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← JEE Advanced 2021 · Paper 1
PhysicsInteger+4Question 18 of 57

JEE Advanced 2021 · Paper 1, Physics question 18

A thin rod of mass MM and length aa is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass MM and of radius a/4a/4 is pivoted on this rod with its center at a distance a/4a/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\Omega and the disc rotating about its vertical axis with angular velocity 4Ω4\Omega. The total angular momentum of the system about the point O is (Ma2Ω48)n\left(\dfrac{Ma^2\Omega}{48}\right)n.

The value of nn is ____.

A horizontal rod of mass M and length a pivoted at its left end O about a vertical axis, rotating with angular velocity Ω. A disc of mass M lies flat on the rod, pivoted at a point a/4 from the rod's free end, spinning about its own vertical axis with angular velocity 4Ω.
Answer and worked solution

Official answer: 49

Each contribution about O.

  • Rod: 13Ma2Ω\frac13Ma^2\Omega.
  • Disc's centre, at 34a\frac34a from O: M(3a4)2Ω=916Ma2ΩM\left(\frac{3a}4\right)^2\Omega=\frac9{16}Ma^2\Omega.
  • Disc's spin about its own axis at 4Ω4\Omega: 12M(a4)2(4Ω)=18Ma2Ω\frac12M\left(\frac a4\right)^2(4\Omega)=\frac18Ma^2\Omega.

Total.
L=Ma2Ω(16+27+648)=Ma2Ω48⋅49,n=49.L=Ma^2\Omega\left(\frac{16+27+6}{48}\right)=\frac{Ma^2\Omega}{48}\cdot49,\qquad n=49.