Question
A uniform solid sphere of radius R rolls without slipping on a rough horizontal surface with a forward velocity v of its center. On its way, it suddenly encounters a small step of height 0.2 R as shown. The angular velocity of the sphere just after the impact is [given that the sphere does not bounce back, rather it goes ahead up the step]
Options
Solution
[Key Concept: During a sudden impact at the step edge P, the impulsive force passes through P, so angular momentum about P is conserved.]
$$\begin{matrix} & \text{~About P~} \\ & L_{i} = L_{F} \\ & MV(0.8R) + \frac{2}{5}MR^{2}\left( \frac{V}{R} \right) = \left( \frac{2}{5}MR^{2} + MR^{2} \right)\omega \\ & 0.8VR + 0.4VR = 1.4R^{2}\omega \\ & \frac{1.2V}{1.4R} = \omega \\ & \omega = \frac{6V}{7R} \\ & \text{~option (c)~} \end{matrix}$$
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