Question:

A particle starting from rest, moves in a circle of radius 'r'. It attains a velocity of V0 m/s in the $n^{th}$ round. Its angular acceleration will be :-

Updated On: Jul 25, 2024
  • $\frac{\vee_{0}}{n}rad / s^{2}$
  • $\frac{\vee^{2}_{0}}{2\pi nr^{2}}rad / s^{2}$
  • $\frac{\vee^{2}_{0}}{4\pi nr^{2}}rad / s^{2}$
  • $\frac{\vee^{4}_{0}}{4\pi nr}rad / s^{2}$
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The Correct Option is C

Solution and Explanation

$\theta = (2\pi n), \omega_0 = 0, \omega = V_0/r$
$\alpha = \frac{\omega^2 - \omega_0^2}{2\theta} = \frac{(V_0/r)^2 - 0}{2(2\pi n)}$
$= \frac{V_0^2}{4\pi nr^2}$
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Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration