A thin uniform rod of mass M and length L has its moment of inertia I₁ about its perpendicular bisector. The rod is bend in the form of a semi-circular arc. Now its moment of inertia through the centre of the semi-circular arc and perpendicular to its plane is I₂. The ratio of I₁: I₂ will be:
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Question

A thin uniform rod of mass M and length L has its moment of inertia I₁ about its perpendicular bisector. The rod is bend in the form of a semi-circular arc. Now its moment of inertia through the centre of the semi-circular arc and perpendicular to its plane is I₂. The ratio of I₁: I₂ will be:

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G1 Physics · Accepted Answer

For the straight rod I₁=ML²/12. Bending into a semicircle keeps arc length L=πR, so I₂=MR²=ML²/π². Since 1/12 < 1/π², I₁/I₂<1. Hence A. Use I=∫r²dm. For a shifted parallel axis use I=I about CM+Md²; for a plane lamina use I about z-axis=Iₓ+Iᵧ..