A hoop of radius 2m, weight 100 kg. It rolls along horizontal floor so that its center of mass has a speed of `20 cm/s`. How much work has to be done to stop it?

Explanation

Work done to stop the hoop = Total kinetic energy of the hoop.
Total kinetic energy of the hoop,K - Translational K.E. + Rotational K.E.
`K=(1)/(2) mv_(cm)^(2)+(1)/(2) I_(cm) omega^(2)`
where `v_(cm)` is the linear velocity of centre of mass and `omega` is the angular velocity of the hoop. For a hoop.
`I_(cm)=mR^(2)`
`:. K=(1)/(2) mv_(cm)^(2)+(1)/(2) mR^(2) omega^(2)`
But `R omega=v_(cm)`
`:. K=(1)/(2) mv_(cm)^(2)+(1)/(2) mv_(cm)^(2) = mv_(cm)^(2)`
Here `m=100 kg`
`v_(cm)=20 cms^(-1)=0.2 ms^(-1)`
`:. K=100xx(0.2)^(2)=4J`.
So, work of 4 J must be done to stop the hoop.