First slide

Rolling motion

Question

The rotational kinetic energy of a solid sphere of mass 3 kg and radius 0.2 m rolling down an inclined plane of height 7 m is :

Moderate

Solution

$\begin{array}{l} m g h = K E_{T} + K E_{R} \\ = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} \\ = \frac{1}{2} m v^{2} + \frac{1}{2} \frac{2}{5} m R^{2} \frac{v^{2}}{R^{2}} \\ = \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} \\ = \frac{7}{10} m v^{2} \\ m v^{2} = \frac{10}{7} \times 3 \times 10 \times 7 = 300 \\ k E_{R} = \frac{1}{2} I ω^{2} = \frac{1}{5} m v^{2} = 60 J \end{array}$

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