Torque and angular acceleration (AQA A-Level Physics): Revision Notes

Torque and angular acceleration

Introduction to rotational dynamics

When a force acts on an object at some distance from an axis of rotation, it produces a rotational effect. To understand how objects spin and accelerate rotationally, we need to connect three key quantities: the applied torque, the object's moment of inertia, and the resulting angular acceleration. This relationship forms the rotational equivalent of Newton's second law.

Torque on a single particle

Let's start by considering a simple case: a single particle of mass $m$ located at a perpendicular distance $r$ from an axis of rotation. When a torque $T$ acts on this particle, it produces an angular acceleration $\alpha$. The relationship between these quantities is:

$T = mr^2 \times \alpha$

This equation emerges from applying rotational concepts to Newton's second law ($F = ma$). Here, the quantity $mr^2$ represents the moment of inertia ($I$) of the particle - a measure of how difficult it is to change the particle's rotational motion.

For a single particle:

• Moment of inertia: $I = mr^2$
• Unit: $\text{kg} \cdot \text{m}^2$

Extending to rigid objects

Real objects aren't just single particles - they're made up of many particles distributed throughout their volume. To find the total torque acting on an extended rigid object, we need to consider what happens to each particle within it.

When an external torque acts on a solid object, we can think of it as applying individual torques $T_i$ to each particle. The total torque is the sum of all these individual torques:

$\text{Total } T = \sum(m_i \times r_i^2 \times \alpha)$

Since $\alpha$ is constant throughout the object, we can factor it out:

$T = \left(\sum m_i \times r_i^2\right) \times \alpha$

The summation $\sum(m_i \times r_i^2)$ represents the total moment of inertia $I$ of the entire object. This is calculated by adding up the moments of inertia of all individual particles that make up the object:

$I = \sum(mr^2)$

This means the moment of inertia of an extended object accounts for how all its mass is distributed relative to the axis of rotation. Mass located further from the axis contributes more to the total moment of inertia.

Newton's second law for rotational motion

By recognizing that the summation of all the individual contributions gives us the total moment of inertia, we arrive at a beautifully simple equation that governs rotational motion:

$T = I \times \alpha$

The equation shows that:

• Torque ($T$) is measured in newton metres (Nm)
• Moment of inertia ($I$) is measured in kilogram metres squared ($\text{kg} \cdot \text{m}^2$)
• Angular acceleration ($\alpha$) is measured in radians per second squared ($\text{rad} \cdot \text{s}^{-2}$)

We can rearrange this equation to find the angular acceleration when we know the applied torque and moment of inertia:

$\alpha = \frac{T}{I}$

This tells us that angular acceleration is directly proportional to the applied torque but inversely proportional to the moment of inertia. Double the torque, and you double the angular acceleration. Double the moment of inertia, and you halve the angular acceleration.

Understanding the units

It's worth verifying that our units are consistent. From $\alpha = T/I$, the unit of moment of inertia would be $(\text{Nm})/(\text{rad} \cdot \text{s}^{-2})$. How does this relate to $\text{kg} \cdot \text{m}^2$, the unit we initially defined for $I$?

Worked example: calculating angular acceleration

Let's apply these concepts to a practical problem.