The pulley in ?Fig. P9.76? has radius 0.160 m and moment of inertia The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor.

Solution 84P Step 1: Initially all the blocks and the pulley were not moving, so the total energy of the system was only the potential energy of the block of mass 4 kg at a height of 5 meters. The total energy was, E = mgh = 4 × 9.8 × 5 = 196 J . tot Step 2: Later when the system evolved in time, the total energy is, E = kinetic energy of the blocks + rotational energy of the pulley + P.E of the 2 kg block tot 1 2 1 2 1 2 = 2× 4 × V + 2 × 2 × V + 2 I + 2 × 9.8 × 5 2 = 2V + V + 2 0.380 2 + 98 2 r Where and rare the angular velocity and the radius of the pulley. So, E = [2 + 1 + 0.19] V + 98 tot 0.160 2 2 = [2 + 1 + 7.421] V + 98 = 10.421 V + 98 Step 4: The law of conservation of energy tells that, 2 10.421 V + 98 = 196 J 2 10.421 V = 196 98 = 98 V = 98 / 10.421 = 9.404 V = 9.404 = 3.066 m/s. Conclusion: The whole system moves with the same velocity so as the 4 kg block. The velocity just before hitting the ground will be 3.066 m/s.