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 $4\, cm$ 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 disc?

• A. $A$
• B. $B$
• C. Both $A$ and $B$
• D. Neither $A$ nor $B$

Answer 1

Answer: (A)

Here, $\mu$ $=0.15,$
$v=33\frac{1}{3}rpm$ $=\frac{100}{3}rpm$ $=\frac{100}{3\times60}rps$ $=\frac{5}{9}rps$
$\therefore \quad$ $\omega=2\pi\upsilon$ $=2\times\frac{22}{7}\times\frac{5}{9}=\frac{220}{63}\,rad$ $s^{-1}$
The coin will revolve with the disc, if the force of friction is enough to provide the necessary centripetal force.
i.e. $\frac{mv^{2}}{r}$ $ \le \mu mg$
As $v=r\omega$, $\therefore $ $m\omega^{2}r$ $\le$ $\mu mg$
For the given $\mu$ and $\omega$, the condition is
$r \le \frac{\mu g}{\omega^{2}}$ $\quad\ldots\left(i\right)$
As $\frac{\mu g}{\omega^{2}}$ $=\frac{0.15\times10}{\left(\frac{220}{63}\right)^{2}}$ $\approx$ $12\, cm$
For coin $A$, $r = 4 \,cm$
The above condition is satisfied, therefore coin $A $ will revolve with the disc.
For coin $B$, $r = 14\, cm$
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.