A coin placed on a rotating table just slips | vedclass

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(D) The condition for the coin to just slip on a rotating table is that the required centripetal force is provided by the maximum static friction forc

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$4$

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(D) The condition for the coin to just slip on a rotating table is that the required centripetal force is provided by the maximum static friction force.$f_{s,max} = m \omega^2 R$Since $f_{s,max} = \mu mg$,we have:$\mu mg = m \omega^2 R$$R = \frac{\mu g}{\omega^2}$This shows that the distance $R$ is inversely proportional to the square of the angular velocity,i.e.,$R \propto \frac{1}{\omega^2}$.Given initial conditions: $R_1 = 1\,cm$ and $\omega_1 = \omega$.New conditions: $\omega_2 = \frac{\omega}{2}$ and $R_2 = ?$.Using the proportionality: $\frac{R_2}{R_1} = \left( \frac{\omega_1}{\omega_2} \right)^2$$\frac{R_2}{1} = \left( \frac{\omega}{\omega/2} \right)^2 = (2)^2 = 4$$R_2 = 4\,cm$.Therefore,the coin will just slip when placed at a distance of $4\,cm$ from the center.

$A$ coin placed on a rotating table just slips when it is placed at a distance of $1\,cm$ from the center. If the angular velocity of the table is halved,it will just slip when placed at a distance of $............\,cm$ from the center.

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