Question

A particle is moving in $x-y$ plane according to $\vec{r}=b \cos \omega t \hat{i}+b \sin \omega t \hat{j}$. Where $\omega$ is constant. Which of the following statement(s) is / are true?

• A. $\frac{E}{\omega}$ is a constant where $E$ is the total energy of the particle
• B. The trajectory of the particle in $x-y$ plane is a circle
• C. In $a_x-a_y$ plane, trajectory of the particle is an ellipse ( $a_x, a_y$ denotes the components of acceleration)
• D. $\vec{a}=\omega^2 \vec{v}$

Answer 1

Answer: (A), (B)

Hint: $\vec{r}=b \cos \omega t i+b \sin \omega t j$
$ \vec{v}=\frac{d \vec{r}}{d t}=-b \omega \sin \omega t+b \omega \cos \omega t j $
$ (\vec{v})=\sqrt{b^2 \omega^2 \sin ^2 \omega t+b^2 \omega^2 \cos ^2 \omega t} $
$(\vec{v})=b \omega \rightarrow \text { constant } \therefore E=\frac{1}{2} m v^2=\text { constant } $
$E=\frac{1}{2} m\left[b^2 \omega^2\right] \Rightarrow \frac{E}{\omega}=\frac{1}{2} m b^2 \cdot \omega=\text { constant }$
$ x=b \cos \omega t $
$ y=b \sin \omega t $
$ x^2+y^2=b^2 \rightarrow \text { equation of circle } \therefore \vec{a} \perp \vec{v}$
So $\vec{a} \neq \omega^2 \vec{v}$