Question

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

• A. $\frac{M L^{2}}{12}$
• B. $\frac{M L^{2}}{54}$
• C. $\frac{M L^{2}}{162}$
• D. $\frac{M L^{2}}{108}$

Answer 1

Answer: (B)

Moment of inertia of any one side about its perpendicular bisector is
$I_{cm}=\frac{1}{2}\left(\frac{M}{3}\right)\left(\frac{L}{3}\right)^{2}$
Moment of inertia of this rod about about $O$
$I=I_{cm}+Md^{2}$
$I=\left[\frac{1}{12} \left(\frac{M}{3}\right) \left(\frac{L}{3}\right)^{2} + \left(\frac{M}{3}\right) \left(\frac{\frac{L}{3}}{2 \sqrt{3}}\right)^{2}\right]$
$I=\frac{ML^{2}}{162}$
Moment of Inertia of a triangle $=3I=\frac{ML^{2}}{54}$