Motion in a Plane - Result Question 50

53. 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: 53. (b)

Solution:

  1. (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} $



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