Question

A particle moves so that its position vector varies with time as $\vec{r} = A \cos ω t \hat{i} + A \sin ω t \hat{j}$ . Find the,

$(a)$ initial velocity of the particle,
$(b)$ angle between the position vector and velocity of the particle at any time, and
$(c)$ speed at any instant.

Accepted Answer

$(a)$ The position of the particle at any time $t$ is,

$\vec{r} = A \cos ω t \hat{i} + A \sin ω t \hat{j}$ .

Instantaneous velocity of the particle is given as, $\vec{v} = \frac{d \vec{r}}{d t}$ .

We have, $\vec{v} = A \frac{d}{d t} (\cos ω t) \hat{i} + A \frac{d}{d t} (\sin ω t) \hat{j}$ ,

$\Rightarrow \vec{v} = - A ω \sin ω t \hat{i} + A ω \cos ω t \hat{j}$ .

For initial velocity,

$t = 0 \Rightarrow v = - A ω \sin 0 \hat{i} + A ω \cos 0 \hat{j} = A ω \hat{j}$ .

$(b)$ For calculating the angle between two vectors, we use the concept of dot product of the vectors. The angle $θ$ between $\vec{r}$ and $\vec{v}$ can be given as,

$θ = \cos^{- 1} \frac{\vec{r} \cdot \vec{v}}{| \vec{r} ‖ \vec{v} |}$

where,

$\vec{r} \cdot \vec{v} = (A \cos ω t \hat{i} + A \sin ω t \hat{j}) \cdot (- A ω \sin ω t \hat{i} + A ω \cos ω t \hat{j})$ ,

$\Rightarrow \vec{r} \cdot \vec{v} = ω A^{2} (- \cos ω t \sin ω t + \sin ω t \cos ω t) = 0$ .

Hence, $θ = \cos^{- 1} 0 = \frac{π}{2}$ .
That means, $\vec{v} ⊥ \vec{r}$ .

$(c)$ Speed at any time is the magnitude of instantaneous velocity, i.e.,

$v = | \vec{v} | = \sqrt{(- A ω \sin ω t)^{2} + (A ω \cos ω t)^{2}} = A ω$ .

B M Sharma Solutions for Chapter: Kinematics I, Exercise 12: ILLUSTRATION 4.12

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 12: ILLUSTRATION 4.12

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