Question

A particle of mass $m$ moves with constant speed $v$ on a circular path of radius $r$ as shown in figure. The average force on it during its motion from $A$ to $B$ is

• A. $\frac{\sqrt{3}m{v}^2}{2\pi r}$
• B. $\frac{m{v}^2}{r}$
• C. $\frac{2\sqrt{3}m{v}^2}{\pi r}$
• D. $\frac{3\sqrt{3}m{v}^2}{4\pi r}$

Answer 1

The correct option is D $\frac{3\sqrt{3}m{v}^2}{4\pi r}$

$Themagnitudeofavgforce\overrightarrow{F_{av}}isgivenby\overrightarrow{F_{av}}=\frac{\overrightarrow{\Delta p}}{\Delta t}\overrightarrow{v_f}=vcos\theta \hat{i}+vsin\theta (-\hat{j})\overrightarrow{v_i}=v\hat{j}\overrightarrow{\Delta p}=\overrightarrow{p_f}-\overrightarrow{p_i}=m\overrightarrow{v_f}-m\overrightarrow{v_i}=mvcos\theta \hat{i}+(-mvsin\theta -mv)\hat{j}=\frac{mv\sqrt{3}}{2}\hat{i}-\frac{3mv}{2}\hat{j}\overrightarrow{\Delta p}=mv\sqrt{\frac{3}{4}+\frac{9}{4}}=mv\sqrt{3}\Delta t=\left(\frac{2\pi r}{v}\right)\frac{2}{3}\overrightarrow{F_{av}}=\frac{mv\sqrt{3}}{\left(\frac{2\pi r}{v}\right)\frac{2}{3}}=\frac{3\sqrt{3}}{4\pi }\frac{m{v}^2}{r}$