A block (P) is rotating in contact with the v | vedclass

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(C) For the block $P$ not to slide down,the frictional force $f$ must balance the weight $mg$,so $f = mg$.Since $f \leq \mu N$,we have $mg \leq \mu N$

Vedclass · Accepted Answer

$\omega_C < \omega_B < \omega_A$

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(C) For the block $P$ not to slide down,the frictional force $f$ must balance the weight $mg$,so $f = mg$.Since $f \leq \mu N$,we have $mg \leq \mu N$,where $N$ is the normal force provided by the wall.The normal force $N$ is the centripetal force,$N = m \omega^2 R$.Thus,for the limiting case of not sliding,$mg = \mu m \omega^2 R$,which implies $\omega^2 R = 	ext{constant}$ (assuming $\mu$ and $m$ are the same for all cases).Therefore,$\omega^2 R = C$ (a constant),which means $\omega \propto \frac{1}{\sqrt{R}}$.Given the radii $R_A  \frac{1}{\sqrt{R_B}} > \frac{1}{\sqrt{R_C}}$.Consequently,the angular velocities must satisfy $\omega_A > \omega_B > \omega_C$.

$A$ block $(P)$ is rotating in contact with the vertical wall of a rotor as shown in figures $A$,$B$,and $C$. Find the relation between angular velocities $\omega_A, \omega_B$,and $\omega_C$ such that the block does not slide down. ($R_A < R_B < R_C$ are the radii).

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