A stone tied to 180 cm long string at its end is making 28 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is 1936/x ms⁻². The value of x ________. (Take π=22/7 )

Solution

Physics sir · 3 steps

Turn revolutions per minute into ω in rad/s, put the length in metres, and use a = ω²r.

1. Only centripetal. The stone goes round at a steady rate, so it has no tangential acceleration. Its whole acceleration is centripetal, a = ω² r, with r = 180 cm = 1.8 m.
2. Find ω. 28 revolutions in 60 s, and each revolution is 2π radians. So ω = (2π × 28)/60 = (2 × 22 × 4)/60 = 44/15 rad s⁻¹.

   a = ω² r = (44/15)² × 1.8 = 1936/225 × 9/5 = 1936/125 m s⁻²

3. So. Matching with 1936/x gives x = 125.

The common slip: 1.25. Leaving the radius as 180 cm instead of 1.8 m makes the acceleration 100 times too big and gives x = 1.25, not 125.

The idea to keep. Any time a circular-motion question gives revolutions per minute, first write ω = 2π n/60 in rad/s. Then a = ω² r with r in metres.

Answer from the NTA final answer key. Solution by PrepWiser.