Question

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

• A. Both will revolve
• B. The nearer one will revolve and the farther one will not
• C. The farther one will revolve and the nearer one will not
• D. Both will not revolve

Answer 1

Answer: (B)

For the coin to revolve with the disc, the force of friction should be enough to provide the necessary
centripetal force i.e. $\frac{m v^{2}}{r} \leq \mu m g $ Now $v=r \omega,$ where
$\omega=\frac{2 \pi}{T}$ is the angular frequency of the disc. For a given
$\mu$ and $\omega$, the condition is $r \leq \mu g / \omega^{2}$
Here, $\omega=2 \pi\left(\frac{100}{3}\right) \times \frac{1}{60} \frac{ rad }{ s }=\frac{10 \pi}{9} \frac{ rad }{ s }$
and $\frac{\mu g}{\omega^{2}}=\frac{0.15 \times 10}{\left(\frac{10 \pi}{9}\right)^{2}} \approx 0.12 m =12 \,cm$
$\Rightarrow $ If $r<\, 12\, cm$, the coin will revolve with the record