JEE Advanced 2013 · Paper 1 · Physics Q19

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PhysicsInteger+4 / -1Question 19 of 60

A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s $^{-1}$ about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s $^{-1}$ ) of the system is

Your answer

Answer and worked solution

Official answer: 8

Moments of inertia. Disc: $\frac12(50)(0.4)^2=4$ kg m². Each ring lies flat with its centre 0.2 m from the axis: $mr^2+md^2=6.25(0.04)+6.25(0.04)=0.5$ kg m². Two rings: 1 kg m².

Angular momentum. $4\times10=(4+1)\omega$ , so $\omega=8$ rad/s.

Answer: 8

Topics:Angular momentum and its conservation Parallel and perpendicular axis theorems

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JEE Advanced 2013 · Paper 1, Physics question 19