Equations of rotational dynamics Revision Notes for AQA A-Level Physics

Equations of rotational dynamics

Introduction to rotational motion equations

In rotational dynamics, we can describe the motion of rotating objects using equations that directly parallel those used for linear motion. Just as we have equations for objects moving in straight lines with constant acceleration, we have equivalent equations for objects rotating with constant angular acceleration.

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The beauty of rotational dynamics lies in its structural similarity to linear dynamics. Once you understand linear motion equations, you already know the form of rotational equations - you just need to substitute the appropriate rotational variables!

The relationship between torque, moment of inertia, and angular acceleration mirrors Newton's second law from linear dynamics. Where linear motion uses force, mass, and acceleration, rotational motion uses their counterparts: torque, moment of inertia, and angular acceleration.

Newton's second law for rotational motion

Newton's second law for rotational motion states that a torque applied to an object produces an angular acceleration that is proportional to the torque and inversely proportional to the moment of inertia.

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The Fundamental Equation of Rotational Dynamics

$T = I\alpha$

where:

$T$ is the torque in newton metres ( $\text{N m}$ )
$I$ is the moment of inertia in kilogram metres squared ( $\text{kg m}^2$ )
$\alpha$ is the angular acceleration in radians per second squared ( $\text{rad s}^{-2}$ )

This equation is the rotational equivalent of $F = ma$ from linear dynamics.

The torque produces angular acceleration in exactly the same way that force produces linear acceleration. The moment of inertia plays the role of mass, representing the object's resistance to rotational acceleration.

The four rotational equations of motion

When dealing with constant angular acceleration, we can use four equations that describe rotational motion. These equations are directly analogous to the linear equations of motion (often called SUVAT equations).

First equation: $\omega_2 = \omega_1 + \alpha t$

This relates final angular velocity to initial angular velocity, angular acceleration, and time.

Second equation: $\theta = \frac{1}{2}(\omega_1 + \omega_2)t$

This gives angular displacement when you know both initial and final angular velocities and the time taken.

Third equation: $\theta = \omega_1 t + \frac{1}{2}\alpha t^2$

This allows calculation of angular displacement from initial angular velocity, angular acceleration, and time.

Fourth equation: $\omega_2^2 = \omega_1^2 + 2\alpha\theta$

This connects the initial and final angular velocities with angular acceleration and angular displacement, without requiring time.

Comparison with linear equations

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The One-to-One Correspondence

The table below shows how each rotational equation corresponds to its linear counterpart. Notice the structure of each equation remains identical - we simply substitute the rotational variable for its linear counterpart.

Linear equation	Rotational equation
$v = u + at$	$\omega_2 = \omega_1 + \alpha t$
$s = \frac{1}{2}(u + v)t$	$\theta = \frac{1}{2}(\omega_1 + \omega_2)t$
$s = ut + \frac{1}{2}at^2$	$\theta = \omega_1 t + \frac{1}{2}\alpha t^2$
$v^2 = u^2 + 2as$	$\omega_2^2 = \omega_1^2 + 2\alpha\theta$

Variables in rotational motion

Understanding the relationship between linear and rotational variables is essential for working with these equations. Each linear quantity has a rotational equivalent:

Linear variable	Unit	Rotational variable	Unit
Displacement, $s$	$\text{m}$	Angle turned through, $\theta$	$\text{rad}$
Initial velocity, $u$	$\text{m s}^{-1}$	Initial angular velocity, $\omega_1$	$\text{rad s}^{-1}$
Final velocity, $v$	$\text{m s}^{-1}$	Final angular velocity, $\omega_2$	$\text{rad s}^{-1}$
Time, $t$	$\text{s}$	Time, $t$	$\text{s}$
Acceleration, $a$	$\text{m s}^{-2}$	Angular acceleration, $\alpha$	$\text{rad s}^{-2}$

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Problem-Solving Strategy

Notice that time remains the same in both linear and rotational motion. This makes sense because time is independent of the type of motion being considered.

A useful strategy when solving rotational problems is to first think about which linear equation you would use, then substitute the rotational variables to get the appropriate rotational equation.

Deriving the rotational equations

We can derive the rotational equations using the same approach as for linear equations. Let's start with the definition of angular acceleration.

Angular acceleration ( $\alpha$ ) is defined as the rate of change of angular velocity:

$\alpha = \frac{\Delta\omega}{\Delta t}$

Over a time interval $t$ , starting with initial angular velocity $\omega_1$ and reaching final angular velocity $\omega_2$ :

$\alpha = \frac{\omega_2 - \omega_1}{t}$

Rearranging this equation by multiplying both sides by $t$ :

$\alpha t = \omega_2 - \omega_1$

Therefore:

$\omega_2 = \omega_1 + \alpha t$

This gives us the first rotational equation of motion. The other three equations can be derived using similar mathematical methods, following the same logic as the derivations for linear motion equations.

