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From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?
A
$\frac{13MR^2}{32}$
B
$\frac{11MR^2}{32}$
C
$\frac{9MR^2}{32}$
D
$\frac{15MR^2}{32}$
Explanation
Subtract the hole's MI (via parallel axis theorem) from the full disc.
Detailed Solution
Mass per unit area $\rho = \frac{M}{\pi R^2}$; mass of removed part $M' = \rho\pi\left(\frac{R}{2}\right)^2 = \frac{M}{4}$
By the parallel axis theorem, MI of removed part about the centre: $I_1 = \frac{1}{2}M'\left(\frac{R}{2}\right)^2 + M'\left(\frac{R}{2}\right)^2 = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{3MR^2}{32}$
MI of the full disc about the centre $I = \frac{MR^2}{2}$
$I_{net} = I - I_1 = \frac{16MR^2}{32} - \frac{3MR^2}{32} = \frac{13MR^2}{32}$