Question

A particle moves along a circle of radius $\frac{40}{\pi }m$ with constant tangential acceleration. If the velocity of the particle is $100m/s$ at the end of the $4th$ revolution after motion has begin, the tangential acceleration is

• A. $\frac{125}{8}$
• B. $\frac{35}{6}$
• C. $\frac{26}{3}$
• D. $\frac{5}{4}$

Answer 1

Answer: (A)

As we know, $\omega =\frac{v}{r}=\frac{100}{40}\times \pi =\frac{5\pi }{2}$
$\theta =2\pi n=2\pi \times 4=8\pi$
$\because \alpha =\frac{{\omega }^2}{2\theta }$
$\alpha =\frac{{\left(\frac{5}{2}\pi \right)}^2}{2\times 8\pi }$
$\alpha =\frac{25}{64}\pi$
$a_T=\alpha r$
$=\frac{25}{64}\pi \times \frac{40}{\pi }$
$a_r=\frac{125}{8}$