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ExamsJEE AdvancedPhysics

A pendulum bob is held with its string horizontal and released from rest. As the bob swings downward, at what angle alpha (measured from the vertical) is the net acceleration of the bob directed purely horizontally?

  1. 0
  2. 90 deg
  3. cos⁻¹(1/sqrt(3))
  4. sin⁻¹(1/sqrt(3))

Correct answer: cos⁻¹(1/sqrt(3))

Solution

Setting the vertical component of the total acceleration (centripetal upward component minus tangential downward component) to zero yields 2*cos²(alpha) = sin²(alpha), giving tan²(alpha) = 2, so alpha = cos⁻¹(1/sqrt(3)).

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