There 1.2 kg masses are located at the vertices of an equilateral triangle 40 cm on a side made by rods of negligible mass. Find the rotational inertia of this system about an axis through the centre of the triangle and perpendicular to its plane.

Options

A. `0.064 kg-m^(2)`
B. `0.192 kg-m^(2)`
C. `0.138 kg- m^(2)`
D. `0.016 kg-m^(2)`

Answer

B

Explanation

`I=m(l^(2))/3xx3 =ml^(2)=1.2 v(0.4)^(2)`