A particle moves around a circular path of radius ' $r$ ' with uniform speed ' $V$ '. After moving half the…
A particle moves around a circular path of radius ' $r$ ' with uniform speed ' $V$ '. After moving half the circle, the average acceleration of the particle is
$\frac{\mathrm{V}^2}{\mathrm{r}}$
$\frac{2 \mathrm{~V}^2}{\mathrm{r}}$
$\frac{2 \mathrm{~V}^2}{\pi \mathrm{r}}$
$\frac{\mathrm{V}^2}{\pi \mathrm{r}}$
Solution
At end points of the half revolution magnitude of the velocity is same but it directs in opposite direction.
$\begin{array}{ll}
\therefore & \Delta V=V-(-V) \\
\therefore & \Delta V=2 V
\end{array}$
Time taken to complete the half revolution is $\mathrm{t}=\frac{\pi \mathrm{r}}{\mathrm{V}}$
Average acceleration is, $a=\frac{\Delta V}{t}=\frac{2 \mathrm{~V}}{\frac{\pi \mathrm{r}}{\mathrm{V}}}$
$\therefore \quad a=\frac{2 V^2}{\pi r}$