Question

Three point masses each of mass m are placed at the corner on an equivalent triangle of side l find the moment of inertia of the system about an axis passing through one of the corners perpendicular to the plane of the triangle Solve this equation by drawing triangle​

Answer 1

The correct equation is option c.

To find the moment of inertia of the system of three point masses about an axis passing through one of the corners and perpendicular to the plane of the triangle, you can use the parallel axis theorem and treat each mass as a point mass.

Moment of inertia of each of the point masses (m) at B and C about the side BC= m(0)2 = 0.

Moment of inertia of point mass m at A about the side BC = m(AD)².

Now, AD = I sin 60°

1 √3 /2 -(height of equilateral triangle)

.. Moment of inertia of the system about side BC

=0+0+m = (1 √3 /2))²

3ml)²/ 4