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(A) When a coin is placed on a rotating turntable,the necessary centripetal force for circular motion is provided by the static frictional force betwe

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(A) When a coin is placed on a rotating turntable,the necessary centripetal force for circular motion is provided by the static frictional force between the coin and the surface of the turntable.For the coin to just slip,the centripetal force must equal the maximum static frictional force:$m r \omega^2 = \mu m g$where $m$ is the mass of the coin,$r$ is the distance from the centre,$\omega$ is the angular velocity,$\mu$ is the coefficient of friction,and $g$ is the acceleration due to gravity.Since $m, \mu$,and $g$ are constant,we have:$r \omega^2 = 	ext{constant}$This implies $r \propto \frac{1}{\omega^2}$.Therefore,the ratio of distances for two different angular velocities is:$\frac{r_2}{r_1} = \left( \frac{\omega_1}{\omega_2} ight)^2$Given $r_1 = 4 \ cm$ and $\omega_2 = 2 \omega_1$,we substitute these values:$\frac{r_2}{4} = \left( \frac{\omega_1}{2 \omega_1} ight)^2 = \left( \frac{1}{2} ight)^2 = \frac{1}{4}$$r_2 = 4 	imes \frac{1}{4} = 1 \ cm$.

$A$ 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: (in $cm$)

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