Question

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

• A. $40m/s^2$
• B. $640m/s^2$
• C. $160\pi m/s^2$
• D. $40\pi m/s^2$

Answer 1

The correct option is A $40m/s^2$

Given:
$r=\frac{20}{\pi }m,v=80m/s,\theta =2rev=4\pi rad$.
From equation ${\omega }^2={\omega }_0^2+2\alpha \theta ({\omega }_0=0)$
$\Rightarrow {\omega }^2=2\alpha \theta (\omega =\frac{v}{r}anda=r\alpha )$
$\Rightarrow a=\frac{v^2}{2r\theta }=40m/s^2$
Hence, the correct answer is option (a).