Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A block of mass $10 \text{ kg}$ is in contact with the inner wall of a hollow cylindrical drum of radius $1 \text{ m}$ . The coefficient of friction between the block and the inner wall of the cylinder is $0.1$ . The minimum angular velocity needed for the cylinder, which is vertical and rotating about its axis, will be: ( $g=10 \text{ m/s}^2$ )

$10 \pi \text{ rad/s}$

$\sqrt{10} \pi \text{ rad/s}$

$\frac{10}{2\pi} \text{ rad/s}$

$10 \text{ rad/s}$

Step-by-Step Solution

System Analysis: The block is rotating with the cylinder. The normal reaction ( $N$ ) from the wall provides the necessary centripetal force. The friction force ( $f$ ) acts vertically upwards to balance the weight ( $mg$ ) of the block and prevent it from sliding down.
Equations of Motion:

Horizontal (Centripetal) Force: $N = m \omega^2 R$
Vertical Equilibrium: $f = mg$

Condition for No Sliding: The static friction must be sufficient to hold the weight. The limiting friction is $f_{max} = \mu N$ . For the block to remain stationary relative to the wall, required friction $f \le f_{max}$ . $mg \le \mu N$ Substitute $N = m \omega^2 R$ : $mg \le \mu (m \omega^2 R)$ $g \le \mu \omega^2 R$
Minimum Angular Velocity: $\omega^2 \ge \frac{g}{\mu R}$ $\omega_{min} = \sqrt{\frac{g}{\mu R}}$
Calculation: Given $g = 10 \text{ m/s}^2$ , $\mu = 0.1$ , $R = 1 \text{ m}$ . $\omega_{min} = \sqrt{\frac{10}{0.1 \times 1}} = \sqrt{100} = 10 \text{ rad/s}$ (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, Section 5.10 Circular Motion and Section 5.9 Friction).

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