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A hoop of radius 2m weights 100 kg. It r...

A hoop of radius 2m weights 100 kg. It rolls along a horizontal floor so that its centre of mass has speed of 20 cm/s. How much work has to be done to stop it ?

Text Solution

Verified by Experts

Moment of inertia of hoop about its center
`I = MR^(2)`
and energy of loop = translational kinetic energy of CM and rotational kinetic energy about axis through CM.
`= (1)/(2) mv^(2)._(CM) + (1)/(2) I omega^(2)" "[I = mR^(2) and omega = (v)/(R)]`
`= (1)/(2) mv^(2)._(CM) + (1)/(2) mR^(2) xx (v^(2)._(CM))/(R^(2)) = mv^(2)._(CM)`
Work done is stopping the hoop
= total KE of hoop `= mv^(2)._(CM) = 100 xx (0.2)^(2) = 4J`
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