Question

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 4cm and 14 cm away from the centre 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 road?

• A. A
• B. B
• C. Both A and B
• D. neither A nor B

Answer 1

The correct option is A A

Here, $mu$=0.15,

$v=33\frac{1}{3}rpm=\frac{100}{3}rpm=\frac{100}{3\times 60}rps=\frac{5}{9}rps$

$\therefore \omega =2\pi v=2\times \frac{22}{7}\times \frac{5}{9}=\frac{220}{63}rad{s}^{-1}$

The coin will revolve with the disc, if the force of friction is enough to provide the necessary centripetal force.
i.e. $\frac{m{v}^2}{r}\le \mu mg$

As $v=r\omega ,\therefore m{\omega }^{2}r\le \mu mg$
For the given $\mu$ and $\omega$, the condition is

$r\le \frac{\mu g}{{\omega }^2}$

As $\frac{\mu g}{{\omega }^2}=\frac{0.15\times 10}{(\frac{220}{63}{)}^2}=12cm$

For coin A, $r=4cm$
The above condition is satisfied, therefore coin A will revolve with the disc.

For coin B, $r=14cm$

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

Note: we have nothing to do with the radius of the disc.