Net External Force on Disk in Rolling Spring System

To find the net external force acting on the disk when its centre of mass is at a displacement $x$ from its equilibrium position, we analyze the translational and rotational dynamics of the rolling disk.

1. System Parameters and Kinematics

• Mass of the disk = $M$
• Radius of the disk = $R$
• Moment of inertia of the uniform disk about its center of mass:
  $I = \frac{1}{2} M R^2$
• The disk is attached to two identical springs of spring constant $k$ connected in parallel. Thus, the effective spring constant is:
  $k_{\text{eff}} = 2k$
• The springs are attached symmetrically to the axle passing through the center of mass (CM). Therefore, the restoring force due to the springs acts through the CM and exerts zero torque about the CM.

Let $x$ be the horizontal displacement of the center of mass from its equilibrium position, and $a = \ddot{x}$ be its linear acceleration.

For rolling without slipping on the horizontal surface, the relation between the linear acceleration $a$ and angular acceleration $\alpha$ about the CM is:
$\alpha = -\frac{a}{R}$
(where a positive displacement $x$ and acceleration $a$ are directed away from the wall, and counter-clockwise rotation is taken as positive).

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2. Equations of Motion

Translational Motion of the CM:

The forces acting on the disk horizontally are the restoring force of the two springs and the static friction force $f$ at the contact point with the ground.
$F_{\text{net}} = M a = F_{\text{spring}} + f$

Since $F_{\text{spring}} = -2kx$:
$M a = -2kx + f \quad \text{--- (Equation 1)}$

Rotational Motion about the CM:

Taking torque about the center of mass, only the friction force $f$ provides a torque:
$\tau_{\text{CM}} = -f R$

Using Newton's second law for rotation ($\tau_{\text{CM}} = I \alpha$):
$-f R = I \alpha$

Substitute $I = \frac{1}{2} M R^2$ and $\alpha = -\frac{a}{R}$:
$-f R = \left(\frac{1}{2} M R^2\right) \left(-\frac{a}{R}\right)$
$-f R = -\frac{1}{2} M R a \implies f = -\frac{1}{2} M a \quad \text{--- (Equation 2)}$

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3. Solving for the Net External Force

Substitute $f = -\frac{1}{2} M a$ from Equation 2 into Equation 1:
$M a = -2kx - \frac{1}{2} M a$

Rearranging the terms:
$M a + \frac{1}{2} M a = -2kx$
$\frac{3}{2} M a = -2kx$
$M a = -\frac{4kx}{3}$

Since the net external force acting on the system is $F_{\text{net}} = M a$:
$F_{\text{net}} = -\frac{4kx}{3}$

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4. Alternative Verification using Energy Method

The total mechanical energy $E$ of the system is the sum of translational kinetic energy, rotational kinetic energy, and potential energy of the springs:
$E = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 + \frac{1}{2} (2k) x^2$

Substituting $I = \frac{1}{2} M R^2$ and $v = \omega R$:
$E = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2 + k x^2$
$E = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 + k x^2 = \frac{3}{4} M v^2 + k x^2$

Since mechanical energy is conserved in pure rolling ($\frac{dE}{dt} = 0$):
$\frac{dE}{dt} = \frac{3}{4} M (2 v a) + 2 k x v = 0$
$\left(\frac{3}{2} M a + 2 k x\right) v = 0$

For $v \neq 0$:
$\frac{3}{2} M a = -2 k x \implies F_{\text{net}} = M a = -\frac{4kx}{3}$

Thus, the net external force acting on the disk is $-\frac{4kx}{3}$.

Correct Option: D