NEET-XII-Physics
07: Circular Motion
- #9A 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 cmdigAnsr: aAns : (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,
`` \frac{r\text{ ' }}{r}={\left({\displaystyle \frac{\omega }{\omega \text{ ' }}}\right)}^{2}``
`` \Rightarrow r\text{ ' }=1\,\mathrm{\,cm \,}``
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