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 rotational 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 rotational inertia (I) is calculated using I = m1*r1^2 + m2*r2^2, where r1 and r2 are the distances of m1 and m2 from the center of mass. The center of mass is closer to the heavier mass (m2). Calculating the center of mass position and distances, and then using them in the formula gives I = 0.48 kg⋅m².

Two Particles (M1 = 0