A particle moves along a circle of radius $\frac{20}{\pi} ext{ m}$ with constant tangential acceleration. If

Question

A particle moves along a circle of radius $\frac{20}{\pi} 	ext{ m}$ with constant tangential acceleration. If the velocity of the particle is $80 	ext{ m/s}$ at the end of the second revolution after motion has begun, the tangential acceleration is:

Accepted Answer

Let the tangential acceleration of the particle be $a_t$.
Given:
Initial velocity, $u = 0$ (starts from rest)
Final velocity, $v = 80 	ext{ m/s}$
Radius, $r = \frac{20}{\pi} 	ext{ m}$
Number of revolutions, $n = 2$

The total linear distance $s$ covered by the particle along the circular path in 2 revolutions is:
$s = n 	imes 2\pi r = 2 	imes 2\pi 	imes \left(\frac{20}{\pi}ight) = 80 	ext{ m}$

Using the third equation of motion ($v^2 = u^2 + 2a_ts$):
$(80)^2 = 0^2 + 2 	imes a_t 	imes 80$
$6400 = 160 a_t$
$a_t = \frac{6400}{160} = 40 	ext{ m/s}^2$.

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