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