The Question is from NEET guide , and it was created by the author P.NARASIMHA

Question

A particle moves in a vertical circle attached to a string of length 'l'.  At the lowest point, its speed is 'u'. What is the minimum value of 'u' required for the string to remain taut at the highest point?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option D
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Energy and Circular Motion
	**Related Topics**: Centripetal Force, Potential Energy, Kinetic Energy
	**Prerequisites**: Basic understanding of energy conservation and circular motion.
## SOLUTION PATHWAY
**Given Information**:
	* Length of the string: l
	* Speed at the lowest point: u
	* String must remain taut at the highest point.

**Method & Steps**:
	1.  **Determine the condition for the string to remain taut at the highest point**.
		Reasoning: For the string to remain taut, the tension in the string must be greater than or equal to zero at the highest point. The minimum speed will occur when the tension is exactly zero.
		Formula/Rule: At the highest point, the centripetal force is provided by the weight of the particle and the tension in the string. $T + mg = \frac{mv^2}{l}$. For minimum speed, we have T = 0, so $mg = \frac{mv^2}{l}$, and $v = \sqrt{gl}$, where 'v' is the speed at the highest point.
	2.  **Apply the conservation of energy between the lowest and highest points**.
		Working: The total energy at the lowest point (kinetic energy only) equals the total energy at the highest point (kinetic and potential energy).  $KE_{lowest} = KE_{highest} + PE_{highest}$. So, $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mg(2l)$. Substituting $v = \sqrt{gl}$ into the energy equation: $\frac{1}{2}mu^2 = \frac{1}{2}m(gl) + 2mgl$. Dividing by m and multiplying by 2: $u^2 = gl + 4gl = 5gl$.
		Key Point: The height difference between the lowest and highest points is 2l.
	3.  **Solve for u**.
		Result: $u = \sqrt{5gl}$
		Verification: The units are consistent, and the derived expression makes physical sense.
 ## OPTION ANALYSIS
[A] :x: This would be the speed required at the highest point to just maintain circular motion.
[B] :x: This is not the correct value for the speed at the lowest point for the string to remain taut at the highest point.
[C] :x: This is also not the correct value.
[D] ✓: This is the correct minimum speed required at the lowest point for the string to remain taut at the highest point.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Forgetting the potential energy at the highest point**: Remember that the particle gains potential energy as it moves from the lowest to the highest point.
		2. **Incorrectly using the centripetal force equation**:  Ensure that the centripetal force is directed towards the center of the circle, and consider the tension in the string.
	**Quick Tips**:
		1. **Draw a diagram**: Visualizing the problem can help in understanding the forces acting on the particle.
		2. **Use energy conservation**: It's often easier to solve problems involving circular motion using energy conservation.

P.NARASIMHA · Answer

$\sqrt{gl}$

P.NARASIMHA · Answer

$\sqrt{2gl}$

P.NARASIMHA · Answer

$\sqrt{3gl}$

P.NARASIMHA · Answer

The correct Option is "$\sqrt{5gl}$"

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