A thin wire of length l and mass m is bent in the form of a semicircle as shown in the figure. 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 in the figure. Its moment of inertia about an axis joining its free ends will be (A) m l2 (B) zero (C) m l2 / π2 (D) m l2 / 2 π2

Tardigrade Admin · Accepted Answer

As π r=l ∴ r=(l/π) Moment of inertia of a ring about its diameter =(1/2) m r2 ∴ Moment of inertia of semicircle =(1/2)[m((l/π))2]=(m l2/2 π2)

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

$m l^{2}$

zero

$m l^{2} / π^{2}$

$m l^{2} /2 π^{2}$

As $π r = l ∴ r = \frac{l}{π}$
Moment of inertia of a ring about its diameter $= \frac{1}{2} m r^{2}$
$∴$ Moment of inertia of semicircle $= \frac{1}{2} [m (\frac{l}{π})^{2}] = \frac{m l ^{2}}{2 π ^{2}}$

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

ml2

zero

ml2/π2

ml2/2π2

As πr=l∴r=πl​ Moment of inertia of a ring about its diameter =21​mr2 ∴ Moment of inertia of semicircle =21​[m(πl​)2]=2π2ml2​

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

$m l^{2}$

$m l^{2} / π^{2}$

$m l^{2} /2 π^{2}$

As $π r = l ∴ r = \frac{l}{π}$
Moment of inertia of a ring about its diameter $= \frac{1}{2} m r^{2}$
$∴$ Moment of inertia of semicircle $= \frac{1}{2} [m (\frac{l}{π})^{2}] = \frac{m l ^{2}}{2 π ^{2}}$