JEE Mains · Physics · STD 11 - 4.2 friction
A coin is placed on a disc. The coefficient of friction between the coin and the disc is \(\mu\). If the distance of the coin from the center of the disc is \(r\), the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is _______.
- A \(\frac{\mu g}{r}\)
- B \(\sqrt{\frac{\mathrm{r}}{\mu g}}\)
- C \(\sqrt{\frac{\mu g}{r}}\)
- D \(\frac{\mu}{\sqrt{\mathrm{rg}}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{\mu g}{r}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{N}=\mathrm{mg}\) \(\mathrm{f}=\mathrm{m} \omega^2 \mathrm{r}\) \(\mathrm{f}=\mu \mathrm{N}\) \(\mu \mathrm{mg}=\mathrm{mr} \omega^2\) \(\omega=\sqrt{\frac{\mu g}{r}}\)
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