A disc revolves with a speed $33\frac{1}{3}$ rev/min and has a radius of 15 cm. Two coins A and B are placed at 4 cm and 14 cm away from the center of the disc. If the coefficient of friction between the coins and the disc is 0.15, which of the coins will revolve with the disc? 

Here, $\mu =0.15$

         $$\begin{gathered} \nu =33\frac{1}{3}rpm \\ =\frac{100}{3}rpm \\ =\frac{100}{3\times 60}rps \\ =\frac{5}{9}rps \end{gathered}$$

$$\begin{gathered} \omega =2\pi \nu \\ =2\times \times \frac{22}{7}\times \frac{5}{9} \\ =\frac{220}{63}rad s^{-1} \end{gathered}$$

The coin will revolve with the disc, if the force of friction is enough to provide the necessary centripetal force.

i.e., $$\begin{gathered} \frac{m{v}^2}{r}\le \mu mg \\ As v=r\omega , \\ \therefore m{\omega }^{2}r\le \mu mg \end{gathered}$$

For the given $\mu$ and $\omega$, the condition is 

$$\begin{gathered} r\le \frac{\mu g}{{\omega }^2} \\ As \frac{\mu g}{{\omega }^2}=\frac{0.15\times 10}{{\left({\displaystyle \frac{220}{63}}\right)}^2}\approx 12 cm \\ For coin A, r=4 cm \\ The above condition is satisfied, therefore, coin A will revolve with the disc. \\ For coin B, r= 14 cm \end{gathered}$$

The above condition is not satisfied, therefore coin B will not revolve with the disc.