First slide

Moment of intertia

Question

Moment of inertia of the isosceles right angled triangle ABC about the side BC is I. Then its moment of inertia about side AC is

Difficult

A

$\frac{I}{2}$
B

$\frac{2 I}{3}$
C

$\frac{I}{\sqrt{2}}$
D

$\frac{I}{3}$

Solution

$\begin{array}{l} I = I_{B C} = \frac{1}{6} . m {(A B)}^{2} = \frac{1}{6} m a^{2} [A B = B C = a] \\ ∴ I_{A C} = \frac{1}{6} . m . {(A B \sin 45^{o})}^{2} = (\frac{1}{6} . m . a^{2}) \frac{1}{2} = \frac{I}{2} \end{array}$

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