Method 1: Using Conservation of Energy

1. Identify the energies involved. The total mechanical energy of the system consists of three parts:

   • Translational kinetic energy of the disk's center of mass (CM), $K_{trans} = \frac{1}{2}MV_{CM}^2$.
   • Rotational kinetic energy of the disk about its CM, $K_{rot} = \frac{1}{2}I_{CM}\omega^2$.
   • Potential energy stored in the two springs, $U_{spring} = 2 \times (\frac{1}{2}kx^2) = kx^2$, where $x$ is the displacement of the CM from the equilibrium position.

2. Use the rolling without slipping condition. For a disk rolling without slipping, the velocity of the center of mass $V_{CM}$ and the angular velocity $\omega$ are related by $V_{CM} = R\omega$. The moment of inertia of a uniform disk about its CM is $I_{CM} = \frac{1}{2}MR^2$. We can express the rotational kinetic energy in terms of $V_{CM}$: $K_{rot} = \frac{1}{2} \left( \frac{1}{2}MR^2 \right) \left( \frac{V_{CM}}{R} \right)^2 = \frac{1}{4}MV_{CM}^2$

3. Write the total energy of the system. The total kinetic energy is $K_{total} = K_{trans} + K_{rot} = \frac{1}{2}MV_{CM}^2 + \frac{1}{4}MV_{CM}^2 = \frac{3}{4}MV_{CM}^2$. The total mechanical energy $E$ is: $E = K_{total} + U_{spring} = \frac{3}{4}MV_{CM}^2 + kx^2$

4. Apply the principle of conservation of energy. The static friction force does no work, and the spring forces are conservative. Thus, the total mechanical energy of the system is conserved. This means its time derivative is zero: $\frac{dE}{dt} = 0$ $\frac{d}{dt} \left( \frac{3}{4}MV_{CM}^2 + kx^2 \right) = 0$

5. Differentiate and solve for acceleration. Using the chain rule, and noting that $V_{CM} = \frac{dx}{dt}$ and the acceleration of the CM is $a_{CM} = \frac{dV_{CM}}{dt}$: $\frac{3}{4}M(2V_{CM})\frac{dV_{CM}}{dt} + k(2x)\frac{dx}{dt} = 0$ $\frac{3}{2}MV_{CM}a_{CM} + 2kxV_{CM} = 0$ For a non-zero velocity $V_{CM}$, we can divide by it: $\frac{3}{2}Ma_{CM} + 2kx = 0$ $\frac{3}{2}Ma_{CM} = -2kx$ $a_{CM} = -\frac{4kx}{3M}$

6. Calculate the net external force. According to Newton's second law, the net external force $F_{net}$ acting on an object is equal to its mass times the acceleration of its center of mass: $F_{net} = M a_{CM}$ Substituting the expression for $a_{CM}$ we found: $F_{net} = M \left( -\frac{4kx}{3M} \right) = -\frac{4kx}{3}$

Method 2: Using Newton's Laws of Motion

1. Identify forces and set up equations. Let's consider the disk at a displacement $x > 0$. The springs pull it to the left.

   • The total spring force is $F_{sp} = -2kx$.
   • There is a static friction force $f$ acting at the point of contact with the ground. The equation for linear motion of the CM in the horizontal direction is: $F_{sp} + f = M a_{CM} \implies -2kx + f = M a_{CM} \quad (1)$ The equation for rotational motion about the CM is $\tau_{CM} = I_{CM} \alpha$. The spring force acts through the CM, so it produces no torque. The friction force produces a torque $\tau_f = -fR$ (assuming $f$ points right, torque is clockwise/negative). $-fR = I_{CM} \alpha = \left(\frac{1}{2}MR^2\right) \alpha \quad (2)$

2. Use the rolling without slipping condition. For rolling to the right, $V_{CM} > 0$ and $\omega < 0$. The relation is $V_{CM} = -R\omega$. Differentiating with respect to time gives $a_{CM} = -R\alpha$.

3. Solve the system of equations. From the rolling condition, $\alpha = -a_{CM}/R$. Substitute this into the torque equation (2): $-fR = \frac{1}{2}MR^2 \left( -\frac{a_{CM}}{R} \right) = -\frac{1}{2}MR a_{CM}$ $f = \frac{1}{2}M a_{CM}$ Now substitute this expression for friction $f$ back into the linear motion equation (1): $-2kx + \left(\frac{1}{2}M a_{CM}\right) = M a_{CM}$ $-2kx = M a_{CM} - \frac{1}{2}M a_{CM} = \frac{1}{2}M a_{CM}$ This gives a different result $a_{CM} = -4kx/M$. Let's recheck the sign convention carefully. Let's assume right is positive for translation and counter-clockwise is positive for rotation. $a_{CM}=R\alpha$ is for when the cylinder rolls such that $v$ increases as $\omega$ (CCW) increases. In our case, for $v>0$ (right), $\omega$ must be clockwise, so $\omega<0$. The condition is $v=-R\omega$, so $a=-R\alpha$. My initial use in this method was wrong. Let's correct it. The restoring force is $-2kx$, so $a_{CM}$ should be negative for $x>0$. For $a_{CM}<0$, we need $\alpha = -a_{CM}/R > 0$ (counter-clockwise). This requires a positive (CCW) torque. Friction must cause this torque. Torque is $\tau = fR$. So $f$ must be positive (to the right). So equations are: Linear: $-2kx + f = M a_{CM}$ (1) Torque: $fR = I_{CM} \alpha = \frac{1}{2}MR^2 \alpha$ (2) Constraint: $a_{CM} = -R\alpha \implies \alpha = -a_{CM}/R$ (3) Substitute (3) into (2): $fR = \frac{1}{2}MR^2(-a_{CM}/R) \implies f = -\frac{1}{2}Ma_{CM}$. Substitute this into (1): $-2kx - \frac{1}{2}Ma_{CM} = Ma_{CM}$. $-2kx = \frac{3}{2}Ma_{CM} \implies a_{CM} = -\frac{4kx}{3M}$. This matches the energy method.

4. Calculate the net external force. The net external force is the sum of all external forces acting on the disk. In the horizontal direction, this is $F_{net} = F_{sp} + f$. $F_{net} = -2kx + f = -2kx + (-\frac{1}{2}Ma_{CM})$. Substitute $a_{CM} = -\frac{4kx}{3M}$: $F_{net} = -2kx - \frac{1}{2}M(-\frac{4kx}{3M}) = -2kx + \frac{2kx}{3} = -\frac{4kx}{3}$. Alternatively, using Newton's second law directly: $F_{net} = Ma_{CM} = M(-\frac{4kx}{3M}) = -\frac{4kx}{3}$.

Both methods yield the same result.

Conclusion

The net external force acting on the disk when its center of mass is at displacement x is $-\frac{4kx}{3}$. This corresponds to option D.