Question

A disc of mass $m$ and radius $R$ has a concentric hole of radius $r$. Its moment of inertia about an axis through its centre and perpendicular to its plane is :

• A. $\frac{1}{2}m(R-r)^2$
• B. $\frac{1}{2}m(R^2-r^2)$
• C. $\frac{1}{2}m(R+r)^2$
• D. $\frac{1}{2}m(R^2+r^2)$

Answer 1

Answer: (D)

$\frac{1}{2}m(R^2+r^2)$

Answer 2

The moment of inertia of a disc with a hole is the difference between the M.I. of a solid disc of radius $R$ and a solid disc of radius $r$, assuming uniform mass density $\sigma$. $I = \frac{1}{2}\sigma\pi R^4 - \frac{1}{2}\sigma\pi r^4 = \frac{1}{2}\sigma\pi(R^4-r^4)$. The mass of the disc with the hole is $m = \sigma\pi(R^2-r^2)$, so $\sigma\pi = \frac{m}{R^2-r^2}$. Substituting $\sigma\pi$, we get $I = \frac{1}{2}m\frac{R^4-r^4}{R^2-r^2} = \frac{1}{2}m(R^2+r^2)$.