Q. A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 s, what is the magnitude and direction of acceleration of the stone ?
A) (π²/4) m s⁻² and direction along the radius towards the centre
B) 2π² m s⁻² and direction along the radius away from centre
C) π² m s⁻² and direction along the radius towards the centre
D) 4π² m s⁻² and direction along the tangent to the circle
Explanation
[Solution from source] f = 22/44 = 0.5 rev/s, so ω = 2πf = π rad/s. a = ω²r = π² × 1 = π² m s⁻², directed along the radius towards the centre (centripetal). Acceleration of uniform circular motion is never tangential or outward. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.