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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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Answer» tion:Radius of the HOOP, r = 2 mMass of the hoop, m = 100 kgVelocity of the hoop, v = 20 cm/s = 0.2 m/sTotal energy of the hoop = Translational K.E. + Rotational K.E.ET = (1/2)mv2 + (1/2) I ω2Moment of inertia of the hoop about its centre, I = mr2ET = (1/2)mv2 + (1/2) (mr2)ω2But we have the relation, v = rω∴ ET = (1/2)mv2 + (1/2)mr2ω2= (1/2)mv2 + (1/2)mv2 = mv2The 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. |
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