The total angular momentum of the system about the point O is the sum of the angular momentum of the rod and the angular momentum of the disc. Since all rotations occur about parallel vertical axes, the angular momentum vectors are collinear, and we can add their magnitudes algebraically. $L_{total} = L_{rod} + L_{disc}$

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

The rod has a mass M and length a. It rotates about a vertical axis passing through its end O with an angular velocity $\Omega$.

• The moment of inertia of a thin rod about an axis passing through its end and perpendicular to its length is $I_{rod} = \frac{1}{3}Ma^2$.
• The angular momentum of the rod is given by: $L_{rod} = I_{rod} \Omega = \frac{1}{3}Ma^2\Omega$

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

The angular momentum of the disc about point O is the sum of its orbital angular momentum (due to the motion of its center of mass around O) and its spin angular momentum (due to its rotation about its own center of mass). $L_{disc} = L_{orbital} + L_{spin}$

• Orbital Angular Momentum ($L_{orbital}$): The disc has mass M. Its center is located at a distance $r_{cm}$ from the pivot O. The disc's center is at a distance a/4 from the free end of the rod. The rod's length is a.

  • The distance of the disc's center from the pivot O is $r_{cm} = a - \frac{a}{4} = \frac{3a}{4}$.
  • The center of mass of the disc revolves around O with the same angular velocity as the rod, $\Omega$. The orbital angular momentum is the angular momentum of the disc treated as a point mass at its center of mass. $L_{orbital} = I_{orbital} \Omega = (M r_{cm}^2) \Omega$ $L_{orbital} = M \left(\frac{3a}{4}\right)^2 \Omega = M \frac{9a^2}{16} \Omega = \frac{9}{16}Ma^2\Omega$

• Spin Angular Momentum ($L_{spin}$): The disc has mass M and radius $R = a/4$. It rotates about its own vertical axis.

  • The moment of inertia of the disc about an axis through its center and perpendicular to its plane is $I_{cm} = \frac{1}{2}MR^2$. $I_{cm} = \frac{1}{2}M\left(\frac{a}{4}\right)^2 = \frac{1}{2}M\frac{a^2}{16} = \frac{Ma^2}{32}$
  • The problem states that a stationary observer finds the disc rotating about its vertical axis with angular velocity $4\Omega$. This implies the absolute angular velocity of the disc's spin is $\omega_{disc} = 4\Omega$.
  • The spin angular momentum is $L_{spin} = I_{cm} \omega_{disc}$. $L_{spin} = \left(\frac{Ma^2}{32}\right) (4\Omega) = \frac{4}{32}Ma^2\Omega = \frac{1}{8}Ma^2\Omega$

• Total Angular Momentum of the Disc: $L_{disc} = L_{orbital} + L_{spin} = \frac{9}{16}Ma^2\Omega + \frac{1}{8}Ma^2\Omega$ To add these, we find a common denominator (16). $L_{disc} = \frac{9}{16}Ma^2\Omega + \frac{2}{16}Ma^2\Omega = \frac{11}{16}Ma^2\Omega$

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

Now, we sum the angular momenta of the rod and the disc. $L_{total} = L_{rod} + L_{disc} = \frac{1}{3}Ma^2\Omega + \frac{11}{16}Ma^2\Omega$ To add these fractions, we find a common denominator, which is $3 \times 16 = 48$. $L_{total} = \left(\frac{16}{48} + \frac{33}{48}\right)Ma^2\Omega = \frac{16+33}{48}Ma^2\Omega = \frac{49}{48}Ma^2\Omega$

4. Finding the value of n

The problem states that the total angular momentum is given by the expression: $L_{total} = \left( {{{M{a^2}\Omega } \over {48}}} \right)n$ Comparing our calculated result with the given expression: $\frac{49}{48}Ma^2\Omega = \frac{n}{48}Ma^2\Omega$ From this comparison, we can see that: $n = 49$