Q. If a cycle wheel of radius 4 m completes one revolution in two second, the acceleration of the cycle in $ms^{- 2}$ is
A) $4$
B) $4\pi^{2}$
C) $2\pi^{2}$
D) $\pi^{2}$
Explanation
(b) Here, $r = 4m,\ T = 2s,\ a = ?$
$a = r\omega^{2} = r\left( \frac{2\pi}{T} \right)^{2} = \frac{4\pi^{2}r}{T^{2}} = 4\pi^{2} \times \frac{4}{2^{2}} = 4\pi^{2}ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.