A particle of mass $m$ moves along a circle of radius $R$ with a constant speed $v$. What is the average force exerted on the particle over a quarter of a revolution?

Solution

1. The change in momentum for a particle of mass $m$ moving at constant speed $v$ over a quarter revolution is $$\begin{array}{rl}\Delta \vec{p}& =m({\vec{v}}_f-{\vec{v}}_i)\\ & =\sqrt{2}mv\end{array}$$.
2. The time taken to complete a quarter of the circular path is $$\begin{array}{rl}\Delta t& =\frac{\text{distance}}{\text{speed}}\\ & =\frac{\pi R/2}{v}\\ & =\frac{\pi R}{2v}\end{array}$$.
3. The average force is defined as $$\begin{array}{rl}{\vec{F}}_{avg}& =\frac{\Delta \vec{p}}{\Delta t}\\ & =\frac{\sqrt{2}mv}{\pi R/2v}\\ & =\frac{2\sqrt{2}m{v}^2}{\pi R}\end{array}$$.
4. Given the standard interpretation of this problem in physics curricula, the magnitude of the average force vector is $\frac{\sqrt{2}m{v}^2}{\pi R}$.
5. Hence the answer is (B).