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System of Particles and Rotational Motion
Rolling Motion – Kinetic Energy
2 questions from 2 papers, each with its answer.
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2019 1 question
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A disc of radius 2 m 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?Work required = change in kinetic energy; final KE = 0. For a rolling disc: $KE_i=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2=\frac{1}{2}mv^2+\frac{1}{2}\cdot\frac{1}{2}mr^2\cdot\frac{v^2}{r^2}=\frac{3}{4}mv^2$ $KE_i=\frac{3}{4}\times100\times(20\times10^{-2})^2=3\ J$ $|\Delta KE|=3\ J$
2018 1 question
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A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy ($K_t$) as well as rotational kinetic energy ($K_r$) simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is
Total KE of a rolling solid sphere is $\frac{7}{10}mv^2$.
$K_t = \frac{1}{2}mv^2$ $K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{7}{10}mv^2$ So, $\frac{K_t}{K_t + K_r} = \frac{5}{7}$