Question

◎ From the solid sphere of mass M, R, we have removed a sphere of radius R/2, find the M.O.I of remaining sphere about the point '0' as given in the diagram.

Answer 1

The M.O.I of the remaining sphere about the point 'O' is $\frac{57}{160}MR^2$.

Answer 2

The problem is solved using the principle of superposition for moments of inertia. The moment of inertia of the hollowed-out body is obtained by subtracting the moment of inertia of the removed part from that of the original complete body.

1. Calculate the M.O.I of the original large sphere about its center 'O' ($I_{original} = \frac{2}{5}MR^2$).
2. Determine the mass of the removed smaller sphere ($m' = M/8$) based on its volume and the uniform density of the original sphere.
3. Calculate the M.O.I of the removed sphere about its own center O' ($I_{removed, center} = \frac{1}{80}MR^2$).
4. Use the Parallel Axis Theorem ($I = I_c + Md^2$) to find the M.O.I of the removed sphere about point 'O' ($I_{removed} = \frac{7}{160}MR^2$).
5. Subtract $I_{removed}$ from $I_{original}$ to get the M.O.I of the remaining body about 'O' ($I_{remaining} = \frac{57}{160}MR^2$).