Vertical Circular Motion PYQ | Minimum Speed at Lowest Point | JEE Main 2026 January

Minimum Speed at the Lowest Point So That Tension Becomes Zero at Point A | JEE Main Physics

This problem is based on the concepts of Vertical Circular Motion, Centripetal Force, and the Conservation of Mechanical Energy. The particle is projected from the lowest point of a vertical circle, and we need to determine the minimum speed required so that the tension in the string just becomes zero at point A.

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Question

A particle attached to a light string of length l is projected from the lowest point of a vertical circle. The string makes an angle of 60° with the upward vertical at point A. Find the minimum speed at the lowest point so that the tension in the string becomes zero at point A.

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Given Data

• Length of string = l
• Angle at point A = 60° with upward vertical
• Tension at point A = 0
• Acceleration due to gravity = g

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Concepts Used

• Condition for minimum speed in vertical circular motion.
• Centripetal force provided by the radial component of gravity.
• Conservation of Mechanical Energy.

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Important Formulae

Radial Equation:

T + mg cosθ = mv²/l

Conservation of Mechanical Energy:

½mv₀² = ½mv² + mgh

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Step 1: Apply the Condition at Point A

At the required minimum speed, the string is just about to become slack at point A. Therefore,

T = 0

Hence the radial force equation becomes

mg cos60° = mv²/l

Cancelling the mass from both sides,

v² = gl cos60°

Since,

cos60° = 1/2

Therefore,

v² = gl/2

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Step 2: Find the Height of Point A

Take the lowest point of the circle as the reference level.

The height of point A above the lowest point is

h = l + l cos60°

= l + l/2

h = 3l/2

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Step 3: Apply Conservation of Mechanical Energy

At the lowest point,

½mv₀² = ½mv² + mgh

Substitute

• v² = gl/2
• h = 3l/2

v₀² = gl/2 + 2g(3l/2)

= gl/2 + 3gl

= 7gl/2

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Final Answer

v₀ = √(7gl/2)

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Why Tension Becomes Zero?

The string can only pull the particle and cannot push it. The minimum speed occurs when the string is just about to lose contact with the particle. At this instant, the tension becomes zero, and gravity alone provides the required centripetal force.

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Key Formulae for Revision

• T + mg cosθ = mv²/l
• At limiting condition, T = 0
• Height at angle θ from upward vertical: h = l(1 + cosθ)
• ½mv₁² + mgh₁ = ½mv₂² + mgh₂

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Common Mistakes

• Using sin60° instead of cos60° in the radial equation.
• Forgetting that the tension is zero at the limiting condition.
• Calculating the height incorrectly from the lowest point.
• Ignoring the conservation of mechanical energy while relating the two speeds.
• Using an incorrect sign convention for gravitational potential energy.

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Exam Tips

• Whenever the question mentions the string becomes slack, immediately write T = 0.
• Always draw the free-body diagram to identify the radial forces.
• Measure the height carefully with respect to the chosen reference point.
• Remember that energy conservation is the quickest method to connect speeds at different points.
• Questions on vertical circular motion are frequently asked in JEE Main and should be practiced thoroughly.