Q: Three identical thin rods, each of mass m and length L are joined to form equilateral triangle. Find the moment of inertia of the triangle about one of its sides.
(a) $\displaystyle \frac{m L^2}{6}$
(b) $\displaystyle \frac{3 m L^2}{2}$
(c) $\displaystyle \frac{m L^2}{3}$
(d) $\displaystyle \frac{m L^2}{2}$
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Ans: (d)
Sol: Using formula , $\displaystyle I = \frac{m L^2}{3}sin^2 \alpha $
Required Moment of Inertia , $\displaystyle I = \frac{m L^2}{3}sin^2 60 + \frac{m L^2}{3}sin^2 60 $
$\displaystyle I = 2 \times \frac{m L^2}{3} \times \frac{3}{4} $
$\displaystyle I = \frac{m L^2}{2} $