First slide

Moment of intertia

Question

The moment of inertia of a cylinder about its own axis is equal to its moment of inertia about an axis passing through its centre and perpendicular to its length. The ratio of length to radius is

Difficult

A

1: 2
B

$\sqrt{2} : 1$
C

$1 : \sqrt{3}$
D

$\sqrt{3} : 1$

Solution

Let the moment of inertia of the cylinder about its own axis AB is I₁, and moment of inertia about an axis passing through its centre and perpendicular to its length be I₂.Then
$I_{1} = I_{2}$
or $\frac{{MR}^{2}}{2} = \frac{{MR}^{2}}{4} + \frac{{ML}^{2}}{12} or \frac{R^{2}}{4} = \frac{L^{2}}{12}$

or $\frac{L^{2}}{R^{2}} = \frac{12}{4} = 3 or \frac{L}{R} = \frac{\sqrt{3}}{1}$

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