Question

Three thin rods each of mass M and length 1 are welded so as to form as equilateral triangle. The moment of inertia of the system about an axis passing through the centre of mass and perpendicular to its plane is:

• A. $M{L}^2$
• B. $\frac{M{L}^2}{2}$
• C. $\frac{M{L}^2}{3}$
• D. $\frac{M{L}^2}{4}$

Answer 1

The correct option is B $\frac{M{L}^2}{2}$

The moment of inertia of a single rod about an axis passing through its centre and perpendicular to it is $\frac{M{L}^2}{12}$

That of each side of the equilateral triangle about an axis passing through the triangle's centre and perpendicular to its plane is$:-$

$\frac{1}{12}M{L}^2+M\left(\frac{L}{2\sqrt{3}}\right)=\frac{1}{6}M{L}^2$

$($by the parallel axis theorem$)$

The moment of inertia of triangle about this axis is then

$3\times \frac{1}{6}M{L}^2=\frac{M{L}^2}{2}$

Hence,

option $\left(B\right)$ is correct answer.