Question

A uniform semi-circular disc of mass $M$ and radius $R$ lies in the $X-Y$ plane with its centre at the origin as shown in the figure. Also, axis $PQ$ is in the $X_Y$ plane. Then, Match for the moment of inertia of semi-circular disc about different axis.

Answer 1

Let us take a small element $dr$ of the semicircular disc.

Then mass $dm$ of the element

$dm=\frac{M}{area}\times areaofdrelement$

$=\frac{M}{\frac{1}{2}\pi R^2}\times \pi rdr$

Moment of Inertia $I={\int }_0^{R}dm{r}^2$

$=\frac{2M}{R^2}{\int }_0^R{r}^{3}dr$

$=\frac{2M}{R^2}\times \frac{R^4}{4}$

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

$I_Z=I_O=\frac{M{R}^2}{2}$

$I_Z=\frac{M{R}^2}{2}$

$I_X=I_Y$

By Perpendicular Axis theorem,

$I_Z=I_X+I_Y$

$I_Z=2I_X$

$\frac{M{R}^2}{2}=2I_X$

$I_X=\frac{M{R}^2}{4}$

Similarly, $I_Y=\frac{M{R}^2}{4}$

Also, $I_{PQ}=I_O=\frac{M{R}^2}{2}$