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 (48Ma2Ω)n. The value of n is ____.
Answer: 49
Solution
Angular momentum has three contributions:
Rod rotating about O: L1=IrodΩ=3Ma2Ω
Disc's center of mass rotating about O at distance 43a: L2=M(43a)2Ω=169Ma2Ω
Disc spinning about its own axis: L3=Idisc⋅4Ω=2M(a/4)2⋅4Ω=8Ma2Ω
Total: L=L1+L2+L3=3Ma2Ω+169Ma2Ω+8Ma2Ω
Finding common denominator (48): L=4816Ma2Ω+27Ma2Ω+6Ma2Ω=4849Ma2Ω
So n=49.
JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Integer answer