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 centre 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

Step-by-step solution

Angular momentum of the system about O , L → = L → Rod + L → Disc ⇒ L → = L → Rod + L → C M Disc + r → C M × M D i s c V → C M ⇒ L = M a 2 Ω 3 + M a 2 8 Ω + M 3 a 4 Ω 3 a 4 ⇒ L = M a 2 3 Ω + M a 2 8 Ω + 9 M a 2 16 Ω ⇒ L = 16 M a 2 Ω + 6 M a 2 Ω + 27 M a 2 Ω 48 ⇒ L = 49 48 M a 2 Ω Compare it with n M a 2 48 Ω , we get n = 49 Hence, n = 49