A coin placed on a rotating table just slips | vedclass

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Question

(C) The coin placed on the rotating table slips when the centripetal force $mr\omega^2$ is greater than or equal to the maximum static frictional forc

Vedclass · Accepted Answer

$r$

Vedclass · Answer

(C) The coin placed on the rotating table slips when the centripetal force $mr\omega^2$ is greater than or equal to the maximum static frictional force $f_{max} = \mu mg$.The condition for the coin to just slip is given by:$mr\omega^2 = \mu mg$This simplifies to:$r = \frac{\mu g}{\omega^2}$From this relation,we can see that $r \propto \frac{1}{\omega^2}$.Let the initial distance be $r_1 = 4r$ at angular velocity $\omega_1 = \omega$.Let the new distance be $r_2$ at angular velocity $\omega_2 = 2\omega$.Using the proportionality $r_1 \omega_1^2 = r_2 \omega_2^2$:$(4r)(\omega)^2 = r_2 (2\omega)^2$$4r \cdot \omega^2 = r_2 \cdot 4\omega^2$$r_2 = r$Thus,the coin will just slip when it is at a distance $r$ from the centre.

$A$ coin placed on a rotating table just slips if it is placed at a distance $4r$ from the centre. On doubling the angular velocity of the table,the coin will just slip when at a distance from the centre equal to

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