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Angular velocity

The angular velocity increases by a constant value at the end of every unit time interval, when angular velocity and acceleration act in the same direction. On the other hand, angular velocity decreases, when angular velocity and acceleration act in the opposite direction.

To appreciate this, we consider a circular motion of a particle whose initial angular velocity is 0.3 rad/s. The motion is subjected to an angular acceleration of magnitude 0.1 rad s 2 in the opposite direction to the initial velocity. In the table here, we calculate angular velocity of the particle, using relation, ω = ω 0 + α t , at the end of every second and plot the data (for first 5 seconds) to understand the variation of angular velocity with time.

----------------------------------------------------- Time Angular acceleration Angular velocity(s) (rad/s.s) (rad/s) -----------------------------------------------------0 -0.1 0.3 1 -0.1 0.3 - 0.1 x 1 = 0.22 -0.1 0.3 - 0.1 x 2 = 0.1 3 -0.1 0.3 - 0.1 x 3 = 0.04 -0.1 0.3 - 0.1 x 4 = -0.1 5 -0.1 0.3 - 0.1 x 5 = -0.2-----------------------------------------------------

Here, the particle stops at the end of 3 seconds. The particle then reverses its direction (clockwise from anti-clockwise) and continues to move around the axis. The angular velocity – time plot is as shown here :

Angular velocity - time plot

We observe following aspects of the illustrated motion with constant acceleration :

  1. The angular velocity decreases at uniform rate and the angular velocity – time plot is straight line. Positive angular velocity becomes less positive and negative angular velocity becomes more negative.
  2. The slope of the plot is constant and negative.

Problem : A particle at the periphery of a disk at a radial distance 10 m from the axis of rotation, uniformly accelerates for a period of 5 seconds. The speed of the particle in the meantime increases from 5 m/s to 10 m/s. Find angular acceleration.

Solution : The initial and final angular velocities are :

ω 0 = 5 10 = 0.5 rad / s ω = 10 10 = 1 rad / s t = 5 s

Using equation of motion, ω = ω 0 + α t , we have :

1 = 0.5 + α X 5 α = 0.5 5 = 0.1 rad / s 2

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Angular displacement

The angular displacement is given as :

Δ θ = θ - θ 0 = ω 0 t + 1 2 α t 2

If we start observation of motion as t = 0 and θ° = 0, then displacement (θ) is :

θ = ω 0 t + 1 2 α t 2

The particle covers greater angular displacement for every successive time interval, when angular velocity and acceleration act in the same direction and the particle covers smaller angular displacement, when angular velocity and acceleration act in the opposite direction.

To appreciate this, we reconsider the earlier case of a circular motion of a particle whose initial angular velocity is 0.3 rad/s. The motion is subjected to an angular acceleration of magnitude 0.1 rad / s 2 in the opposite direction to the initial velocity. In the table here, we calculate angular displacement of the particle, using relation, θ = ω 0 t + 1 2 α t 2 , at the end of every second and plot the data (for first 7 seconds) to understand the variation of angular displacement with time.

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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