Question

A disc revolves with a speed $33$ $\frac{1}{3}$ rev/ min, and has a radius of $15cm$. Two coins $A$ and $B$ are placed at $4cm$ and $14cm$ 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, $\mu$ $=0.15,$
$v=33\frac{1}{3}rpm$ $=\frac{100}{3}rpm$ $=\frac{100}{3\times 60}rps$ $=\frac{5}{9}rps$
$\therefore$ $\omega =2\pi \upsilon$ $=2\times \frac{22}{7}\times \frac{5}{9}=\frac{220}{63}rad$ $s^{-1}$
The coin will revolve with the disc, if the force of friction is enough to provide the necessary centripetal force.
i.e. $\frac{m{v}^2}{r}$ $\le \mu mg$
As $v=r\omega$, $\therefore$ $m{\omega }^{2}r$ $\le$ $\mu mg$
For the given $\mu$ and $\omega$, the condition is
$r\le \frac{\mu g}{{\omega }^2}$ $\dots \left(i\right)$
As $\frac{\mu g}{{\omega }^2}$ $=\frac{0.15\times 10}{{\left(\frac{220}{63}\right)}^2}$ $\approx$ $12cm$
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.