A solid sphere of mass M and radius R is divided into two unequal parts. The smaller part, of mass M/8, is recast into a sphere of radius r, and the larger part is recast into a circular disc of thickness t and radius 2R. If I₁ is the moment of inertia of the sphere of radius r about an axis throug…

Question

A solid sphere of mass M and radius R is divided into two unequal parts. The smaller part, of mass M/8, is recast into a sphere of radius r, and the larger part is recast into a circular disc of thickness t and radius 2R. If I₁ is the moment of inertia of the sphere of radius r about an axis through its centre and I₂ is the moment of inertia of the disc about its diameter, then (I₂)/(I₁) is

Accepted Answer

(2) 70. First find r. Recasting keeps the density, so mass goes as volume, which goes as the cube of the radius. (M/8)/M=(r³)/(R³), so r³=(R³)/8 and r=R/2. Sphere about a central axis: I₁=2/5(M/8)(R/2)²=2/5·M/8·(R²)/4=(MR²)/(80). The larger part has mass M-M/8=(7M)/8 and becomes a disc of radius 2R. Disc about a diameter: I₂=1/4((7M)/8)(2R)²=1/4·(7M)/8·4R²=(7MR²)/8. (I₂)/(I₁)=(7MR²/8)/(MR²/80)=7/8×80=70. The thickness t never enters: it is fixed by the mass and radius, and a disc's moment of inertia about a diameter does not depend on it.

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