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. How much work has to be done to stop it?


What we want to find is work done. We know that the work done to stop a moving body is the total kinetic energy of the body.


Work Done = KE = KErot + KEtran


Where KErot is rotational kinetic energy and


KEtran is translational kinetic energy


KErot = 1/2 Iω2


KEtran = 1/2 Mv2


Where M is mass of hoop


I is moment of inertia of hoop


v is velocity of centre of mass


ω is angular velocity


Now the given values are,


Mass of hoop, M = 100 Kg


Radius of hoop, R = 2 m


Speed of centre of mass of hoop, v = 20 Cm s-1


= .2 m s-1


The axis of rotation of hoop passes through point B, we know that velocity of point O (centre of loop) is v. If ω is angular velocity of hoop then,


v = R ω (R is Distance between axis and centre(O))



Where ω is angular velocity.


i.e. ω = v/R = 0.2/2 = .1 s-1


Moment of inertia of hoop, I = MR2


Where, M is mass of hoop


R is Radius of hoop


I = 100(2)2


I = 400 Kg m2


Total Kinetic Energy of hoop,


KE = KErot + KEtran


KErot = 1/2 400 × (.1)2


= 2 J


KEtran = 1/2 100 × (.2)2


= 2 J


Total Kinetic energy,


KE = 2 + 2 = 4 J


Work done to stop the moving hoop is equal to its KE which is 4 J


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