Question

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 $\Omega$ and the disc rotating about its vertical axis with angular velocity $4\, \Omega$. The total angular momentum of the system about the point $O$ is $\left(\frac{ Ma ^{2} \Omega}{48}\right) n$. The value of $n$ is ______

Answer 1

Correct Answer: 49

Given:

• Thin rod of mass M and length a rotating about point O with angular velocity Ω
• A thin disc of mass M and radius a/4 is attached at distance a/4 from the free end of the rod
• The disc is rotating with angular velocity 4Ω about its own vertical axis
• We are to find the total angular momentum of the system about point O in terms of (M a² Ω / 48)·n

Step 1: Angular momentum of the rod about point O

Moment of inertia of thin rod about one end (axis perpendicular to length): 
I_rod = (1/3) M a² 
Angular momentum = I × Ω = (1/3) M a² Ω

Step 2: Angular momentum of the disc

Total angular momentum of the disc has two parts:

1. Due to rotation of the disc's center about O:
2. Due to spinning of the disc about its own axis:

Total disc angular momentum = (9/16 + 1/8) M a² Ω

Convert to common denominator: (9/16 + 2/16) = 11/16 
So, L_disc = (11/16) M a² Ω

 

Step 3: Total angular momentum about O

L_total = L_rod + L_disc = (1/3) M a² Ω + (11/16) M a² Ω

Take LCM: 
(1/3) = (16/48), (11/16) = (33/48) 
⇒ L_total = (16 + 33)/48 M a² Ω = (49/48) M a² Ω

Final Answer: n = 49