Two particles (m₁ = 0.20 kg,m₂ = 0.30 kg)are positioned at the ends of a 2.0-m long rod of negligible mass.What is the moment of inertia of this rigid body about an axis perpendicular to the rod and through the center of mass? A)0.48 kg.m² B)0.50 kg.m² C)1.2 kg.m² D)0.80 kg.m² E)0.70 kg.m²

Question

Quizplus · Accepted Answer

The moment of inertia of a rigid body about an axis perpendicular to the rod and through the center of mass is given by:$$I = m_1r_1^2 + m_2r_2^2$$where m_1 and m_2 are the masses of the particles, and r_1 and r_2 are their distances from the axis of rotation.In this case, m_1 = 0.20 kg, m_2 = 0.30 kg, r_1 = 1.0 m, and r_2 = 1.0 m.Therefore,$$I = (0.20 kg)(1.0 m)^2 + (0.30 kg)(1.0 m)^2 = 0.50 kgm^2$$So, the correct answer is B) 0.50 kgm^2.

Two Particles (M1 = 0

Two Particles (M₁ = 0