A rod of length L and mass M is bent to form a semicircular ring as shown in figure. The moment of inertia about XY is

Question

A rod of length L and mass M is bent to form a semicircular ring as shown in figure. The moment of inertia about XY is (A) (1/4) (M L2/π2) (B) (2 M L2/3 π2) (C) (M L2/4) (D) (M L2/2)

Tardigrade Admin · Accepted Answer

Here, L=π R or R=(L/π) ∴ Moment of inertia about X Y=(1/2)((1/2) M R2) =(1/4) (M L2/--π2)

Q. A rod of length $L$ and mass $M$ is bent to form a semicircular ring as shown in figure. The moment of inertia about $X Y$ is

$\frac{1}{4} \frac{M L ^{2}}{π ^{2}}$

$\frac{2 M L ^{2}}{3 π ^{2}}$

$\frac{M L ^{2}}{4}$

$\frac{M L ^{2}}{2}$

Here,
$L = π R$ or $R = \frac{L}{π}$
$∴$ Moment of inertia about
$X Y = \frac{1}{2} (\frac{1}{2} M R^{2})$
$= \frac{1}{4} \frac{M L ^{2}}{-- π ^{2}}$

Q. A rod of length L and mass M is bent to form a semicircular ring as shown in figure. The moment of inertia about XY is

41​π2ML2​

3π22ML2​

4ML2​

2ML2​