Question

A particle moves along a circle of radius $r$ with constant tangential acceleration. If the velocity of the particle is $v$ at the end of second revolution after the revolution has started then the tangential acceleration is

• A. $\frac{v^{2}}{8 \pi r}$
• B. $\frac{v^{2}}{6 \pi r}$
• C. $\frac{v^{2}}{4 \pi r}$
• D. $\frac{v^{2}}{10 \pi r}$

Answer 1

Answer: (A)

Using $v^{2}-u^{2}=2 a s$ and $u=u_{1}$ and $s=4 \pi r$
$\therefore 2 a s=v^{2} \Rightarrow \frac{v^{2}}{2 s}=a=\frac{v^{2}}{2 \times 4 \pi r}=\frac{v^{2}}{8 \pi r}$