Q. A particle moves along a circle of radius $\frac{20}{\pi}m$ with constant tangential acceleration. If the velocity of the particle is $80\ m/s$ at the end of the second revolution after motion has begin, the tangential acceleration is
A) $640\ \pi\ m/s^{2}$
B) $160\ \pi\ m/s^{2}$
C) $40\ \pi\ m/s^{2}$
D) $40\ m/s^{2}$
Explanation
(d) $\omega = \frac{v}{r} = \frac{80}{20/\pi} = 4\pi,\omega_{0} = 0,\theta = 2\pi(n) = 4\pi(As\ n = 2)$
$\omega^{2} = \omega_{0}^{2} + 2\alpha\theta \Rightarrow \alpha = \frac{\omega^{2} - \omega_{0}^{2}}{2\theta} = \frac{16\pi^{2}}{2 \times 4\pi} = 2\pi$
Tangential acceleration $a_{t} = r\alpha = \frac{20}{\pi} \times 2\pi = 40m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.