(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 $mg \leq \mu N$

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$\omega_C < \omega_B < \omega_A$

(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 $\omega^2 R = \text{constant}$ (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 $\omega_A > \omega_B > \omega_C$.

Class 11 Physics Newton's Laws of Motion and Friction Circular motion with Friction

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$A$ 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).

AP EAMCET 2024 All AP EAMCET

A
$\omega_A < \omega_B < \omega_C$
B
$\omega_A = \omega_B = \omega_C$
C
$\omega_C < \omega_B < \omega_A$
D
$\omega_C = \omega_A + \omega_B$

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Newton's Laws of Motion and Friction

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Circular motion with Friction

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$A$ 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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$A$ vehicle is moving with speed $v$ on a curved road of radius $r$. The coefficient of friction between the vehicle and the road is $\mu$. The angle $\theta$ of banking needed is given by

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