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. $\frac{v^2}{2\pi r}$

Answer 1

Answer: (A)

$\frac{v^2}{8\pi r}$

Initial speed of the particle $u=0$

Final speed of the particle is $v$

Distance covered in 2 revolutions $S=2\left(2\pi r\right)=4\pi r$

Using $v^2-u^2=2aS$

where $a$ is the tangential acceleration

$\therefore$ $v^2-0=2a\left(4\pi r\right)$

$⟹$ $a=\frac{v^2}{8\pi r}$