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