Question

The position vector of a particle $\overrightarrow R $ as a function of time is given by,
$\overrightarrow R = 4\sin \left( {2\pi t} \right)\widehat i + 4\cos \left( {2\pi t} \right)\widehat j,$
where R is in meters, t is in seconds and $\widehat i$ and $\widehat j$ denote unit vectors along X-axis and Y-axis directions respectively. Which one of the following statements is wrong for the motion of particles?

A. Path of the particle is a circle of radius 4 m
B. Acceleration vector is along $ - \overrightarrow R $
C. Magnitude of the acceleration vector is $\dfrac{{{v^2}}}{R}$, where v is the velocity of particle
D. Magnitude of the velocity of particle is 8 m/s

Answer 1

Hint: On converting the vector equation in Cartesian form we can easily determine the locus of $\overrightarrow R $. On differentiating $\overrightarrow R $ with respect to time we can get velocity vector and differentiating velocity vector with respect to time we can easily get the acceleration vector as well as their direction and magnitude.

Complete step-by-step answer:
According to the question,
The position vector of the particle is given by,
$\overrightarrow R = 4\sin \left( {2\pi t} \right)\widehat i + 4\cos \left( {2\pi t} \right)\widehat j$
Suppose,
$x = 4\sin \left( {2\pi t} \right) \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)$
$y = 4\cos \left( {2\pi t} \right) \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)$
On squaring and adding (1) and (2) we get
${x^2} + {y^2} = {\left( 4 \right)^2}$
The above equation represents the circle of radius 4 m.
We know that velocity vector given by,
$\overrightarrow v = \dfrac{{d\overrightarrow R }}{{dt}}$
$\therefore \overrightarrow v = \dfrac{d}{{dt}}\left( {4\sin \left( {2\pi t} \right)\widehat i + 4\cos \left( {2\pi t} \right)\widehat j} \right)$
On solving we get
$\therefore \overrightarrow v = 4 \times 2\pi \left( {\cos \left( {2\pi t} \right)\widehat i - \sin \left( {2\pi t} \right)\widehat j} \right)$
Magnitude of velocity vector given by
$v = \sqrt {{{\left( {8\pi } \right)}^2} + {{\left( {8\pi } \right)}^2}} \Rightarrow 8\pi \sqrt 2 $ m/s
Also we know that in a circular motion, the centripetal acceleration vector is always towards the center of the circle which is equal to $\dfrac{{{v^2}}}{R}$. The direction of position vector $\overrightarrow R $ is always away from the center.
Therefore finally we conclude that the direction of acceleration vector is opposite to $\overrightarrow R $ i.e. acceleration vector is along $ - \overrightarrow R $.
Hence A, B and C are correct statements and D is the wrong statement.
Therefore the correct option is D.

Note: In order to find the direction and magnitude of acceleration we can also differentiate the velocity vector to get its magnitude as well as direction. When we will find the direction of acceleration vector then we will get that the direction of acceleration vector is along $ - \overrightarrow R $ i.e. towards the center. The magnitude of centripetal acceleration depends on the velocity as well as radius of the circle.