As illustrated in the figure, a pulley possesses a radius of 20 cm and a moment of inertia of 0.2 kg-m ^2 . A string passing over it connects to a vertical spring (spring constant 50 N/m ) fixed from below at one end, and holds a 1 kg mass at the other. With the spring initially at its natural length, the system is released from rest. Determine the speed of the block once it has descended through 10 cm . Assume g = 1

Step-by-step solution

By the principle of conservation of mechanical energy, the loss in gravitational potential energy of the block is equal to the gain in kinetic energy of the block and the pulley, plus the gain in elastic potential energy of the spring. Let v be the speed of the block and be the angular speed of the pulley after the block has descended by a distance h . Since the string does not slip, we have = v R . The energy conservation equation is: mgh = 1 2 mv^2 + 1 2 I ^2 + 1 2 kh^2 Substituting = v R : mgh = 1 2 mv^2 + 1 2 I