Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm. When released from rest the heavier mass is observed to fall 81 cm in 9 s. The rotational inertia of the pulley is _ _ _ _ kg m ² . ( g =9.8 ~m / s ² )

Step-by-step solution

From kinematics with s = 0.81 m and t = 9 s: a = 2s/t^2 = 0.02 m/s². For the pulley system, applying Newton's second law to both masses and torque equation: (m₁ - m₂)g - (m₁ + m₂)a = Ia/R^2 . Substituting values: 0.05 9.8 - 0.75 0.02 = I 50 gives 0.49 - 0.015 = 50I . Therefore I = 9.5 10⁻³ kg·m²