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Question
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?
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Solution 1
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 KE + Rotational KE
`E_r = 1/2mv^2 + 1/2 Iomega^2`
Moment of inertia of the hoop about its centre, I = mr2
`E_r = 1/2 mv^2 + 1/2(mr^2)omega^2`
But we have the relation, `v = romega`
`:.E_r = 1/2 mv^2 + 1/2 mr^2omega^2`
`=1/2mv^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 = mv2 = 100 × (0.2)2 = 4 J
Solution 2
Here, R = 2 m, M = 100 kg
v = 20 cm/s = 0.2 m/s
Total energy of the hoop =1/2Mv2 + 1/2Iw2
=1/2Mv2 + 1/2(MR2)w2
=1/2Mv2 +1/2Mv2 =Mv2
Work required to stop the hoop = total energy of the hoop W = Mv2 = 100 (0.2)2= 4 Joule
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