A very thin horizontal,2.00-m long,5.00-kg uniform beam that lies along the east-west direction is acted on by two forces.At the east end of the beam,a 200-N force pushes downward.At the west end of the beam,a 200-N force pushes upward.What is the angular acceleration of the beam? A) 240 rad/s² north B) 1.33 × 10² rad/s² north C) zero D) 240 rad/s² south E) 1.33 × 10² rad/s² south

Question

Quizplus · Accepted Answer

The net torque on the beam is 400 N·m, and the moment of inertia of the beam about an axis through its center of mass and perpendicular to its length is (1/12)ML² = (1/12)(5.00 kg)(2.00 m)² = 1.67 kg·m². Therefore, the angular acceleration of the beam is α = τ/I = (400 N·m)/(1.67 kg·m²) = 240 rad/s² north.

A Very Thin Horizontal,2