Angular Momentum of Rotating Rod and Disc System About Fixed Axis

To find the total angular momentum of the system about the fixed vertical axis passing through point $O$, we can compute the angular momentum of the rod and the disc separately and sum them up using the principle of superposition of angular momentum.

1. Angular Momentum of the Rod ($L_{\text{rod}}$)

The thin uniform rod has mass $M$ and length $a$. It rotates about a fixed vertical axis passing through one of its ends (point $O$) with an angular velocity $\Omega$.

The moment of inertia of the rod about this axis is: $I_{\text{rod}} = \frac{1}{3} M a^2$

Therefore, the angular momentum of the rod about point $O$ is: $L_{\text{rod}} = I_{\text{rod}} \Omega = \frac{1}{3} M a^2 \Omega$

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2. Angular Momentum of the Disc ($L_{\text{disc}}$)

The total angular momentum of a rotating rigid body about a fixed point $O$ can be decomposed into two components: $\vec{L}_{\text{disc}} = \vec{L}_{\text{orbital}} + \vec{L}_{\text{spin}}$

(a) Orbital Angular Momentum ($\vec{L}_{\text{orbital}}$):

The center of mass of the circular disc is located at a distance from point $O$ equal to: $r_{\text{cm}} = a - \frac{a}{4} = \frac{3a}{4}$

Since the center of mass rotates about point $O$ with angular velocity $\Omega$, the magnitude of the orbital angular momentum of the disc's center of mass is: $L_{\text{orbital}} = M r_{\text{cm}}^2 \Omega = M \left(\frac{3a}{4}\right)^2 \Omega = \frac{9}{16} M a^2 \Omega$

(b) Spin Angular Momentum ($\vec{L}_{\text{spin}}$):

The disc has a radius $R = \frac{a}{4}$ and mass $M$. Its moment of inertia about its own vertical center of mass axis is: $I_{\text{cm}} = \frac{1}{2} M R^2 = \frac{1}{2} M \left(\frac{a}{4}\right)^2 = \frac{M a^2}{32}$

As observed by a stationary external observer, the disc spins about its vertical axis with an angular velocity of $4\Omega$ in the same direction as $\Omega$. Therefore, the spin angular momentum is: $L_{\text{spin}} = I_{\text{cm}} (4\Omega) = \left(\frac{M a^2}{32}\right) (4\Omega) = \frac{1}{8} M a^2 \Omega$

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3. Total Angular Momentum of the System ($L_{\text{total}}$)

Since all rotation vectors are aligned along the same vertical axis, the net angular momentum $L_{\text{total}}$ about point $O$ is the scalar sum of all the individual terms:

$L_{\text{total}} = L_{\text{rod}} + L_{\text{orbital}} + L_{\text{spin}}$

$L_{\text{total}} = \left(\frac{1}{3} + \frac{9}{16} + \frac{1}{8}\right) M a^2 \Omega$

Finding a common denominator of $48$: $L_{\text{total}} = \left(\frac{16 + 27 + 6}{48}\right) M a^2 \Omega = \left(\frac{49}{48}\right) M a^2 \Omega$

Comparing this expression to the given form $\left(\frac{M a^2 \Omega}{48}\right) n$: $n = 49$