A disc of radius 2m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?

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Detailed Solution

Work required = change in kinetic energy

Final KE = 0
Initial $KE = \frac{1}{2} {mv}^{2} + \frac{1}{2} {lω}^{2} = \frac{1}{2} {mv}^{2} + \frac{1}{2} {mv}^{2} (\frac{K^{2}}{R^{2}}) = \frac{3}{4} {mv}^{2}$

$= \frac{3}{4} x 100 x {(20 x 10^{- 2})}^{2} = 3 J$

$|∆ KE| = 3 J$

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