First slide

Moment of intertia

Question

A thin wire of mass M and length L is turned into a ring. The moment of inertia about its will be :

Moderate

A

$\frac{1}{4 π^{2}} {ML}^{2}$
B

$\frac{1}{12} {ML}^{2}$
C

$\frac{1}{3 π^{2}} {ML}^{2}$
D

$\frac{1}{π^{2}} {ML}^{2}$

Solution

When the wire is bend to form a circular ring, length of wire is same as circumference of ring, $2 πR = L$
Radius of ring is $R = \frac{L}{2 π}$
Moment of inertia of ring about its axis is given as
$I = {MR}^{2} = M {(\frac{L}{2 π})}^{2}$
$I = \frac{1}{4 π^{2}} {ML}^{2}$ .

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