Question
A thin uniform circular ring of mass m is rolling without slipping down an inclined plane of inclination 30° with the horizontal. The coefficient of friction between the ring and the surface is μ. The correct statement is :
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Solution
[Key Concept: For rolling without slipping, a = g sin θ/(1 + I/mR2) and the friction needed is mg sin θ/(1 + mR2/I), limited by μN.]
$$a = \frac{gsin\theta}{1 + \frac{I}{{mR}^{2}}}$$
$$\begin{matrix} & a = \frac{g\left( \frac{1}{2} \right)}{1 + \frac{{mR}^{2}}{{mR}^{2}}} = \frac{g}{4} \\ & f = \frac{{mg}^{\sin\theta}}{1 + \frac{{mR}^{2}}{I}} \\ & f = \frac{mg\left( \frac{1}{2} \right)}{1 + 1} \\ & f = \frac{mg}{4} \rightarrow \text{~ans~}(b) \end{matrix}$$
$$\begin{matrix} & \frac{mg}{4} \leq \mu mg\frac{\sqrt{3}}{2} \\ & \mu_{\min} \geq \frac{1}{2\sqrt{3}} \end{matrix}$$
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