First slide

Angular velocity, Angular acceleration

Question

The position vector of a particle $\vec{R}$ as a function of time is given by
$\vec{R} = 4 \sin (2 π t) \hat{i} + 4 \cos (2 π t) \hat{j}$
Where R is in meters, t is in seconds and $\hat{i} and \hat{j}$ denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?

Moderate

A

Magnitude of the velocity of particle is 8 meter/second.
B

Path of the particle is a circle of radius } 4 \text { meter.
C

$Acceleration vector is along - \vec{R} .$
D

Magnitude of acceleration vector is $\frac{v^{2}}{R}$ , where v is the velocity of particle.

Solution

$Here, \vec{R} = 4 \sin (2 π t) \hat{i} + 4 \cos (2 π t) \hat{j}$
The velocity of the particle is
$\vec{v} = \frac{d \vec{R}}{d t} = \frac{d}{d t} [4 \sin (2 π t) \hat{i} + 4 \cos (2 π t) \hat{j}]$
$= 8 π \cos (2 π t) \hat{i} - 8 π \sin (2 π t) \hat{j}$
its magnitude is
$| \vec{v} | = \sqrt{(8 π \cos (2 π t))^{2} + (- 8 π \sin (2 π t))^{2}}$
$= \sqrt{64 π^{2} \cos^{2} (2 π t) + 64 π^{2} \sin^{2} (2 π t)}$
$= \sqrt{64 π^{2} [\cos^{2} (2 π t) + \sin^{2} (2 π t)]}$
$= \sqrt{64 π^{2}}$ $(as \sin^{2} θ + \cos^{2} θ = 1)$
$= 8 π m / s$

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