Q. Three point masses each of mass $m$ are placed at the corners of an equilateral triangle of side ‘$a$’. Then the moment of inertia of this system about an axis passing along one side of the triangle is
A) $ma^{2}$
B) $3\ ma^{2}$
C) $3/4\ ma^{2}$
D) $2/3\ ma^{2}$
Explanation
(c) From the triangle $BCD$
$CD^{2} = BC^{2} - BD^{2} = a^{2} - \left( \frac{a}{2} \right)^{2}$
$x^{2} = \frac{3a^{2}}{4} \Rightarrow x = \frac{\sqrt{3}\ a}{2}$
Moment of inertia of system along the side $AB$
$I_{system} = I_{1} + I_{2} + I_{3} = m \times (0)^{2} + m \times (x)^{2} + m \times (0)^{2}$
$= mx^{2} = m\left( \frac{\sqrt{3}\ a}{2} \right)^{2} = \frac{3ma^{2}}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.