Two blocks having masses of 400 g and 200 g are connected by a light string passing over a pulley that is free to rotate about its axis. The pulley possesses a moment of inertia of 1.6 10⁻⁴ kg-m ^2 and a radius of 2.0 cm . Determine the speed of the blocks when the 400 g block falls through a distance of 50 cm .

Step-by-step solution

Using the principle of conservation of mechanical energy, the loss in potential energy of the system is equal to the gain in kinetic energy of the system. (m₁ - m₂)gh = 1 2 m₁v^2 + 1 2 m₂v^2 + 1 2 I ^2 Since the string does not slip over the pulley, = v R . Substituting this into the equation gives: (m₁ - m₂)gh = 1 2 (m₁ + m₂ + I R^2 ) v^2 Given values are m₁ = 0.4 kg , m₂ = 0.2 kg , h = 0.5 m , R = 0.02 m , I = 1.6 10⁻⁴ kg-m ^2 , and taking g = 9.8 m/s ^2 : (0.4 - 0.2) 9.8 0.5 = 1 2 (0.4 + 0.2 + 1.6 10⁻⁴ (0.02)^2