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A coin sits at distance $r$ from the centre of a turntable rotating with angular speed $\omega$. The **maximum $\omega$** before the coin slips (friction coefficient $\mu$) is:
A$\sqrt{\mu g/r}$
B$\mu g r$
C$\sqrt{\mu g r}$
D$g/(\mu r)$
Answer & Solution
Correct answer: A. $\sqrt{\mu g/r}$
Friction supplies centripetal force: $\mu mg \ge m\omega^2 r$ ⇒ $\omega_{\max} = \sqrt{\mu g/r}$.
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