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