### Q.19:- 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?

**Answer:-**

Radius of the hoop, *r* = 2 m

Mass of the hoop, *m* = 100 kg

Velocity of the hoop, *v* = 20 cm/s = 0.2 m/s

Total energy of the hoop = Translational K.E. + Rotational K.E.

*E*_{T} = (1/2)*mv*^{2} + (1/2) *I* ω^{2}

Moment of inertia of the hoop about its centre, *I *= *mr*^{2}

*E*_{T} = (1/2)*mv*^{2} + (1/2) (*mr*^{2})ω^{2}

But we have the relation, v = rω

∴ *E*_{T} = (1/2)*mv*^{2} + (1/2)*mr*^{2}ω^{2}

= (1/2)*mv*^{2} + (1/2)*mv*^{2 }= *mv*^{2}

The work required to be done for stopping the hoop is equal to the total energy of the hoop.

∴ Required work to be done, *W* = *mv*^{2 }= 100 × (0.2)^{2 }= 4 J.