Q. A particle moves along a circle of radius $\left( \frac{20}{\pi} \right)$ m with constant tangential acceleration. If the velocity of the particle is 80 $ms^{- 1},$ at the end of seconds revolution after motion has begun, the tangential acceleration is
A) 40 $ms^{- 2}$
B) 640 $\pi\ ms^{- 2}$
C) 1609$\ \pi\ ms^{- 2}$
D) 40 $\pi\ ms^{- 2}$
Explanation
(a) Initial angular velocity $\omega_{0} = 0$. Final angular velocity $\omega = \frac{v}{r} = \frac{80}{(20/\pi)} = 4\pi$ rad $s^{- 1}$
angle described, $\theta = 4\pi$ rad
$\therefore$ Angular acceleration, $\alpha = \frac{\omega^{2} - \omega_{0}^{2}}{2\theta}$
$= \frac{(4\pi)^{2} - 0}{2 \times 4\pi} = 2\pi$ rad $s^{- 2}$
Linear acceleration, $a = \alpha r$
$= 2\pi \times \frac{20}{\pi} = 40\ ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.