Question

A cylinder with radius $R$ spins about its horizontal axis with angular speed $\omega .$ There is a small block lying on the inner surface of the cylinder. The coefficient of friction between the block and the cylinder is $\mu$. If the block does not slip w.r.t. cylinder surface.

• A. The normal force on block will be $m{\omega }^{2}R-mgsin\theta$
• B. The frictional force on block must be $mg\sin \theta$
• C. Minimum angular velocity ${\omega }_{min}$ should be, $\sqrt{\frac{g\sqrt{1+{\mu }^2}}{R\mu }}$
• D. Minimum angular velocity ${\omega }_{min}$ should be, $\sqrt{\frac{g\sqrt{1+{\mu }^2}}{2R\mu }}$
  

Answer 1

Answer: (A), (C)

The correct options are
A The normal force on block will be $m{\omega }^{2}R-mgsin\theta$
C Minimum angular velocity ${\omega }_{min}$ should be, $\sqrt{\frac{g\sqrt{1+{\mu }^2}}{R\mu }}$

Consider the block at any position $\theta$ as shown in figure. [in the reference frame of cylinder]



Normal force, $N=m{\omega }^{2}R-mgsin\theta \to \left(1\right)$

Force of friction, $f=mgcos\theta \to \left(2\right)$

Since, $f\le \mu N$

$\Rightarrow mgcos\theta \le \mu m{\omega }^{2}R-\mu mgsin\theta$
$\Rightarrow g[cos\theta +\mu sin\theta ]\le \mu {\omega }^{2}R$

For all values of $\theta ,$ maximum value of function $[acos\theta +bsin\theta ]$ is ${[a^2+b^2]}^{\frac{1}{2}}$

Maximum value of $[cos\theta +\mu sin\theta ]$ can be determined to be ${[1+{\mu }^2]}^{\frac{1}{2}}$
$\therefore g\sqrt{1+{\mu }^2}\le \mu {\omega }^{2}R$
$\Rightarrow \omega \ge \sqrt{\frac{g\sqrt{1+{\mu }^2}}{R\mu }}$
$\Rightarrow {\omega }_{min}=\sqrt{\frac{g\sqrt{1+{\mu }^2}}{R\mu }}$