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?

Asked by Sudhanshu | 1 year ago |  147

1 Answer

Solution :-

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

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

ET = (\( \dfrac{1}{2}\))mv2 + (\( \dfrac{1}{2}\)) I ω2

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

ET = (\( \dfrac{1}{2}\))mv2 + (\( \dfrac{1}{2}\)) (mr22

Also, we know v = rω

∴ ET = (\( \dfrac{1}{2}\) )mv2 + (\( \dfrac{1}{2}\))mr2ω2

=( \( \dfrac{1}{2}\))mv2 + (\( \dfrac{1}{2}\))mv= mv2

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

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

Answered by Sudhanshu | 1 year ago

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