Part (a): Maximum Compression of the Spring

Maximum compression occurs when the relative angular velocity between the two discs is zero. At this instant, both discs rotate with a common angular velocity $\omega_c$.

1. Conservation of Angular Momentum:

Since there are no external torques on the system (friction is absent), the total angular momentum is conserved.

$$ L_i = L_f $$

$$ I \omega_0 + I(0) = (I + I) \omega_c $$

$$ \omega_c = \frac{\omega_0}{2} $$

2. Conservation of Energy:

The loss in kinetic energy is stored as potential energy in the spring.

$$ \Delta K.E. = \frac{1}{2} k x_{max}^2 $$

$$ \frac{1}{2}I\omega_0^2 – \frac{1}{2}(2I)\omega_c^2 = \frac{1}{2} k x_{max}^2 $$

Substitute $\omega_c = \omega_0/2$ and $I = \frac{1}{2}mr^2$:

$$ \frac{1}{2}I\omega_0^2 – I(\frac{\omega_0}{2})^2 = \frac{1}{2}I\omega_0^2 – \frac{1}{4}I\omega_0^2 = \frac{1}{4}I\omega_0^2 $$

$$ \frac{1}{4}(\frac{1}{2}mr^2)\omega_0^2 = \frac{1}{2} k x_{max}^2 $$

$$ \frac{1}{8}mr^2\omega_0^2 = \frac{1}{2} k x_{max}^2 \implies x_{max}^2 = \frac{mr^2\omega_0^2}{4k} $$

$$ x_{max} = \frac{\omega_0}{2} \sqrt{\frac{mr^2}{k}} $$

Part (b): Angular Velocities when Relaxed

When the spring relaxes, it returns all the stored potential energy back to the kinetic energy of the discs. This process is analogous to a perfectly elastic collision between two identical masses.

In a 1D elastic collision between identical masses where one is moving and the other is stationary, the velocities are exchanged.

• Upper Disc Initial: $\omega_0$
• Lower Disc Initial: $0$

After the interaction (velocities swapped):

• Upper Disc Final: $0$
• Lower Disc Final: $\omega_0$