Disk

To find the maximum initial velocity $V_0$ for which the disk rolls without slipping, we analyze the equations of motion for the disk undergoing Simple Harmonic Motion (SHM).

1. Equations of Motion

Let $x$ be the displacement of the center of mass of the disk from its equilibrium position. Two springs, each of constant $k$, are attached to the disk, giving a total spring force of $-2kx$.

Let $f_s$ be the static friction force acting on the disk at the contact point in the direction of motion.

For the translational motion of the center of mass:
$M a = -2kx + f_s \quad \text{--- (1)}$

For the rotational motion about the center of mass:
$\tau = -f_s R = I \alpha$

Since the disk is a uniform thin cylinder, its moment of inertia is $I = \frac{1}{2} M R^2$. Substituting this gives:
$-f_s R = \left(\frac{1}{2} M R^2\right) \alpha \implies \alpha = -\frac{2 f_s}{M R}$

For pure rolling without slipping, the condition connecting linear acceleration $a$ and angular acceleration $\alpha$ is:
$a = \alpha R$

Substituting $\alpha$:
$a = -\frac{2 f_s}{M} \implies f_s = -\frac{M a}{2} \quad \text{--- (2)}$

2. Angular Frequency of SHM

Substitute $f_s = -\frac{M a}{2}$ into equation (1):
$M a = -2kx - \frac{M a}{2}$
$\frac{3}{2} M a = -2kx \implies a = -\left(\frac{4k}{3M}\right) x$

This represents SHM of the form $a = -\omega^2 x$, where the angular frequency $\omega$ is:
$\omega = \sqrt{\frac{4k}{3M}}$

3. Maximum Friction Force

From equation (2), the magnitude of the static friction force at a displacement $x$ is:
$|f_s| = \frac{M |a|}{2} = \frac{M}{2} \left(\frac{4k}{3M} |x|\right) = \frac{2k}{3} |x|$

The friction force reaches its maximum magnitude when the displacement $x$ reaches the amplitude $A$ of the motion:
$(f_s)_{\max} = \frac{2k}{3} A$

4. Condition for Rolling Without Slipping

To prevent slipping, the maximum friction force required must not exceed the maximum possible static friction force:
$(f_s)_{\max} \le \mu N = \mu M g$
$\frac{2k}{3} A \le \mu M g \implies A \le \frac{3 \mu M g}{2k}$

5. Maximum Initial Velocity $V_0$

At the equilibrium position ($x = 0$), the speed of the disk is maximum and equals the amplitude times the angular frequency:
$V_0 = \omega A \implies A = \frac{V_0}{\omega} = V_0 \sqrt{\frac{3M}{4k}}$

Substituting $A$ into the condition for no slipping:
$V_0 \sqrt{\frac{3M}{4k}} \le \frac{3 \mu M g}{2k}$

Solving for $V_0$:
$V_0 \le \frac{3 \mu M g}{2k} \sqrt{\frac{4k}{3M}}$
$V_0 \le \frac{3 \mu M g}{2k} \cdot \frac{2}{\sqrt{3}} \sqrt{\frac{k}{M}}$
$V_0 \le \mu g \sqrt{\frac{3M}{k}}$

Thus, the maximum value of $V_0$ for which the disk will roll without slipping is:
$(V_0)_{\max} = \mu g \sqrt{\frac{3M}{k}}$

Correct Answer: C