The pulley has mass 12.0 kg, outer radius Ro=250 mm, inner radius Ri=200 mm, and radius of gyration kO=231 mm. Cylinder A weighs 71.0 N. Assume there is no friction between the pulley and its axle and that the rope is massless. At the instant when ω=69.0 rad/s clockwise, what is the kinetic energy of the system?

2213.7 J

Explanation:

The total kinetic energy is the sum of the kinetic energies of the pulley and the weight.

The kinetic energy of the weight is:

Ew = 1/2 * m * v^2

The mass is:

m = P / g = 71 / 9.81 = 7.24 kg

We need its speed, this is:

w * Ri = 69 * 0.2 = 13.8 m/s

So:

Ew = 1/2 * 7.24 * 13.8^2 = 689.4 J

The kinetic energy of the pulley is:

Ep = 1/2 * J * w^2

The polar moment is

J = m * k0^2

Then

Ep = 1/2 * m * k0^2 * w^2

Ep = 1/2 * 12 * 0.231^2 * 69^2 = 1524.3 J

So, the kinetic energy is

689.4 + 1524.3 = 2213.7 J