First slide

Moment of intertia

Question

A rod of mass M kg and length L m is bent in the form of an equilateral triangle as shown in Fig. The moment of inertia of the triangle about a vertical axis perpendicular to the plane of the triangle and passing through the centre O (in units of kg $m^{2}$ ) is

Moderate

A

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

$\frac{{ML}^{2}}{54}$
C

$\frac{{ML}^{2}}{162}$
D

$\frac{{ML}^{2}}{108}$

Solution

Mass of each side = $\frac{M}{3}$
$x = \frac{1}{3} [\frac{L}{3} \cos 30 °] = \frac{\frac{L}{3}}{2 \sqrt{3}}$
moment of inertia of one side about O:
$I = [\frac{1}{12} \frac{M}{3} {(\frac{L}{3})}^{2} + \frac{M}{3} x^{2}] = \frac{{ML}^{2}}{162}$
moment of inertia of whole triangle about O : 3I = $\frac{{ML}^{2}}{54}$

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