First slide

Moment of intertia

Question

A uniform cylinder has a radius R and length L. If the moment of inertia of this cylinder about an axis passing through its centre and normal to its circular face is equal to the moment of inertia of the same cylinder about an axis passing through its centre and normal to its length;
then:

Moderate

A

L = R
B

L = $\sqrt{3} R$
C

L = $\frac{R}{\sqrt{3}}$
D

L = 0

Solution

Given $I_{1} = I_{2}$
$\Rightarrow \frac{1}{2} {mR}^{2} = \frac{{mL}^{2}}{12} + \frac{{mR}^{2}}{4}$
solve to get L = $\sqrt{3} R$

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