Question

The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire and passing through its centre is:

• A. $M{r}^2$
• B. $\frac{1}{2}M{r}^2$
• C. $\frac{1}{4}M{r}^2$
• D. $\left(\frac{2}{5}\right)M{r}^2$

Answer 1

The correct option is A $M{r}^2$

consider small element of mass $dm$ on the semicircular wire.

moment of inertial about a point at distance r is given by:

$dI=dm{r}^2$

to find total moment of inertia, integrate both sides

$\int dI=r^2\int dm$

$\int dm=M$, total mass of body.

$I=M{r}^2$.

so The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire and passing through its centre is $M{r}^2$