Question

Three particles, each of mass 200 g, are kept at the corners of an equilateral triangle of side 10 cm. Find the moment of inertia of the system about an axis
(a) joining two of the particles and
(b) passing through one of the particles and perpendicular to the plane of the particles.

Answer 1

Therefore, the ⊥ distance from the axis (AD) = √3 / 2 x10 = 5 √3 cm.
Therefore moment of inertia about the axis BC will be
I = mr² = 200 K (5 √3)² = 200 × 25 × 3
= 15000 gm – cm² = 1.5 × 10^–3 kg – m².
b) The axis of rotation let pass through A and ⊥ to the plane of triangle
Therefore the torque will be produced by mass B and C
Therefore net moment of inertia = I = mr² + mr²
= 2 × 200 ×10² = 40000 gm-cm² = 4 ×10^–3 kg-m².