Motion in a Plane - Result Question 41
44. A particle moves so that its position vector is given by $\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}$. Where $\omega$ is a constant. Which of the following is true?[2016]
(a) Velocity and acceleration both are perpendicular to $\vec{r}$
(b) Velocity and acceleration both are parallel to $\vec{r}$
(c) Velocity is perpendicular to $\vec{r}$ and acceleration is directed towards the origin
(d) Velocity is perpendicular to $\vec{r}$ and acceleration is directed away from the origin
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Answer:
Correct Answer: 44. (c)
Solution:
- (c) Given: Position vector
$\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}$
$\therefore \quad$ Velocity vector, $\vec{v}=-\omega \sin \omega t \hat{x}+\omega \cos \omega t$ $\hat{y}$ acceleration vector,
$\vec{a}=-\omega^{2} \cos \omega t \hat{x}-\omega^{2} \sin \omega t \hat{y}=-\omega^{2} \vec{r}$
$\vec{r} \cdot \vec{v}=0$ hence $\vec{r} \perp \vec{v}$ and $\vec{a}$ is directed towards the origin.
Nothing actually moves in the direction of the angular velocity vector $\vec{\omega}$. The direction of $\vec{\omega}$ simply represents that the circular motion is taking place in a plane perpendicular to it.
