Question

A particle is moving with a constant speed $v$ in a circle of radius $R$. What is the magnitude of average acceleration after half revolution?

• A. $\frac{v^2}{2R}$
• B. $\frac{2v^2}{\pi R}$
• C. $\frac{v^2}{R}$
• D. $\frac{v^2}{\pi R}$

Answer 1

The correct option is B $\frac{2v^2}{\pi R}$

First find the change in momentum

$Changeinmomentum=finalmomentum-initialmomentum$

$=mv-(-mv)$

$=2mv$

Now

$Timetaken=\frac{distance}{speed}$

$=\frac{\pi r}{v}$

And

$averageforce=\frac{Changeinmomentum}{timetaken}$

$=\frac{2m{v}^2}{\pi r}$

Since, Force$=ma$, this implies that the average acceleration will be

$\begin{array}{c}ma=\frac{2m{v}^2}{\pi r}\\ a=\frac{2v^2}{\pi r}\end{array}$

Hence, the correct option is $B$