Question

A coin placed on a rotating turntable just slips. If it is placed at a distance of 4 cm from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of
(a) 1 cm
(b) 2 cm
(c) 4 cm
(d) 8 cm

Answer 1

(a) 1 cm

Let the force of friction between the coin and the rotating turntable be F.
For the coin to just slip, we have:
$F=m{\omega }^{2}r$
Here, $m{\omega }^{2}r$is the centrifugal force acting on the coin.
For constant F and m, we have:
$r\propto \frac{1}{{\omega }^2}$
Therefore,
$$\begin{gathered} \frac{r\text{'}}{r}={\left({\displaystyle \frac{\omega }{\omega \text{'}}}\right)}^2 \\ \Rightarrow r\text{'}=1\mathrm{cm} \end{gathered}$$