A thin rod of mass M and length 𝑎 is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass M and of radius 𝑎/4 is pivoted on this rod with its center at a distance 𝑎/4 from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity Ω and the disc rotating about its vertical axis with angular velocity 4Ω. The total angular momentum of the system about the point O is $$\left(\frac{M a^2 Ω}{48}\right)n$$.
The value of n is .........

The vertical axis through the point O acts as the reference (z-axis). All angular momenta are required about this axis.

1. Angular momentum of the rod
For a uniform thin rod of length $$a$$ lying radially in the horizontal plane and rotating about a vertical axis through one end, the moment of inertia is
$$I_{\text{rod}} = \int_0^{a} \left(\frac{M}{a}\right)x^{2}\,dx = \frac{M a^{2}}{3}.$$
Hence the rod’s angular momentum is
$$L_{\text{rod}} = I_{\text{rod}}\; \Omega = \frac{M a^{2} \Omega}{3}.$$

2. Orbital (centre-of-mass) angular momentum of the disc
The disc’s centre is fixed on the rod at a distance $$a/4$$ from the free end, so its distance from O is
$$R = a - \frac{a}{4} = \frac{3a}{4}.$$
Treating the disc as a point mass M at radius R, the orbital angular momentum is
$$L_{\text{orb}} = M R^{2}\, \Omega = M\left(\frac{3a}{4}\right)^{2}\Omega = \frac{9}{16}\,M a^{2}\Omega.$$

3. Spin angular momentum of the disc about its own centre
For a thin uniform disc about its symmetry axis,
$$I_{\text{disc}} = \frac{1}{2}\,M r^{2},\qquad r = \frac{a}{4}.$$
Thus
$$I_{\text{disc}} = \frac{1}{2}\,M\left(\frac{a}{4}\right)^{2} = \frac{M a^{2}}{32}.$$
The disc spins with angular speed $$4\Omega$$ relative to the inertial frame, so
$$L_{\text{spin}} = I_{\text{disc}}\,(4\Omega) = \frac{M a^{2}}{32}\,(4\Omega) = \frac{M a^{2}\Omega}{8}.$$

4. Total angular momentum about O
$$\begin{aligned} L_{\text{total}} &= L_{\text{rod}} + L_{\text{orb}} + L_{\text{spin}}\\ &= \frac{M a^{2}\Omega}{3} + \frac{9}{16}\,M a^{2}\Omega + \frac{M a^{2}\Omega}{8}\\ &= M a^{2}\Omega\left( \frac{1}{3} + \frac{9}{16} + \frac{1}{8} \right). \end{aligned}$$
Converting to a common denominator 48:
$$\frac{1}{3} = \frac{16}{48},\qquad \frac{9}{16} = \frac{27}{48},\qquad \frac{1}{8} = \frac{6}{48}.$$
Sum = $$(16 + 27 + 6)/48 = 49/48.$$

Therefore
$$L_{\text{total}} = \left(\frac{M a^{2}\Omega}{48}\right)\,49.$$

Comparing with the given form $$\left(\frac{M a^{2} \Omega}{48}\right)n$$, we obtain

$$n = 49.$$