Question

A uniform thin bar of mass $3 \,kg$ and length $0.9 \,m$ is bent to make an equilateral triangle. The moment of inertia of the triangle through its centre of mass and perpendicular to plane of triangle is :

• A. $ 0.030\, kg-m^{2} $
• B. $ 0.015\, kg-m^{2} $
• C. $ 0.090\,kg-m^{2} $
• D. $ 0.045\, kg-m^{2} $

Answer 1

Answer: (D)

From parallel axis theorem, $I=I_{C M}+M a^{2}$
where $I_{C M}$ is moment of inertia about $COM$ and a is distance of mass from axis of rotation.
Moment of inertia of thin bar is $\frac{M l^{2}}{12} $
$\therefore I=\frac{M l^{2}}{12}+M a^{2}$
From $\Delta O A O, \tan 30^{\circ}=\frac{a}{l / 2} $
$\Rightarrow a=\frac{l}{2} \tan 30^{\circ} $
$\therefore I=\frac{M l^{2}}{6}$
$ \therefore $ Total $M I=\frac{M l^{2}}{6}+\frac{M l^{2}}{6}+\frac{M l^{2}}{6}$
$=0.045\, kg - m ^{2}$