A hollow vertical cylinder of radius R is rot | vedclass

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(D) For the mass $M$ to remain suspended on the inner wall of the rotating cylinder,the upward frictional force $f_s$ must balance the downward gravit

Vedclass · Accepted Answer

$\mu = \frac{g}{\omega^2 R}$

Vedclass · Answer

(D) For the mass $M$ to remain suspended on the inner wall of the rotating cylinder,the upward frictional force $f_s$ must balance the downward gravitational force $mg$.$f_s \geqslant mg$Since the frictional force $f_s = \mu N$,where $N$ is the normal force provided by the wall of the cylinder,we have:$\mu N \geqslant mg$The normal force $N$ provides the necessary centripetal force for the mass $M$ to move in a circle of radius $R$ with angular velocity $\omega$:$N = M R \omega^2$Substituting this into the inequality:$\mu (M R \omega^2) \geqslant Mg$$\mu \geqslant \frac{g}{R \omega^2}$Therefore,the minimum coefficient of static friction is:$\mu_{\min} = \frac{g}{R \omega^2}$

$A$ hollow vertical cylinder of radius $R$ is rotated with angular velocity $\omega$ about an axis through its center. What is the minimum coefficient of static friction necessary to keep the mass $M$ suspended on the inside of the cylinder as it rotates?

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