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

Position vector of the particle
(t) = (a cos $ \omega t) \widehat{i} + (a \, sin \omega t) \widehat{j}$
velocity vector
$\overrightarrow{v} = \frac{d \overrightarrow{r}}{dt} = ( - a \omega sin\, \omega t) \widehat{i} + (a \omega \, cos\, \omega t ) \widehat{j} $
= $ \omega [ ( - a \,sin\, \omega t) \widehat{i} + ( a\, cos\, \omega t ) \widehat{j} ] $
$ \overrightarrow{v} \cdot \overrightarrow{r} = \omega [ ( - a \, sin \, \omega t ) \widehat{i} + (a \, cos\, \omega t) \widehat{ j} ) ] \cdot [ ( a \, cos \,\omega t) \widehat{ i} + (a \, sin \omega t) \widehat{j} ) ] $
= $ \omega [ - a^2 \, sin \, \omega t \, cos \, \omega t + a^2 \, cos \, \omega t \, sin \, \omega t ] = 0 $
Therefore velocity vector is perpendicular to the
displacement vector.