A coin is kept at distance of 10 cm from the center of a circular turn table. If µ =0.8 , the frequency of rotation at which the coin just begins to slip is

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32.4 rpm

62.8 rpm

54.6 rpm

84.54 rpm

answer is B.

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Detailed Solution

$\begin{array}{l} {mω}^{2} r = μmg; ω = \sqrt{\frac{μ_{sg}}{r}} = \sqrt{\frac{0.8 \times 9.8}{0.1}} rad \sec^{- 1} \\ = \sqrt{\frac{0.8 \times 9.8}{0.1}} \times \frac{60}{2 π} = 84.54 rpm \end{array}$

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