A 2 kg ball is attached to a 0.80 m string and whirled in a horizontal circle at a constant speed of 6 m/s. The work done on the ball during each revolution is: * 1 point (A) 90 J (B) 72 J (D) 16 J (D) zero

The answer should be 36 Joules.

Explanation:

This can be solved as a angular motion.

Angular velocity w = V/r

Where V is the linear velocity 6 m/s

And r is the radius of rotation which is equal to the lenght of the string 0.8 m

w = V/r = 6/0.8 = 7.5 rad/s

Moment of inertia of rotation I = mr^2

Where m is the mass of the ball 2 kg

I = 2 x 0.8^2 = 1.28 kg-m2

From energy conservation, work done is equal to the kinetic energy of the angular motion.

KE = 0.5Iw^2

= 0.5 x 1.28 x 7.5^2 = 36 Joules.