MCQ
A particle moves along a circle if 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
  • A
    $640\,\pi \,m/{s^2}$
  • B
    $160\,\pi \,m/{s^2}$
  • C
    $40\,\pi \,m/{s^2}$
  • $40\,\,m/{s^2}$

Answer

Correct option: D.
$40\,\,m/{s^2}$
d
(d)$\omega = \frac{v}{r} = \frac{{80}}{{20/\pi }} = 4\pi ,\,\,{\omega _0} = 0,\,\,\theta = 2\pi (n) = 4\pi $ ${\omega ^2} = \omega _0^2 + 2\alpha \theta \Rightarrow \alpha = \frac{{{\omega ^2} - \omega _0^2}}{{2\theta }} = \frac{{16{\pi ^2}}}{{2 \times 4\pi }} = 2\pi $
स्पर्श रेखीय त्वरण ${a_t} = r\alpha = \frac{{20}}{\pi } \times 2\pi = 40m/{s^2}$

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