Properties of particle motion from position vector – Rankers Physics
Topic: Kinematics
Subtopic: Equations of Motion

Properties of particle motion from position vector

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

(2015)

Path of the particle is a circle of radius 4 metre
Acceleration vectors is along \(-vec{R}\)
Magnitude of acceleration vector is \(\frac{V^2}{R}\) where V is the velocity of particle.
Magnitude of the velocity of particle is 8 metre/second

Solution:

From \(\vec{R} = 4sin(2\pi t)\hat{i} + 4cos(2\pi t)\hat{j}\), \(x=4sin(2\pi t)\) and \(y=4cos(2\pi t)\). \(x^2+y^2=16\) implies a circle of radius 4m. \(\vec{V} = 8\pi cos(2\pi t)\hat{i} - 8\pi sin(2\pi t)\hat{j}\). \(|\vec{V}| = 8\pi\text{ m/s}\). \(\vec{a} = -16\pi^2sin(2\pi t)\hat{i} - 16\pi^2cos(2\pi t)\hat{j} = -4\pi^2 \vec{R}\). So \(\vec{a}\) is along \(-\vec{R}\). Also, \(|\vec{a}| = 16\pi^2\) and \(\frac{V^2}{R} = \frac{(8\pi)^2}{4} = 16\pi^2\). Therefore, (a), (b), (c) are correct. (d) is wrong because \(|\vec{V}| = 8\pi\text{ m/s}\), not 8 m/s.

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