Question

Find speed given to particle at lowest point so that tension in string at point A becomes zero. 

• A. $\sqrt{\dfrac{7g\ell}{2}}$
• B. $\sqrt{3g\ell}$
• C. $\sqrt{\dfrac{9g\ell}{4}}$
• D. $\sqrt{\dfrac{g\ell}{2}}$

Hint:

When tension becomes zero in circular motion, gravity alone provides the centripetal force.

Answer 1

The Correct Option is A

Step 1: Identify the physical situation

A particle of mass m is tied to a light string of length ℓ and moves in a vertical circle. At point A, the string makes an angle of 60° with the vertical and the tension in the string becomes zero.

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Step 2: Use radial force balance at point A

When tension is zero, the centripetal force is provided only by the radial component of weight.

Radial component of weight = mg cos 60° = mg / 2

Equating to centripetal force:

mg / 2 = mv² / ℓ

v² = gℓ / 2

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Step 3: Calculate change in height from lowest point to point A

Vertical rise of the particle from the lowest point to point A is:

h = ℓ + ℓ cos 60°

h = ℓ + ℓ / 2 = 3ℓ / 2

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Step 4: Apply work–energy theorem

Initial kinetic energy at the lowest point = Work done against gravity + Kinetic energy at point A

(1/2) m u² = mg (3ℓ / 2) + (1/2) m v²

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Step 5: Substitute known values

(1/2) u² = (3gℓ / 2) + (1/2)(gℓ / 2)

(1/2) u² = (3gℓ / 2) + (gℓ / 4)

(1/2) u² = 7gℓ / 4

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Step 6: Final result

u² = 7gℓ / 2

u = √(7gℓ / 2)