Solveeit Logo
Login

Question

Question: Three point masses each of mass \(m\)are placed at the corners of an equilateral triangle of side a....

Three point masses each of mass mmare 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

ma2ma^{2}

B

3ma23ma^{2}

C

34ma2\frac{3}{4}ma^{2}

D

23ma2\frac{2}{3}ma^{2}

Answer

34ma2\frac{3}{4}ma^{2}

Explanation

Solution

The moment of inertia of system about AB side of triangle I=IA+IB+ICI = I_{A} + I_{B} + I_{C}

=0+0+mx2= 0 + 0 + mx^{2}

=m(a32)2=34ma2= m\left( \frac{a\sqrt{3}}{2} \right)^{2} = \frac{3}{4}ma^{2}