Question

The position vector of a particle is $ \overrightarrow{r} = (a \, \cos \, \omega t) \widehat{i} + (a \sin \omega t) \widehat{j}$. The velocity of the particle is

• A. directed towards the origin
• B. directed away from the origin
• C. parallel to the position vector
• D. perpendicular to the position vector

Answer 1

The Correct Option is D

Let's analyze the given position vector of the particle: \overrightarrow{r} = (a \cos \, \omega t) \widehat{i} + (a \sin \omega t) \widehat{j} .

This position vector indicates that the particle is moving in a circular path, where \(a\) is the radius of the circle, and \( \omega t \) is the angular position of the particle.

1. First, we calculate the velocity of the particle by differentiating the position vector with respect to time \(t\): \begin{align*} \overrightarrow{v} &= \frac{d\overrightarrow{r}}{dt} \\ &= \frac{d}{dt} \left( (a \cos \omega t) \widehat{i} + (a \sin \omega t) \widehat{j} \right) \\ &= \left( -a \omega \sin \omega t \right) \widehat{i} + \left( a \omega \cos \omega t \right) \widehat{j}. \end{align*}
2. The velocity vector is \overrightarrow{v} = (-a \omega \sin \omega t) \widehat{i} + (a \omega \cos \omega t) \widehat{j} .
3. Now, check if the velocity vector is perpendicular to the position vector. For two vectors to be perpendicular, their dot product should be zero.
4. Calculate the dot product of the position vector \overrightarrow{r} and velocity vector \overrightarrow{v} : \begin{align*} \overrightarrow{r} \cdot \overrightarrow{v} &= ((a \cos \omega t) \widehat{i} + (a \sin \omega t) \widehat{j}) \cdot ((-a \omega \sin \omega t) \widehat{i} + (a \omega \cos \omega t) \widehat{j}) \\ &= (a \cos \omega t) (-a \omega \sin \omega t) + (a \sin \omega t) (a \omega \cos \omega t) \\ &= -a^2 \omega \cos \omega t \sin \omega t + a^2 \omega \sin \omega t \cos \omega t \\ &= 0. \end{align*}

Since the dot product is zero, the velocity vector is indeed perpendicular to the position vector.

Conclusion: The correct answer is that the velocity of the particle is perpendicular to the position vector.