First slide

Moment of intertia

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 and axis joining its free ends will be

Moderate

A

$\large ml^2$
B

zero
C

$\large \frac {ml^2}{\pi^2}$
D

$\large \frac {ml^2}{2\pi^2}$

Solution

πr = l ⇒ r = l/π
Semi circular arc was rotated it appeared as rotation of circular ring about diameter.
$\large I_{Dia}=\frac {Mr^2}{2}=\frac {m\times (l/\pi )^2}{2}=\frac {ml^2}{2\pi^2}$

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