A large drum having radius R is spinning around its axis with angular velocity omega$\omega$, as shown in figure. The minimum value of omega$\omega$ so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M is mu$\mu$, is :

### Related Formula N = Momega^2 R$N = M\omega^2 R$ f_textmax = mu N$$f_{\text{max}} = \mu N$$ f = Mg$f = Mg$ ### Core Logic

For the mass M$M$ to remain stuck to the inner wall and not slide down, the upward frictional force must balance the downward gravitational force. f = Mg$f = Mg$ The normal force N$N$ providing the friction is generated by the centrifugal effect (or provides the centripetal acceleration): N = Momega^2 R$N = M\omega^2 R$ The condition for no sliding is that required friction cannot exceed the maximum static friction: Mg le mu N$Mg \le \mu N$ ### Step 1: Final Conclusion Substitute N$N$ into the inequality: Mg le mu (Momega^2 R)$$Mg \le \mu (M\omega^2 R)$$ g le mu omega^2 R$$g \le \mu \omega^2 R$$ omega^2 ge fracgmu R$$\omega^2 \ge \frac{g}{\mu R}$$ For minimum angular velocity: omega_textmin = sqrtfracgmu R$$\omega_{\text{min}} = \sqrt{\frac{g}{\mu R}}$$ ### Pattern Recognition In a "rotor" ride setup, gravity is opposed by friction, and the normal force is solely centripetal. Equating mu F_c$\mu F_c$ to mg$mg$ always yields the minimum spin limit. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion