Question

A rod of mass $Mkg$ 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 $O$
$I=I_{cm}+M{d}^2$
$I=\left[\frac{1}{12}\left(\frac{M}{3}\right){\left(\frac{L}{3}\right)}^2+\left(\frac{M}{3}\right){\left(\frac{\frac{L}{3}}{2\sqrt{3}}\right)}^2\right]$
$I=\frac{M{L}^2}{162}$
Moment of Inertia of a triangle $=3I=\frac{M{L}^2}{54}$