Key Points

The centre of mass is a point representing the mean position of the matter in a body. For a system of particles, its position vector $\mathbf{R}$ is given by $\mathbf{R} = \frac{\sum m_i \mathbf{r}_i}{M}$, where $M$ is the total mass.

The centre of mass of a system moves as if the entire mass of the system were concentrated at that point and all external forces were applied there. The equation of motion is $\mathbf{F}_{\text{ext}} = M \mathbf{A}_{\text{CM}}$.

The total linear momentum of a system of particles is the product of the total mass and the velocity of its centre of mass: $\mathbf{P} = M \mathbf{V}_{\text{CM}}$. Newton's second law for the system is $\mathbf{F}_{\text{ext}} = \frac{d\mathbf{P}}{dt}$.

The vector product (or cross product) of two vectors $\mathbf{a}$ and $\mathbf{b}$ is a vector $\mathbf{c} = \mathbf{a} \times \mathbf{b}$. Its magnitude is $c = ab \sin\theta$ and its direction is perpendicular to the plane of $\mathbf{a}$ and $\mathbf{b}$, given by the right-hand rule.

Torque is the rotational analogue of force and is defined as the vector product of the position vector $\mathbf{r}$ and the force $\mathbf{F}$. It is given by $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$. Its SI unit is newton-metre (N m).

Angular momentum is the rotational analogue of linear momentum. For a particle, it is defined as $\mathbf{l} = \mathbf{r} \times \mathbf{p}$, where $\mathbf{p}$ is the linear momentum. For a system of particles, the total angular momentum is $\mathbf{L} = \sum \mathbf{l}_i$.

The time rate of change of the total angular momentum of a system of particles is equal to the sum of the external torques acting on the system. This is expressed as $\frac{d\mathbf{L}}{dt} = \boldsymbol{\tau}_{\text{ext}}$.

If the total external torque on a system is zero, its total angular momentum remains constant. This is the law of conservation of angular momentum, expressed as $I\omega = \text{constant}$.

A rigid body is in mechanical equilibrium if the net external force is zero (translational equilibrium, $\sum \mathbf{F}_{\text{ext}} = \mathbf{0}$) and the net external torque is zero (rotational equilibrium, $\sum \boldsymbol{\tau}_{\text{ext}} = \mathbf{0}$).

Moment of inertia ($I$) is the rotational analogue of mass, representing an object's resistance to angular acceleration. It is defined as $I = \sum m_i r_i^2$, where $r_i$ is the perpendicular distance of mass $m_i$ from the axis of rotation.

The radius of gyration ($k$) is the distance from the axis of rotation to a point where the entire mass of the body could be concentrated without changing its moment of inertia. It is related by $I = M k^2$.

The kinetic energy of a body rotating about a fixed axis is given by $K = \frac{1}{2} I \omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.

The rotational analogue of Newton's second law ($F = ma$) is $\tau = I \alpha$. This states that the net external torque on a body is equal to the product of its moment of inertia and its angular acceleration.

The linear velocity $\mathbf{v}$ of a particle in a rigid body rotating with angular velocity $\boldsymbol{\omega}$ is given by the vector product $\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$, where $\mathbf{r}$ is the position vector of the particle from the axis.

For uniform angular acceleration $\alpha$, the equations of rotational motion are analogous to linear motion: $\omega = \omega_0 + \alpha t$, $\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2$, and $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$.