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ExamsJEE AdvancedPhysics

A small particle of mass m is given an initial velocity in a horizontal plane and winds its cord around a fixed vertical shaft of radius 1 m. The initial angular velocity of the cord (measured from the tangency point) is 0.8 rad/s when the distance from the particle to the tangency point is 5 m. Find the angular velocity omega (in rad/s) of the cord after it has wound sufficiently (assume the cord length to the tangency point decreases by the circumference wound, i.e. the cord shortens).

  1. 0.2
  2. 0.4
  3. 0.8
  4. 1.6

Correct answer: 1.6

Solution

The tension in the cord passes through the tangency point on the shaft, so there is no torque about the shaft axis. The speed v of the particle remains constant. Initially v = r0 * omega₀ = 5 * 0.8 = 4 m/s. After the cord winds to a new length r, omega = v/r. The problem asks for omega after the cord shortens by winding; the question is incomplete without specifying the final length. However, from the standard version of this problem the cord shortens to r = 5 - 2*pi*1 ~ 2.5 m (after one half-turn), but from the answer options omega = v/r = 4/r must match one option. At r = 2.5: omega = 1.6 rad/s. Answer: 1.6 rad/s.

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