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ExamsJEE AdvancedPhysics

A particle P of mass m is attached to a vertical axis by two strings AP and BP, each of length l. The points A and B on the axis are separated by a distance l (so triangle APB is equilateral). The particle rotates in a horizontal circle about the axis with angular velocity omega. The tensions in strings AP and BP are T1 (upper string) and T2 (lower string) respectively. Under what condition does string BP remain taut?

  1. T1 = T2
  2. T1 + T2 = m*omega²*l
  3. T1 - T2 = 2mg
  4. omega >= sqrt(2g/l)

Correct answer: omega >= sqrt(2g/l)

Solution

From vertical equilibrium: T1 - T2 = 2mg. From horizontal (centripetal): T1 + T2 = m*omega²*l. Adding: 2T1 = 2mg + m*omega²*l. Subtracting: 2T2 = m*omega²*l - 2mg. For T2 >= 0: m*omega²*l >= 2mg, so omega² >= 2g/l, i.e., omega >= sqrt(2g/l).

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