JEE Main · 2005 · 08 May 2005 · Offline · Rotational Mechanics

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The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is

(A) $\frac{1}{4} Mr^{2}$
(B) $\frac{2}{5} Mr^{2}$
(C) $Mr^{2}$
(D) $\frac{1}{2} Mr^{2}$ ✓ Correct

Answer: (D)

Solution

Think of the semicircular disc as half of a full disc of the same radius. A full disc of mass

2 M

has moment of inertia

\frac{( 2 M ) r ^{2}}{2} = M r^{2}

about the axis perpendicular to its plane through the centre. Cutting it into two identical halves shares this equally, so each semicircular disc of mass

M

has

I = \frac{1}{2} M r^{2}

about that axis.

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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