A 50 lb pulley has two weights, A and B, attached at different distances from the center. If the pulley has an angular velocity of 20 rad/s, what is the total kinetic energy of the whole system? Assume the radius of gyration of the pulley is 0.6 ft.

Velocity Diagram

The total kinetic energy is the sum of the kinetic energy of each of the three parts.

1. Kinetic energy of the pulley:
     T₁ = 0.5 (mk²) ω²
         = 0.5(50/32.2)(0.6²) 20² = 111.8 ft-lb
              (recall, moment of inertia is related to
                  the radius of gyration, k, as I = mk²)

2. Kinetic energy of mass A
     V_A = 1.0ω = 20 ft/s
     T_A = 0.5(20/32.2) 20² = 124.2 ft-lb

3. Kinetic energy of mass B
     V_B = 0.5ω= 10 ft/s
     T_B = 0.5(30/32.2) 10² = 46.59 ft-lb

T = T₁ + T_A + T_B = 111.8 + 124.2 + 46.59

T = 282.6 ft lb