A thin wire of length l and mass m is bent in the form of a semicircle as shown. Its moment of inertia about an axis joining its free ends will be,

Question

A thin wire of length l and mass m is bent in the form of a semicircle as shown. Its moment of inertia about an axis joining its free ends will be, <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-wxosctcba4bis6xa.jpg" /> (A) ml2 (B) zero (C) (m l2/π 2) (D) (m l2/2 π 2)

Tardigrade Admin · Accepted Answer

The moment of inertia of semicircular portion is half of the total ring. So, I=(1/2)M textring(R2/2)=(1/2)(2 m)(R2/2)=(m R2/2) and π R=l . So, R=(l/π ) l=(m l2/2 π 2) .

Q. A thin wire of length $l$ and mass $m$ is bent in the form of a semicircle as shown. Its moment of inertia about an axis joining its free ends will be,

$m l^{2}$

zero

$\frac{m l ^{2}}{π ^{2}}$

$\frac{m l ^{2}}{2 π ^{2}}$

The moment of inertia of semicircular portion is half of the total ring.
So, $I = \frac{1}{2} M_{ring} \frac{R ^{2}}{2} = \frac{1}{2} (2 m) \frac{R ^{2}}{2} = \frac{m R ^{2}}{2}$
and $π R = l$ .
So, $R = \frac{l}{π}$
$l = \frac{m l ^{2}}{2 π ^{2}}$ .

Q. A thin wire of length l and mass m is bent in the form of a semicircle as shown. Its moment of inertia about an axis joining its free ends will be,

ml2

zero

π2ml2​

2π2ml2​

The moment of inertia of semicircular portion is half of the total ring. So, I=21​Mring​2R2​=21​(2m)2R2​=2mR2​ and πR=l . So, R=πl​ l=2π2ml2​ .

Q. A thin wire of length $l$ and mass $m$ is bent in the form of a semicircle as shown. Its moment of inertia about an axis joining its free ends will be,

$m l^{2}$

$\frac{m l ^{2}}{π ^{2}}$

$\frac{m l ^{2}}{2 π ^{2}}$

The moment of inertia of semicircular portion is half of the total ring.
So, $I = \frac{1}{2} M_{ring} \frac{R ^{2}}{2} = \frac{1}{2} (2 m) \frac{R ^{2}}{2} = \frac{m R ^{2}}{2}$
and $π R = l$ .
So, $R = \frac{l}{π}$
$l = \frac{m l ^{2}}{2 π ^{2}}$ .