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 centre 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

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Quizplus · Accepted Answer

The moment of inertia of a rigid body about an axis perpendicular to the rod and through the centre of mass is given by:I = m1*r1^2 + m2*r2^2where m1 and m2 are the masses of the particles, r1 and r2 are the distances from the axis to the particles, and I is the moment of inertia.In this case, m1 = 0.20 kg, m2 = 0.30 kg, r1 = 1.0 m, and r2 = 1.0 m.Therefore, I = (0.20 kg)*(1.0 m)^2 + (0.30 kg)*(1.0 m)^2 = 0.50 kg.m^2.

Two Particles (M1 = 0