A 120 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be done on the hoop to stop it?

Answer:W = - 5.808J

Explanation:

Given:

Mass of hoop, m=120kg

Centre of mass speed of hoop, v = 0.220m/s

Rotational inertia, I = mr^2

I=120r^2

Kinetic energy of hoop = Linear kinetic energy + Rotational kinetic energy

K = (1/2mv^2) + (1/2Iw^2)

But w = (v/r) ^2

K = (1/2mv^2) + (1/2*120r^2 (v/r) ^2)

K = (1/2*120*0.220^2) + (1/2*120*0.220^2)

K = 5.808J

To stop the hoop, its final kinetic energy must be zero,

Workdone = Kfinal-Kinitial = 0 - 5.808

W = - 5.808J