Angular Speed of Rotating Disc Driven by Internal Motor

To find the angular speed of the large disc, we apply the principle of Conservation of Angular Momentum about the vertical rotation axis passing through the center of the large disc.

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1. Parameters and Moments of Inertia

• Large Disc:

  • Mass = $M$, Radius = $R$
  • Moment of inertia about its vertical central axis: $I_1 = \frac{1}{2} M R^2$

• Small Disc:

  • Mass = $M$, Radius = $r = \frac{R}{2}$
  • Moment of inertia about its own center of mass: $I_{\text{cm}} = \frac{1}{2} M r^2 = \frac{1}{2} M \left(\frac{R}{2}\right)^2 = \frac{1}{8} M R^2$
  • Distance of the small disc's center of mass from the main vertical axis = $R$.

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2. Angular Momentum Calculation

Let the angular velocity of the large disc about its central axis in the ground frame be $\Omega$.

1. Angular momentum of the large disc: $L_1 = I_1 \Omega = \frac{1}{2} M R^2 \Omega$

2. Angular momentum of the small disc: The motion of the small disc consists of the translation of its center of mass in a circle of radius $R$ about the central axis, plus rotation about its own center of mass.

   • Orbital angular momentum of its center of mass: $L_{\text{orbital}} = M v_{\text{cm}} R = M (\Omega R) R = M R^2 \Omega$
   • Spin angular momentum about its center of mass: $L_{\text{spin}} = I_{\text{cm}} \omega_s = \frac{1}{8} M R^2 \omega_s$ where $\omega_s$ is the absolute angular velocity of the small disc in the ground frame.

   Thus, the total angular momentum of the small disc about the central axis is: $L_2 = L_{\text{orbital}} + L_{\text{spin}} = M R^2 \Omega + \frac{1}{8} M R^2 \omega_s$

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3. Conservation of Angular Momentum

Since no external torque acts on the system about the vertical axis: $L_{\text{total}} = L_1 + L_2 = 0$

$\frac{1}{2} M R^2 \Omega + M R^2 \Omega + \frac{1}{8} M R^2 \omega_s = 0$

$\frac{3}{2} \Omega + \frac{1}{8} \omega_s = 0$

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4. Determining $\omega_s$ and $n$

Case 1: $\omega$ is relative to the motor (large disc)

Since the motor is mounted on the large disc, the uniform angular speed $\omega$ produced by the motor is relative to the large disc. Thus, the absolute angular velocity of the small disc is: $\omega_s = \Omega + \omega$

Substituting $\omega_s$ into the conservation equation: $\frac{3}{2} \Omega + \frac{1}{8}(\Omega + \omega) = 0$

$\left(\frac{3}{2} + \frac{1}{8}\right)\Omega + \frac{1}{8}\omega = 0$

$\frac{13}{8} \Omega = -\frac{1}{8} \omega \implies |\Omega| = \frac{\omega}{13}$

Comparing with $|\Omega| = \frac{\omega}{n}$, we get: $n = 13$

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Case 2: $\omega$ is relative to the ground frame

If $\omega$ represents the absolute angular speed of the small disc ($\omega_s = \omega$): $\frac{3}{2} \Omega + \frac{1}{8} \omega = 0 \implies |\Omega| = \frac{\omega}{12}$

Comparing with $|\Omega| = \frac{\omega}{n}$, we get: $n = 12$

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Final Answer

Taking the physical interpretation of an internal motor driving the shaft relative to its base disc, the value of $n$ is 13 (or 12 if $\omega$ is specified with respect to the ground frame).