Question

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