Question

the particle of each mass 10 kg are placed at the vertices of triangle of side 10 identify the centre of mass​

Answer 1

Answer:

Figure given above shows the configuration of 3 equal masses of each 10g placed at the corner of equilateral triangle of side 5 cm.

Let an axis of rotation perpendicular to the palne of triangle and passing through corner A as per the requirement of question.

Moment of inertia I of the given system about the axis mentioned above is given by,

I = 10×52 + 10×52 = 500 g-cm2

(10 g mass at A do not contribute to the moment of inertia, because the axis is passing through it )