Question

A particle moves, such that its position vector $\overrightarrow r (t) = \cos \omega t\widehat i + \sin \omega t\widehat j$ where $\omega$ is a constant and t is time. Then which of the following statements is true for the vector $\overrightarrow v (t)$ acceleration $\overrightarrow a (t)$ of the particle
(A) $\overrightarrow v$ is perpendicular to $\overrightarrow r$ and $\overrightarrow a$ is directed away from the origin
(B) $\overrightarrow v$ and $\overrightarrow a$ both are parallel to $\overrightarrow r$
(C) $\overrightarrow v$ and $\overrightarrow a$ both are perpendicular to $\overrightarrow r$
(D) $\overrightarrow v$ is perpendicular to $\overrightarrow r$ and $\overrightarrow a$ is directed towards the origin

Answer 1

In order to solve the above problem first we have to see what is given in question. Here position vector $\overrightarrow r (t)$ is given.On differentiating $\overrightarrow r (t)$ with respect to time, we get $\overrightarrow v (t)$ velocity vector.
After then on differentiating $\overrightarrow v (t)$ with respect to time, we will get an acceleration vector $\overrightarrow a (t)$.After calculating $\overrightarrow v (t)$ and $\overrightarrow a (t)$ we will check each option one by one.

Complete step by step answer:
In question given that position vector $\overrightarrow r (t) = \cos \omega t\widehat i + \sin \omega t\widehat j$ …..(1)
We know that velocity vector $\overrightarrow v (t) = \dfrac{{d[\overrightarrow r (t)]}}{{dt}}$
So, $\overrightarrow v (t) = \dfrac{{d[\cos (\omega t)\widehat i + \sin (\omega t)\widehat j]}}{{dt}}$
$\Rightarrow\overrightarrow v (t) = \dfrac{{d[\cos (\omega t)]\widehat i}}{{dt}} + \dfrac{{d[\sin (\omega t)]\widehat j}}{{dt}}$
$\Rightarrow\overrightarrow v (t) = - \omega \sin \omega t\widehat i + \omega \cos \omega t\widehat j$
$\Rightarrow\overrightarrow v (t) = \omega [\cos (\omega t)\widehat j - \sin (\omega t)\widehat i]$ …..(2)
We also know that
Acceleration vector $\overrightarrow a (t) = \dfrac{{d[\overrightarrow v (t)]}}{{dt}}$
From equation 1
$\overrightarrow a (t) = \dfrac{{d[\omega (\cos \omega t\widehat j - \sin \omega t\widehat i)]}}{{dt}}$
$\Rightarrow\overrightarrow a (t) = \omega \left[ {\dfrac{{d(\cos \omega t)}}{{dt}}\widehat j - \dfrac{{d(\sin \omega t)}}{{dt}}\widehat i} \right]$
$\Rightarrow\overrightarrow a (t) = \omega \left[ { - \omega \sin \omega t\widehat j - \omega \cos \omega t\widehat i} \right]$
$\Rightarrow\overrightarrow a (t) = - {\omega ^2}(\cos \omega t\widehat i + \sin \omega t\widehat j)$ ………(3)
Now we will check each option one by one.
(A) If $\overrightarrow v$ and $\overrightarrow r$ perpendicular to each other, So, $(\overrightarrow v \cdot \overrightarrow r )$ Should be zero.
Let’s check $\Rightarrow \overrightarrow v \cdot \overrightarrow r$
From equation 1 and 2
$\Rightarrow \left[ {\omega (\cos \omega t\widehat j - \sin \omega t\widehat i)} \right] \cdot ({\cos ^2}\omega t\widehat i + \sin \omega t\widehat j)$
$\Rightarrow\overrightarrow v \cdot \overrightarrow r = \omega (\cos \omega t\sin \omega t - \sin \omega t\cos \omega t)$
$\Rightarrow \overrightarrow v \cdot \overrightarrow r = 0$ …..(4)
And also given in statement $\overrightarrow a$ is directed away from origin. Let’s check.
So, from equation 1 and 3
$\overrightarrow a = - {\omega ^2}\overrightarrow r$
$\Rightarrow\overrightarrow a = {\omega ^2}( - \overrightarrow r )$ …..(5)
This $- ve$ sign represents that $\overrightarrow a$ is directed towards the origin.
Hence, A is not true.
(B) Now we will check option B
In this statement given that $\overrightarrow v$ and $\overrightarrow a$ both are parallel to $\overrightarrow r$.
But from equation 4 it is clear that $\overrightarrow v$ and $\overrightarrow r$ perpendicular to each other.
Hence, option B is also not true.
(C) In this statement given that $\overrightarrow v$ and $\overrightarrow a$ both are perpendicular to $\overrightarrow r$. From equation 4 it is clear that $\overrightarrow v$ and $\overrightarrow r$ are perpendicular to each other.
Now we will check that $\overrightarrow a \cdot \overrightarrow r$ will be zero or not.
So, from equation 1 and 3
\overrightarrow a \cdot \overrightarrow r = \left\\{ { - {\omega ^2}\left[ {\cos \omega t\widehat i + \sin \omega t\widehat j} \right]} \right\\} \cdot \left( {{{\cos }^2}\omega t\widehat i + \sin \omega t\widehat j} \right)
$\Rightarrow\overrightarrow a \cdot \overrightarrow r = - {\omega ^2}({\cos ^2}\omega t + {\sin ^2}\omega t)$
$\Rightarrow\overrightarrow a \cdot \overrightarrow r = - {\omega ^2}(1)$
$\Rightarrow\overrightarrow a \cdot \overrightarrow r = - {\omega ^2}$
$\therefore\overrightarrow a \cdot \overrightarrow r \ne 0$
So, $\overrightarrow a$ and $\overrightarrow r$ are not perpendicular to each other.
Hence, option C is also not true.
(D) From equation 4 and 5, it is clear that $\overrightarrow v$ and $\overrightarrow r$ are perpendicular to each other and $\overrightarrow a$ is directed towards the origin.

Hence, option D is true.So, option D is the correct answer.

Note: Many time students may get confused between perpendicular and parallel concepts of 2 vectors. If 2 vectors are parallel to each other then their cross product will be zero.
$\overrightarrow A \times \overrightarrow B = 0$
If 2 vectors are perpendicular to each other then their dot product is zero
$\overrightarrow A \cdot \overrightarrow B = 0$