A4-2. Rigid Body Mechanics: Angular KinematicsHL ONLY

1. Angular Displacement, Velocity & Acceleration

A rigid rotating body can be described using angular equivalents to standard linear kinematics. Every point on a rigid body rotates through the exact same angle in the same amount of time, even though points further from the center travel a longer linear distance at a higher linear speed.

Fundamental Definitions:

  • Angular Displacement ($\theta$): The angle through which a point or line has been rotated, measured in radians (rad).
  • Angular Velocity ($\omega$): The rate of change of angular displacement.
    $$\omega = \frac{\Delta \theta}{\Delta t}$$
  • Angular Acceleration ($\alpha$): The rate of change of angular velocity.
    $$\alpha = \frac{\Delta \omega}{\Delta t}$$

Connecting Linear and Angular Motion:

For a specific point located at a distance $r$ from the axis of rotation, its linear (translational) motion is directly tied to the rigid body's angular motion:

Linear Variable Angular Variable Relationship
Distance / Arc Length ($s$) Displacement ($\theta$) $s = r\theta$
Tangential Velocity ($v$) Velocity ($\omega$) $v = r\omega$
Tangential Acceleration ($a_t$) Acceleration ($\alpha$) $a_t = r\alpha$

Crucial Distinction: Tangential acceleration ($a_t = r\alpha$) changes how fast the object is spinning. However, even if the object spins at a constant speed ($\alpha = 0$), it still experiences Centripetal Acceleration ($a_c = r\omega^2 = \dfrac{v^2}{r}$), which simply keeps the mass moving in a circle by constantly changing its direction.

⚠ Graphical Analysis Rules

Graphs of rotational motion operate under the exact same calculus rules as linear motion:

  • The gradient of an angular displacement-time graph gives angular velocity ($\omega$).
  • The gradient of an angular velocity-time graph gives angular acceleration ($\alpha$).
  • The area under an angular velocity-time graph gives angular displacement ($\theta$).

2. Angular Kinematic Equations (Rotational SUVAT)

If a rigid body is undergoing constant angular acceleration, the four kinematic equations for uniform linear acceleration map perfectly to uniform rotational acceleration by simply swapping the English variables for their Greek equivalents.

Linear Equations:

$v = u + at$

$s = ut + \dfrac{1}{2}at^2$

$v^2 = u^2 + 2as$

$s = \dfrac{1}{2}(u + v)t$

Rotational Equivalents:

$\omega_f = \omega_i + \alpha t$

$\theta = \omega_i t + \dfrac{1}{2}\alpha t^2$

$\omega_f^2 = \omega_i^2 + 2\alpha\theta$

$\theta = \dfrac{1}{2}(\omega_i + \omega_f)t$

Example 3: Decelerating Turntable

Problem: The turntable of a record player is spinning at an initial angular velocity of 45 RPM (revolutions per minute). It is turned off and decelerates at a constant rate of $0.8\text{ rad/s}^2$. Determine the total angular displacement (in radians) completed before coming to a complete stop.

Solution:

  • Step 1: Convert RPM to standard rad/s.
    1 revolution = $2\pi$ radians. 1 minute = 60 seconds.
    $\omega_i = \dfrac{45 \times 2\pi}{60} = 1.5\pi\text{ rad/s} \approx 4.71\text{ rad/s}$
  • Step 2: List the knowns.
    $\omega_i = 1.5\pi\text{ rad/s}$, $\omega_f = 0$, $\alpha = -0.8\text{ rad/s}^2$ (negative because it is decelerating). We need $\theta$.
  • Step 3: Select the correct rotational kinematic equation.
    We do not have time ($t$), so we use: $\omega_f^2 = \omega_i^2 + 2\alpha\theta$
  • Step 4: Calculate.
    $0 = (1.5\pi)^2 + 2(-0.8)\theta$
    $0 = 2.25\pi^2 - 1.6\theta$
    $1.6\theta = 2.25\pi^2$
    $\theta = \dfrac{2.25\pi^2}{1.6} \approx \mathbf{13.9\text{ rad}}$