JEE Advanced 2021 Paper 1 · Q18 · Angular Momentum

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 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 (Ma²Ω/48)·n. The value of n is ___.

[Figure: Rod of length a hinged at O at one end, lying in horizontal plane. A disc of radius a/4 is pivoted on the rod at a distance a from O minus a/4 = 3a/4 from O (centre of disc). Rod rotates about O at angular velocity Ω; disc spins about its own vertical axis at 4Ω.]

SolveKar AI · Accepted Answer

Centre of disc is at distance 3a/4 from O. Angular momentum of disc about O = (orbital) + (spin) = M·(3a/4)²·Ω + (½ M (a/4)²)·(4Ω) = M·9a²/16·Ω + ½·M·a²/16·4Ω = (9Ma²Ω/16) + (Ma²Ω/8) = (9Ma²Ω/16) + (2Ma²Ω/16) = 11Ma²Ω/16. Angular momentum of rod about O = (Ma²/3)·Ω. Total L = (Ma²Ω/3) + (11Ma²Ω/16) = Ma²Ω·(16 + 33)/48 = 49 Ma²Ω/48. Hence n = 49.

JEE Advanced 2021 Paper 1 · Q18 · Angular Momentum

Solution