First slide

Moment of intertia

Question

Fig. shows a thin metallic triangular sheet,{BC.
The sides AB and BC are equal of length L. The mass of the sheet is M. The moment of inertia of the sheet about AC is

Easy

A

$\frac{{ML}^{2}}{8}$
B

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

$\frac{{ML}^{2}}{6}$
D

$\frac{{ML}^{2}}{4}$

Solution

The moment of inertia of ABC about AC.
= $\frac{1}{2}$ moment of inertia of ABCD about 4C
$= \frac{1}{2} [2 M] \frac{L^{2}}{12} = \frac{{ML}^{2}}{12}$

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