Motion In A Plane Ques 49
49. 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 begun, the tangential acceleration is
[2003]
(a) $40\ \pi\ \text{m}/\text{s}^2$
(b) $40\ \text{m}/s^{2}$
(c) $640\ \pi\ \text{m}/\text{s}^2$
(d) $160\ \pi\ \text{m}/\text{s}^2$
Show Answer
Answer:
Correct Answer: 49.(b)
Solution:
- (b) Given, $r=\frac{20}{\pi} m$
$v_j=80$ $ m / sec \Rightarrow w_f=\frac{8 \pi}{2} \Rightarrow 4 \pi$
$\theta=2\pi$ radian $N=4$
From the equation,
$w_f^{2}=w_0^{2}+2 \alpha \theta \quad[\because w_0=0]$
$(4 \pi)^{2}=0+2 \alpha \cdot 4 \pi$
$\alpha=2\pi$
tangential acceleration
$ \begin{alignedat} & a_t=\alpha \cdot r \\ & a t=2 \pi \cdot \frac{20}{\pi}=40 \text{ m/s}^2 \end{aligned} $
BETA
