A particle moves so that its position vector is given by, where is a constant. Which of the following is true? [2016]

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Solution

Q.
A particle moves so that its position vector is given by $\vec{r} = \cos ω t \hat{x} + \sin ω t \hat{y},$ where $ω$ is a constant. Which of the following is true [2016]

1 Velocity is perpendicular to $\vec{r}$ and acceleration is directed towards the origin.

2 Velocity is perpendicular to $\vec{r}$ and acceleration is directed away from the origin.

3 Velocity and acceleration both are perpendicular to $\vec{r}$ .

4 Velocity and acceleration both are parallel to $\vec{r}$ .

Ans.
(1)

Given, $\vec{r} = \cos ω t \hat{x} + \sin ω t \hat{y}$

$∴ \vec{v} = \frac{d \vec{r}}{d t} = - ω \sin ω t \hat{x} + ω \cos ω t \hat{y}$

$\vec{a} = \frac{d \vec{v}}{d t} = - ω^{2} \cos ω t \hat{x} - ω^{2} \sin ω t \hat{y} = - ω^{2} \vec{r}$

Since position vector $(\vec{r})$ is directed away from the origin, so, acceleration $(- ω^{2} \vec{r})$ is directed towards the origin.

Also,

$\vec{r} \cdot \vec{v} = (\cos ω t \hat{x} + \sin ω t \hat{y}) \cdot (- ω \sin ω t \hat{x} + ω \cos ω t \hat{y})$

$= - ω \sin ω t \cos ω t + ω \sin ω t \cos ω t = 0$

$\Rightarrow \vec{r} ⊥ \vec{v}$

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