System of Particles and Rotational Dynamics

Kinematics of Rotational Motion about a Fixed Point

Suppose there is a motorcycle riding on a road. It is observed that when the acceleration is more the wheels of the bike spins more and rotates through many revolutions. This only happens when the wheels have angular acceleration for a long time. Let us study more about angular acceleration in detail.

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Kinematics of Rotational Motion about a Fixed Point

We all know that rotational motion and translational motion are analogous to each other. In rotational motion, the angular velocity is ω which is analogous to the linear velocity v in the transitional motion. Let us discuss further the kinematics of rotational motion about a fixed point.

The kinematic quantities in rotational motion like the angular displacement θ, angular velocity ω and angular acceleration α respectively corresponds to kinematic quantities in linear motion like displacement x, velocity v and acceleration a.

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Angular Acceleration

The rate of change of angular velocity is the angular acceleration.

α = \( \frac{dω}{ dt } \)  ( rad/sec²)

Now let us consider a particle P on a rotating object. The object undergoes a rotation motion at the fixed point. The angular displacement of a particle P is θ. Hence in time t = 0, the angular displacement of the particle P is 0. So we can say that in time t, its angular displacement will be equal to θ.

Angular Velocity

Angular velocity is the time rate of change of angular displacement. We can write it as,

ω = \( \frac{dθ}{ dt } \)

As we know that the rotational motion here is fixed, so there is no need to change the angular velocity. We know angular acceleration is α = \( \frac{dω}{ dt } \). So the kinematics equations of linear motion with uniform acceleration is,

v = v0+ at

x = x+ v0t + \( \frac{1}{2} \) at²

v² = v02+ 2ax

Where  x0  is the initial displacement and vis the initial velocity of the particle. Here initial means t = 0. Now, this equation corresponds to the kinematics equation of the rotational motion.

Kinematics Equations for Rotational Motion with Uniform Angular Acceleration

ω = ω0+ αt

θ = θ+ ω0t + \( \frac{1}{2} \) αt²

ω² =  ω0² + 2α (θ – θ0)

Where  θ0  is the initial angular displacement of the rotating body and ωis the initial angular velocity of the particle of the body.

Solved Examples For You

Q1. A wheel rotating with uniform angular acceleration covers 50 rev in the first five seconds after the start. If the angular acceleration at the end of five seconds is x π rad/s² find the value of x.

  1. 4
  2. 8
  3. 6
  4. 10

Solution: B
θ = \( \frac{1}{2} \) αt²
α = \( \frac{2θ}{t²} \) = \( \frac{2(50) (2π)}{5²} \) = 8π rad/s² = 25.14 rad/s²
comparing with α, x = 8 rad/s²

Q2. Starting from rest, a fan takes five seconds to attain the maximum speed of 400 rpm. Assume constant acceleration, find the time taken by the fan attaining half the maximum speed.

  1. 11 s
  2. 2.0 s
  3. 2.5 s
  4. 2.0 s

Solution: D. The maximum angular velocity is given by,
w= 400rpm = 400 × \( \frac{2π}{60} \) = \( \frac{40}{3} \) rad/sec
Initial angular velocity is w= 0
So angular accleration α = \( \frac{w_ m – ω – 0}{t} \) = \( \frac{40/3 – 0}{5} \) = \( \frac{8π}{3} \) rad/sec²
Now ω = w+ at we get ωm/2 = 0 + at = \( \frac{40π/3}{2(8π/3)} \) = 2.5s

Q3. Identify the direction of the angular velocity vector for the second hand of a clock going from 0 to 60 seconds?

  1. outward from clock face
  2. inward towards the clock face
  3. upward
  4. dwonward

Solution : B

Angular Acceleration

Angular velocity = ω × r. The second hand of the clock rotates in clockwise direction. From the above figure, the direction of angular velocity is into the plane of the page that is inward towards the clock face.

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