Question

A thin ring of mass 1kg and radius 1m is rolling at a speed of 1ms^-1 . Its kinetic energy is

Easy

Solution

${KE}_{roll} = \frac{1}{2} m v^{2} (1 + \frac{k^{2}}{r^{2}}) = \frac{1}{2} 1 \times 1^{2} (1 + 1) = 1$

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