A disc of radius $2text{ m}$ and mass $100text{ kg}$ rolls on a horizontal floor. Its centre of mass has speed of $20text{ cm/s}$. How much work is needed to stop it? (2019)
Solution:
Total kinetic energy of the rolling disc is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{3}{4}mv^2$. Substituting $m = 100\text{ kg}$ and $v = 0.2\text{ m/s}$ gives $K = 3\text{ J}$. Work required to stop it is equal to its total kinetic energy, which is $3\text{ J}$.
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