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(D) In uniform circular motion,the magnitude of centripetal acceleration is $a = \frac{v^2}{R}$.At point $P_1$,the acceleration vector $\vec{a}_1$ is

Vedclass · Accepted Answer

$\frac{{{v^2}}}{R} \sqrt 2 $

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(D) In uniform circular motion,the magnitude of centripetal acceleration is $a = \frac{v^2}{R}$.At point $P_1$,the acceleration vector $\vec{a}_1$ is directed towards the center,i.e.,$\vec{a}_1 = \frac{v^2}{R} \hat{i}$.At point $P_2$,the acceleration vector $\vec{a}_2$ is directed towards the center,i.e.,$\vec{a}_2 = \frac{v^2}{R} \hat{j}$.The change in acceleration is $\Delta \vec{a} = \vec{a}_2 - \vec{a}_1 = \frac{v^2}{R} \hat{j} - \frac{v^2}{R} \hat{i}$.The magnitude of the change in acceleration is $|\Delta \vec{a}| = \sqrt{(\frac{v^2}{R})^2 + (-\frac{v^2}{R})^2} = \sqrt{2(\frac{v^2}{R})^2} = \frac{v^2}{R} \sqrt{2}$.

The figure below shows a body of mass $M$ moving with a uniform speed $v$ on a circular path of radius $R$. What is the change in acceleration when moving from point $P_1$ to $P_2$?

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