Question

A disc revolves with a speed of $33\frac{1}{3}rev/min$ and has a radius of $15cm$. Two coins are placed at $4cm$ and $14cm$ away from the centre of the record. If the coefficient of friction between the coins and the record is $0.15$, which of the coin revolves with the record? Answer $1.00$ for the coin placed at $4cm$, $2.00$ for the coin placed at $14cm$ and $3.00$ for both coins.

Answer 1

Given
angular frequency $\omega =33\frac{1}{3}rev/min=\frac{100}{3}\times \frac{1}{60}rev/s$
$\omega =2\times \frac{22}{7}\times \frac{100}{3}\times \frac{1}{60}=3.5\text{rad/s}$
For coin to revolve with the record, the force of friction must be enough to provide necessary centripetal force,
i.e.,
$\frac{m{v}^2}{R}\le \mu mg\Rightarrow m{\omega }^{2}R\le \mu mg$
$\Rightarrow R\le \frac{\mu g}{{\omega }^2}\Rightarrow \frac{0.15\times 10}{(3.5{)}^2}=0.12\text{m or 12 cm}$
The coin should be maximum 12 cm from the centre to revolve with the disc. Hence, the coin placed at $4cm$ from centre will revolve with the disc.