Year 1 Sem 2/LC Classical Mechanics and Relativity 2/Lecture 01

Lecture 1: Mechanics I: Rigid Bodies
Tue 20 Jan 2026 11:00

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Recommended Course Text

Introduction to Classical Mechanics, Oxford University Press

Rotational Mechanics

We’ve been introduced to a number of equations for translational mechanics such as:

$$F = ma$$

$$W = \int \vec{F} \cdot d\vec{r}$$

$$K = \frac{1}{2}mv^2$$

In simple cases that we’ll deal with on this course, these all have equivalent formulae for rotational mechanics. Instead of position $r$, we have angle $\theta$. Instead of velocity $v$, we have angular velocity $\omega$. Instead of acceleration $a$, we have angular acceleration $\alpha$. Instead of mass $m$, we have moment of inertia $I$. Instead of force $F$, we have torque $\tau$. Etc…

The main goal is to discover alternative formulae for rotational quantities, and to understand how they relate to the translational ones.

Centre of Mass

We can consider two possible cases:

A single point like object, such as a particle.
An extended object, such as a rod or a disc, made up of many particles.

Given a system of $i$ discrete masses, $m_i$, located at positions $\vec{r}_i$, the centre of mass $\vec{R}$ is given by:

$$M \vec{R} = \sum_i m_i \vec{r}_i$$

Where $M = \sum_i m_i$ is the total mass. For every component mass $m_i$, we consider two types of net forces acting upon it:

Internal forces, $\vec{f}_i^\text{int}$.
External forces, $\vec{f}_i^\text{ext}$.

The internal forces are those exerted by other masses in the system, and the external forces are those exerted by objects outside the system. Hence:

$$\vec{f}_i^\text{int} = \sum_{j \neq i} \vec{f}_{ij}$$

We want to consider how an object moves, and if we assume the object has a constant mass and is rigid (does not deform) then we can consider the motion of the centre of mass.

$$M \ddot{\vec{R}} = \sum_i m_i \ddot{\vec{r}}_i = \sum_i m_i \vec{a}_i = \sum_i \left(\vec{f}_i^\text{int} + \vec{f}_i^\text{ext}\right)$$

$$= \sum_i \left( \sum_{j \neq i} \vec{f}_{ij} + \vec{f}_i^\text{ext} \right)$$

$$= \sum_i \sum_{j \neq i} \vec{f}_{ij} + \sum_i \vec{f}_i^\text{ext}$$

However, if one internal mass exerts a force on another, then the second mass exerts an equal and opposite force on the first. Hence, the internal forces all nicely cancel out:

$$\sum_i \sum_{j \neq i} \vec{f}_{ij} = 0$$

$$M \vec{\ddot{R}} = \sum_i \vec{f}_i^\text{ext} = \vec{F}^\text{ext}$$

This is the analogue of Newton’s 2nd Law but for a non-point-like object. For a rigid body, the centre of mass moves as if the entirety of the object’s mass were concentrated at that point and all external forces were applied at that point.

For a rigid body, we can break down the motion into two components:

Translational motion of the centre of mass.
Rotational motion about the centre of mass (in the centre of mass frame)

Continuous Objects

However, many extended objects can be thought of as having a continuous distribution of mass, rather than being made up of finite discrete masses.

We define the density $\rho(\vec{r})$, such that $\rho(\vec{r}) dV$ is the mass contained in an infinitesimal volume element $dV$ around the point $\vec{r}$. While a point is massless, a small volume can contain a mass, albeit a tiny mass element. We can therefore get the total mass by integrating across all of these mass elements (effectively, we just swap sums for integrals).

Note that:

$$dV = dx \, dy \, dz \equiv d^3 \vec{r}$$

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