7.19 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?

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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    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 = θ1 m α + α

    Moment of inertia about the centre, I = mr2

    So the total energy = 12 m v2 + 12Iω2

    Since v = r 12 we get

    Total energy = v2 m 12mr2ω2 + ω = 12

    Required work to be done = 100 x 0.2 x 0.2 J = 4 J

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