A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm s-1. How much work has to be done to stop it?

Here, R = 2m, M = 100 kg, v = 20 cm s-1                                 = 20 × 10-2 ms-1 Total kinetic of the loop = KT+ KR        = 12Mv2+12Iω2       [∴ For a hoop, I = MR2]      = 12Mv2 + 12MR2ω2        = 12Mv2 + 12Mv2            [∵ v = Rω]         = Mv2Work required to stop the hoop = Total kinetic energy of the hoop       = Mv2 = (100 kg)(20 × 10-2 ms-1)2 = 4 J