Question

A particle is moving in $x-y$ plane. The positive vector of the particle at time $t$ is $r=b(1-\cos \omega t)\hat{i}+b\sin \omega t\hat{j}$, where $b$ and $\omega$ are positive constants. The acceleration of the particle at instant $t$ is

• A. $a=b{\omega }^2\sin \omega t\hat{i}-b{\omega }^2\cos \omega t\hat{j}$
• B. None of these
• C. $a=b{\omega }^2\cos \omega t\hat{i}-b{\omega }^2\sin \omega t\hat{j}$
• D. zero

Answer 1

The correct option is C $a=b{\omega }^2\cos \omega t\hat{i}-b{\omega }^2\sin \omega t\hat{j}$

Given, displacement
$\overrightarrow{r}=b(1-\cos \omega t)\hat{i}+b\sin \omega t\hat{j}$
velocity,
$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=b\omega \sin \omega t\hat{i}+b\omega \cos \omega t\hat{j}$
Acceleration,
$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=b{\omega }^2\cos \omega t\hat{i}-b{\omega }^2\sin \omega t\hat{j}$