First slide

The two types of acceleration

Question

A particle moves, so that its position vector is given by $r = cosω t \hat{x} + sinω t \hat{y}$ , where $ω$ is a constant. Which of the following is true?

Difficult

A

Velocity and acceleration both are parallel to r.
B

Velocity is perpendicular to r and acceleration is directed towards the origin.
C

Velocity is perpendicular to r and acceleration is directed away from the origin
D

Velocity and acceleration both are perpendicular to r.

Solution

Position vector of the particle is given by
$r = cosω t \overset{⏞}{x} + sinω t \overset{⏞}{y}$
where, $ω$ is a constant.
Velocity of the particle is
$v = \frac{d r}{d t} = \frac{d}{d t} (\cos ω t \hat{x} + \sin ω t \hat{y})$
$= (- \sin ω t) ω \hat{x} + (\cos ω t) ω \hat{y}$
$= - ω (\sin ω t \hat{x} - \cos ω t \hat{y})$
Acceleration of the particle,
$a = \frac{d v}{d t} = \frac{d}{d t} (- ω \sin ω t \hat{x} + ω \cos ω t \hat{y})$
$= - ω^{2} \cos ω t \hat{x} - ω^{2} \sin ω t \hat{y}$
$\Rightarrow a = - ω^{2} r = ω^{2} (- r)$
Assuming the particle is at P, then its position vector is directed as shown in the diagram.
Therefore, acceleration is directed towards -r, i.e. towards O (origin).
$v \cdot r = - ω (\sin ω t \hat{x} - \cos ω t \hat{y}) \cdot (\cos ω t \hat{x} + \sin ω t \hat{y})$
$= - ω [\sin ω t \cdot \cos ω t + 0 + 0 - \sin ω t \cdot \cos ω t]$
$= - ω (0) = 0$
$\Rightarrow v ⊥ r$
Thus, velocity is perpendicular to r.

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