The Question is from neet-physics laws-of-motion , and it was created by the author Mr. ANIL KUMAR.G

Question

A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 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 to keep the block stationary when the cylinder is vertical and rotating about its axis, will be: ${\rm{(g  = 10}}\,{\rm{m/}}{{\rm{s}}^{\rm{2}}})$

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Circular Motion and Friction
	**Related Topics**: Centripetal Force, Static Friction, Newton's Laws of Motion
	**Prerequisites**: Understanding of forces in circular motion and friction.
## SOLUTION PATHWAY
**Given Information**:
	$m = 10$ kg (mass of the block)
	$r = 1$ m (radius of the cylinder)
	$\mu = 0.1$ (coefficient of friction)
	$g = 10$ m/s$^2$ (acceleration due to gravity)
**Method & Steps**:
	1. **Analyze the forces acting on the block**: 
		Reasoning: The block is stationary, so the forces must be balanced.
		Forces involved are: Normal force ($N$) from the cylinder wall (radially inwards), weight ($mg$) downwards, and frictional force ($f$) upwards.
	2. **Relate the normal force to the centripetal force**: 
		Working: For the block to remain stationary, the normal force provides the centripetal force required for circular motion.  $N = m\omega^2r$
		Key Point: The centripetal force is always directed towards the center of the circle.
	3. **Relate the frictional force to the weight**: 
		Working: The frictional force prevents the block from sliding down. The minimum angular velocity corresponds to the limiting case where the static friction equals the weight. $f = \mu N = mg$
	4. **Solve for angular velocity**: 
		Working: Substituting $N$ from step 2 into step 3: $\mu (m\omega^2r) = mg$. Simplifying, we get $\omega^2 = \frac{g}{\mu r}$. Therefore, $\omega = \sqrt{\frac{g}{\mu r}}$
		Result: $\omega = \sqrt{\frac{10}{0.1 	imes 1}} = \sqrt{100} = 10$ rad/s
		Verification: The units are consistent with angular velocity (rad/s).
## OPTION ANALYSIS
[A] :x: 5 rad/s: Incorrect. This would not provide enough centripetal force to keep the block stationary.
[B] ✓: 10 rad/s: Correct. This is the minimum angular velocity required for the block to remain stationary.
[C] :x: 20 rad/s: Incorrect. While the block would remain stationary at this speed, it's not the *minimum* required angular velocity.
[D] :x: 1 rad/s: Incorrect. This is too slow and the block would slide down.
## LEARNING AIDS
	**Common Mistakes**:
		1. Confusing the direction of the normal force: The normal force acts radially inwards, towards the center of the circular path.
			How to avoid: Remember that the normal force is always perpendicular to the surface of contact.
		2. Incorrectly applying the friction condition: The minimum angular velocity corresponds to the maximum static friction, which is equal to $\mu N$.
			How to avoid: Understand the concept of limiting friction.
	**Quick Tips**:
		1. Remember the relationship between centripetal force, angular velocity, and radius: $F_c = m\omega^2r$.
		2. The minimum angular velocity corresponds to the maximum static friction.

Mr. ANIL KUMAR.G · Answer

5 rad/s

Mr. ANIL KUMAR.G · Answer

The correct Option is "10 rad/s"

Mr. ANIL KUMAR.G · Answer

20 rad/s

Mr. ANIL KUMAR.G · Answer

1 rad/s

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