Q. A particle moves along a circle of radius $\left\lbrack \frac{20}{\pi} \right\rbrack$ 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 begun, the tangential acceleration is
A) $40\ ms^{- 2}$
B) 640 $\pi ms^{- 2}$
C) 160 $\pi ms^{- 2}$
D) 40 $\pi ms^{- 2}$
Explanation
(a) Given, $r = (20/\pi)m$
$v = 80\ m/s$
$\theta = 2\ rev = 4\pi\ rad$
$\omega_{0} = 0$
From the equation
$\omega^{2} = \omega_{0}^{2} + 2\alpha\theta,$ we have
$\omega^{2} = 2\alpha\theta$
or $\frac{v^{2}}{r^{2}} = 2 \bullet \frac{a}{r}\theta$
or $a = \frac{v^{2}}{2r\theta} = \frac{(80)^{2}}{2 \times (20/\pi) \times 4\pi}$
$= 40\ ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.