# 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

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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