Question

Three 500-g point masses are at the corners of an equilateral triangle with 50-cm sides. What is the moment of inertia of this system about an axis perpendicular to the plane of the triangle and passing through one of the masses at a corner of the triangle?

Answer 1

0.25 kg m^2

Step-by-step explanation:

mass of each , m = 500 g = 0.5 kg

distance, r = 50 cm = 0.5 m

Moment of inertia about the axis passing through one corner and perpendicular to the plane of triangle

I = mr^2 + mr^2

I = 2 mr^2

I = 2 x 0.5 x 0.5 x 0.5

I = 0.25 kgm^2