Question

State an expression for moment of inertia of a ring about an axis passing through its centre and perpendicular to it.

Answer 1

moment of inertia of ring about an axis passing through its centre and perpendicular to it.

Let mass of ring is M and radius is R.
cut an element , dx at circumference of ring.
so, mass of elementary length of ring, dm is given by $dm=\frac{m}{2\pi R}dx$

now, $I=(dm)R^2$

$I=\frac{m}{2\pi R}dxR^2$

$I=\frac{mR}{2\pi}\int\limits^{2\pi R}_0{dx}$

$I=\frac{mR}{2\pi}2\pi R$

$I=mR^2$

Answer 2

Answer:

Explanation:

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