Question

Calculate the moment of inertia of a ring having mass M, radius R and having uniform mass distribution about an axis passing through the centre of ring and perpendicular to the plane of ring.

Answer 1

Mass of the Ring$=M$

Mass per unit length of ring $\lambda =\frac{m}{2\pi R}$

Consider a small segment $dl$

Mass of $dl=\lambda dl$

Moment of inertia of this segment about axis of rotation.

$dI=\left(\lambda dl\right)R^2$

Moment of inertia of the total ring is

$I=\int dI$

$\Rightarrow =\int \lambda dl{R}^2=\lambda R^2\int dl$

$\Rightarrow I=2\pi R.\lambda R^2$

$\Rightarrow I=2\pi R.\frac{M}{2\pi R}{R}^2$

$\Rightarrow I=M{R}^2$