A 140 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.150 m/s. How much work must be done on the hoop to stop it?
K = 1/2Iomega^2 + 1/2mv^2
The Attempt at a Solution
K translational = 1/2mv^2 = 0.5(140)(0.150)^2 = 1.575 J
Apparently I double this amount to get the total KE, meaning the rotational energy is also 1.575 J, and the total work that needs to be done is 3.15 J. Does this mean that the rotational energy always equals the translational energy? I don't think so. How did this business with doubling come about?