A 150 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.130 m/s. how much work must be done on the hoop to stop it?

The work needed is equal to the KE of the hoop which is = 150 * 0.130^2 / 2 ... (1)

Let the following be:r be the radiuscenter of mass in the moment of inertia is denoted by:

I = 150 r^2 kg m^2.

The angular velocity relative to the center of mass is:

w = 0.130 / r rad / s.

Rotational KE = Iw^2 / 2

= (150 * r^2) (0.130 / r) ^2 / 2

= 150 * 0.130^2 / 2 J ... (2)

Adding (1) and (2):

Total KE = 130 * 0.130^2

= 2.2 J.