A block of mass m is kept on a horizontal tur | vedclass

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(B) For the block not to slip,the required centripetal force must be provided by the static friction force.The condition for the block not to slip is:

Vedclass · Accepted Answer

$\sqrt{\frac{\mu g}{x}}$

Vedclass · Answer

(B) For the block not to slip,the required centripetal force must be provided by the static friction force.The condition for the block not to slip is:$f_{s} \leq \mu N$Here,the centripetal force is $F_{c} = m \omega^2 x$.The normal force $N$ on the horizontal surface is $N = mg$.Therefore,the maximum static friction force is $f_{s,max} = \mu mg$.Equating the centripetal force to the maximum static friction force:$m \omega^2 x = \mu mg$Dividing both sides by $mx$:$\omega^2 = \frac{\mu g}{x}$Taking the square root of both sides,we get the maximum angular speed:$\omega = \sqrt{\frac{\mu g}{x}}$

$A$ block of mass $m$ is kept on a horizontal turntable at a distance $x$ from the center. If the coefficient of friction between the block and the surface of the turntable is $\mu$,what is the maximum angular speed of the table so that the block does not slip?

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