Q. A stone tied to the end of a string $1m$ long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, what is the magnitude and direction of acceleration of the stone
A) $\pi^{2}/4\ ms^{- 2}$ and direction along the radius towards the centre
B) $\pi^{2}ms^{- 2}$ and direction along the radius away from the centre
C) $\pi^{2}ms^{- 2}$ and direction along the radius towards the centre
D) $\pi^{2}ms^{- 2}$ and direction along the tangent to the circle
Explanation
(c) $a = \frac{v^{2}}{r} = \omega^{2}r = 4\pi^{2}n^{2}r = 4\pi^{2}\left( \frac{22}{44} \right)^{2} \times 1 = \pi^{2}m/s^{2}$
and its direction is always along the radius and towards the centre
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.