Question

For a particle in uniform circular motion the acceleration $\overrightarrow{a}$ at a point $P(R,\theta )$on the circle of radius R is (here $\theta$ is measured from the x-axis)

• A. $-\frac{v^2}{R}cos\theta \hat{i}+\frac{v^2}{R}sin\theta \hat{j}$
• B. $-\frac{v^2}{R}sin\theta \hat{i}+\frac{v^2}{R}cos\theta \hat{j}$
• C. $-\frac{v^2}{R}cos\theta \hat{i}+\frac{v^2}{R}sin\theta \hat{j}$
• D. $\frac{v^2}{R}+\frac{v^2}{R}\hat{j}$

Answer 1

The correct option is C

$-\frac{v^2}{R}cos\theta \hat{i}+\frac{v^2}{R}sin\theta \hat{j}$

$\overrightarrow{r}=Rcos\theta \hat{j}$
$\therefore \frac{d\overrightarrow{r}}{dt}=-Rsin\theta \frac{d\theta }{dt}\hat{i}+Rcos\theta \frac{d\theta }{dt}\hat{j}$
$\Rightarrow \overrightarrow{v}=-R\omega sin\theta \hat{i}+R\omega cos\theta \hat{j}$
$(as\omega =\frac{d\theta }{dt}and\frac{d\overrightarrow{r}}{dt}=\overrightarrow{v})$
$\Rightarrow \frac{d\overrightarrow{v}}{dt}=\overrightarrow{a}=R\omega cos\theta \frac{d\theta }{dt}\hat{i}+R\omega (-sin\theta )\frac{d\theta }{dt}\hat{j}$
$\Rightarrow \frac{ds}{dt}\equiv \frac{d\theta r}{dt}=r\frac{d\theta }{dt}=r\omega$
$\therefore \omega or\frac{d\theta }{dt}isconstant$
$\therefore \overrightarrow{v}=-R{\omega }^{2}cos\theta \hat{i}-R{\omega }^{2}sin\theta \hat{j}$
Substituting $\omega =\frac{v}{R}$
$\overrightarrow{a}=\frac{-v^2}{R}cos\theta \hat{i}-\frac{-v^2}{R}sin\theta \hat{j}$