Question

A uniform bar of mass $M$ and Length $L$ is bent in the form of an equilateral triangle. Find the moment of inertia of the triangle about an axis passing through the centre of mass and perpendicular to the plane of the triangle.

• A. $M{L}^2$
• B. $M{L}^2/2$
• C. $M{L}^2/27$
• D. $M{L}^2/54$

Answer 1

Answer: (D)

$x=\frac{L}{6}tan30^{\circ }$
$=\frac{\sqrt{13}}{8}L$
Moment of inertia about C.G of one rod $=I_g+M{x}^2$
$=\frac{\frac{M}{3}\times {\left(\frac{L}{3}\right)}^2}{12}+\frac{M}{3}\times {\left(\frac{\sqrt{3}}{18}L\right)}^2$

Moment of inertia of all $3$ rods
$=\frac{M{L}^2}{162}\times 3$
$=\frac{M{L}^2}{54}$