Question

A wire of length $l$ is bent in the form of a quarter circle. The M.I of the wire about an axis passing through the centre of the quarter circle and touching one end of the wire in the plane is (${\pi ^2} = 10$)
(A) $0.4m{l^2}$
(B) $0.6m{l^2}$
(C) $0.2m{l^2}$
(D) $m{l^2}$

Answer 1

We need to use the moment of inertia of a complete ring (assuming that a complete ring was formed) but divide by 4. The length of the wire will be the circumference of the quarter circle.
Formula used: In this solution we will be using the following formulae;
$I = \dfrac{{M{R^2}}}{2}$ where $I$ is the moment of inertia of a ring about its centre joining the centre touching the circumference, $M$ is the mass of the ring, $R$ is the radius of the ring.
$C = \dfrac{{\pi R}}{2}$ where $C$ is the circumference of a quarter circle, and $R$ is the radius of the circle.

Complete Step-by-Step Solution:
A wire is said to be bent into a quarter circle and the moment of inertial passing through the centre touching one of the ends is to be found.
First, imagine 3 other wires of the same mass were combined with the wire to complete a circle. The moment of a thin ring amount the same axis is given as
$I = \dfrac{{M{R^2}}}{2}$where $M$ is the mass of the ring, $R$ is the radius of the ring.
Then since the mass of the wire is $m$, the mass of the complete ring will be $4m$.
Hence, $I = \dfrac{{4m{R^2}}}{2}$.
Then the moment of inertia of the quarter circle is
${I_q} = \dfrac{I}{4} = \dfrac{{m{R^2}}}{2}$
Now, the circumference of the quarter circle is the length of the wire, i.e.
$C = l = \dfrac{{\pi R}}{2}$
$\Rightarrow R = \dfrac{{2l}}{\pi }$
Hence,
${I_q} = \dfrac{{m{{\left( {\dfrac{{2l}}{\pi }} \right)}^2}}}{2} = \dfrac{{4m{l^2}}}{{2{\pi ^2}}}$
Since, ${\pi ^2} = 10$, by computation, we have
${I_q} = 0.2m{l^2}$

Hence, the correct answer is C

Note For clarity, the moment of inertia of the quarter circle could easily be found by dividing the full circle by four because of the additivity of moments of inertia. That is to say, the moment of inertia of the circle can be said to be the sum of the moments of inertia of the quarter circle as long as it is about the same axis or line. Hence,
$I = {I_q} + {I_q} + {I_q} + {I_q} = 4{I_q}$
$\Rightarrow {I_q} = \dfrac{I}{4}$