Question

A ring 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 1

Given-

The radius of the ring, r = 2 m

Mass of the ring, m = 100 kg

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

 

We have to determine the work done 

 

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

ET = (1/2)mv² + (1/2) I ω²

We know, the moment of inertia of a ring about its centre, I = mr²

ET = (1/2)mv² + (1/2) (mr²)ω²

Also, we know v = rω

∴ ET = ( 1/2 )mv² + ( 1/2 )mr²ω²

=> ( 1/2 )mv² + ( 1/2 )mv² = mv²

Thus the amount of energy required to stop the ring = the total energy of the loop.

∴ The amount of work required, W = mv2 = 100 × (0.2)2 = 4 J.

 

 

Work done = 4J