First slide

Moment of intertia

Question

A circular disc of radius R and thickness (R/6) has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recast into a solid sphere. The moment of inertia of the sphere about its diameter on axis of rotation is

Difficult

A

I
B

$\frac{2 I}{8}$
C

$\frac{I}{5}$
D

$\frac{I}{10}$

Solution

When the disc is melted and recasted into a solid sphere, then,
Volume of disc = Volume of the sphere
$\begin{array}{l} {πR}^{2} \times (R / 6) = (4 / 3) {πr}^{3} \\ ∴ r = R / 2 \end{array}$
Now, $I = \frac{{MR}^{2}}{2} (Fordisc)$
$\begin{aligned} I^{'} = \frac{2}{5} {MR}^{2} (For sphere) \\ ∴ I^{'} = \frac{2}{5} M \times {(\frac{R}{2})}^{2} = \frac{1}{2} {MR}^{2} \times \frac{1}{5} \end{aligned}$
or , $I^{'} = I / 5$

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