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Question

(A, B) Given position vector: $\vec{r} = b \cos \omega t \hat{i} + b \sin \omega t \hat{j}$.Velocity vector: $\vec{v} = \frac{d\vec{r}}{dt} = -b \omeg

Vedclass · Accepted Answer

The trajectory of the particle in $x-y$ plane is a circle.

Vedclass · Answer

(A, B) Given position vector: $\vec{r} = b \cos \omega t \hat{i} + b \sin \omega t \hat{j}$.Velocity vector: $\vec{v} = \frac{d\vec{r}}{dt} = -b \omega \sin \omega t \hat{i} + b \omega \cos \omega t \hat{j}$.Magnitude of velocity: $v = |\vec{v}| = \sqrt{(-b \omega \sin \omega t)^2 + (b \omega \cos \omega t)^2} = b \omega$.Since $v$ is constant, kinetic energy $E = \frac{1}{2}mv^2 = \frac{1}{2}mb^2\omega^2$ is constant. Thus, $\frac{E}{\omega} = \frac{1}{2}mb^2\omega$ is constant. Statement $(A)$ is true.Trajectory: $x = b \cos \omega t$ and $y = b \sin \omega t$. Squaring and adding: $x^2 + y^2 = b^2(\cos^2 \omega t + \sin^2 \omega t) = b^2$. This is a circle. Statement $(B)$ is true.Acceleration: $\vec{a} = \frac{d\vec{v}}{dt} = -b \omega^2 \cos \omega t \hat{i} - b \omega^2 \sin \omega t \hat{j} = -\omega^2 \vec{r}$.Components: $a_x = -b \omega^2 \cos \omega t$ and $a_y = -b \omega^2 \sin \omega t$. Thus, $a_x^2 + a_y^2 = (b \omega^2)^2$, which is a circle in the $a_x-a_y$ plane. Statement $(C)$ is false.Since $\vec{a} = -\omega^2 \vec{r}$ and $\vec{v} \perp \vec{r}$, $\vec{a}$ is not parallel to $\vec{v}$. Statement $(D)$ is false.

$A$ 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 a constant. Which of the following statement$(s)$ is/are true?

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