Important conditions for using these equations

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Critical Restriction: Constant Angular Acceleration Required

These rotational equations of motion can only be used when angular acceleration is constant (uniform). If the angular acceleration varies with time, these equations are not valid, and alternative methods (such as graphical techniques or calculus) must be employed.

This is exactly the same restriction that applies to the linear equations of motion - they only work for constant acceleration scenarios.

Converting between radians and revolutions

When solving problems, you may need to convert between radians and revolutions.

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Unit Conversion: Radians and Revolutions

$2\pi \text{ radians} = 1 \text{ revolution}$

To convert from revolutions to radians, multiply by $2\pi$
To convert from radians to revolutions, divide by $2\pi$

Worked example: accelerating wheel

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Worked Example: Calculating Final Velocity and Displacement

A stationary wheel undergoes a constant angular acceleration of $8.0~\text{rad s}^{-2}$ . This acceleration is maintained for $30~\text{s}$ . Calculate the final angular velocity and the angular displacement in radians. Also determine how many complete revolutions this represents.

Solution:

Step 1: Identify the known values

Initial angular velocity: $\omega_1 = 0~\text{rad s}^{-1}$ (stationary)
Angular acceleration: $\alpha = 8.0~\text{rad s}^{-2}$
Time: $t = 30~\text{s}$

Step 2: Calculate final angular velocity

We need to find $\omega_2$ , so we use: $\omega_2 = \omega_1 + \alpha t$

Since $\omega_1 = 0$ :

$\omega_2 = 0 + (8.0 \times 30) = 240~\text{rad s}^{-1}$

Step 3: Calculate angular displacement

We need to find $\theta$ , so we use: $\theta = \omega_1 t + \frac{1}{2}\alpha t^2$

Since $\omega_1 = 0$ , the first term disappears:

$\theta = \frac{1}{2} \times 8.0 \times 30^2 = \frac{1}{2} \times 8.0 \times 900 = 3600~\text{rad}$

Step 4: Convert to revolutions

$\text{Number of revolutions} = \frac{\theta}{2\pi} = \frac{3600}{2\pi} \approx 570~\text{revolutions}$

Answer: The final angular velocity is 240 rad s⁻¹, the angular displacement is 3600 rad, which is approximately 570 complete revolutions.

Worked example: stopping flywheel

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Worked Example: Finding Required Torque

A flywheel has a moment of inertia of $2.4 \times 10^{-2}~\text{kg m}^2$ and is rotating with an angular velocity of $20~\text{rad s}^{-1}$ . Calculate the torque required to bring the flywheel to rest in exactly five revolutions.

Solution:

Step 1: Identify the known values

Moment of inertia: $I = 2.4 \times 10^{-2}~\text{kg m}^2$
Initial angular velocity: $\omega_1 = 20~\text{rad s}^{-1}$
Final angular velocity: $\omega_2 = 0~\text{rad s}^{-1}$ (comes to rest)
Angular displacement: $\theta = 5.0~\text{revolutions} = 5.0 \times 2\pi = 10\pi~\text{rad}$

Step 2: Calculate angular acceleration

We use the equation that connects initial velocity, final velocity, acceleration, and displacement:

$\omega_2^2 = \omega_1^2 + 2\alpha\theta$

Substituting the values:

0^2 &=& 20^2 + 2\alpha(10\pi) \\ 0 &=& 400 + 20\pi\alpha \\ 20\pi\alpha &=& -400 \\ \alpha &=& -\frac{400}{20\pi} = -\frac{20}{\pi} \\ \alpha &=& -6.37~\text{rad s}^{-2} \text{ (to 3 significant figures)} \end{array}$$ The :highlight[negative sign] indicates this is a deceleration (the angular velocity is decreasing). **Step 3:** Calculate the torque Using Newton's second law for rotational motion: $$\begin{array}{rcl} T &=& I\alpha \\ T &=& (2.4 \times 10^{-2}) \times (-6.37) \\ T &=& -0.153~\text{N m} \\ T &=& -0.15~\text{N m} \text{ (to 2 significant figures)} \end{array}$$ **Answer:** A torque of :success[0.15 N m opposing the direction of motion] is required. The negative sign indicates the torque acts in the opposite direction to the rotation, causing the flywheel to slow down.

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Key Points to Remember:

Newton's second law for rotation: $T = I\alpha$ connects torque, moment of inertia, and angular acceleration
Four rotational equations mirror the linear SUVAT equations exactly - just substitute rotational variables for linear ones ( $\omega$ for $v$ , $\alpha$ for $a$ , $\theta$ for $s$ )
These equations only work for constant angular acceleration - if acceleration varies, use alternative methods
Variable mapping: displacement → angle, velocity → angular velocity, acceleration → angular acceleration, mass → moment of inertia, force → torque
Conversion factor: $2\pi$ radians equals one complete revolution - remember this when converting between units

Equations of rotational dynamics (AQA A-Level Physics): Revision Notes