Question

Calculate the moment of inertia of a thin ring of mass $\mathrm{m}$ and radius $\mathrm{r}$.

Answer 1

Step 1: Given data

A thin ring of mass and radius is,

Mass = $\mathrm{m}$

Radius = $\mathrm{r}$

Step 2: Calculation of moment of inertia

Moment of inertia$\left(\mathrm{I}\right)$ $={\mathrm{MR}}^2$

Step 3: Find the moment of inertia

By using the perpendicular axis theorem,

${\mathrm{I}}_{\mathrm{x}}+{\mathrm{I}}_{\mathrm{y}}={\mathrm{I}}_{\mathrm{z}}$

Since both ${\mathrm{I}}_{\mathrm{X}}$ and ${\mathrm{I}}_Y$ are along the diameter,

${\mathrm{I}}_{\mathrm{X}}={\mathrm{I}}_{\mathrm{y}}$

$2{\mathrm{I}}_{\mathrm{x}}={\mathrm{I}}_{\mathrm{z}}={\mathrm{MR}}^2$

${\mathrm{I}}_{\text{diameter}}=\frac{{\mathrm{MR}}^2}{2}$

Therefore, the moment of inertia is $\frac{{\mathrm{MR}}^2}{2}$.