Generalised Coordinates and Constraints

2.1 Generalised Coordinates

A system with $n$ degrees of freedom can be described by $n$ generalised coordinates $q_1, q_2, \ldots, q_n$Which may be angles, arc lengths, or any other set of parameters that Uniquely determines the configuration.

The Cartesian coordinates are functions of the generalised coordinates (and possibly time):

$\mathbf{r}_i = \mathbf{r}_i(q_1, q_2, \ldots, q_n, t), \quad i = 1, \ldots, N$

The velocities are:

$\dot{\mathbf{r}}_i = \sum_{j=1}^n \frac{\partial \mathbf{r}_i}{\partial q_j}\dot{q}_j + \frac{\partial \mathbf{r}_i}{\partial t}$

Example. A simple pendulum has one degree of freedom. We can use the angle $\theta$ from the Vertical as the generalised coordinate, rather than the Cartesian coordinates $(x, y)$ of the bob.

2.2 Constraints

Holonomic constraints relate the coordinates by equations:

$f(\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N, t) = 0$

A holonomic constraint reduces the number of degrees of freedom.

Non-holonomic constraints involve inequalities or non-integrable differential relations:

$\sum_j a_j(q, t)\, dq_j + a_t(q, t)\, dt = 0$

Which cannot be integrated to yield a relation among the $q_j$ alone.

Scleronomic constraints do not depend explicitly on time. Rheonomic constraints do.

Example. A bead sliding on a fixed wire: the constraint $f(x,y) = 0$ is holonomic and scleronomic. A bead on a wire that moves with time: holonomic and rheonomic.

2.3 Worked Example: Classifying Constraints

Problem. Classify the following constraints: (a) a particle on the surface of a sphere of radius $R$(b) a rolling disk (vertical), (c) a particle constrained to $z \geq 0$(d) a pendulum whose pivot oscillates as $x_0(t) = A\cos(\omega t)$.

2.4 Worked Example: Finding Generalised Coordinates

Problem. A rod of length $l$ and negligible mass has masses $m_1$ and $m_2$ at its ends. The rod slides on a frictionless horizontal table. Find suitable generalised coordinates.

2.5 Virtual Work and D”Alembert’s Principle

A virtual displacement $\delta \mathbf{r}_i$ is an infinitesimal change in position consistent with The constraints at a fixed instant in time ($\delta t = 0$).

Definition (Virtual Work). The virtual work of the forces is:

$\delta W = \sum_{i=1}^N \mathbf{F}_i \cdot \delta \mathbf{r}_i$

Definition (Ideal Constraints). Constraints are ideal if the virtual work of the constraint forces is zero:

$\sum_{i=1}^N \mathbf{C}_i \cdot \delta \mathbf{r}_i = 0$

Where $\mathbf{C}_i$ is the constraint force on particle $i$.

Theorem 2.1 (Principle of Virtual Work). A system is in static equilibrium if and only if the virtual work of the applied forces vanishes for all virtual displacements consistent with the constraints.

Proof. In static equilibrium, $\mathbf{F}_i + \mathbf{C}_i = \mathbf{0}$ for each particle. Therefore:

$\sum_i (\mathbf{F}_i + \mathbf{C}_i) \cdot \delta\mathbf{r}_i = 0$

For ideal constraints, $\sum_i \mathbf{C}_i \cdot \delta\mathbf{r}_i = 0$So $\sum_i \mathbf{F}_i \cdot \delta\mathbf{r}_i = 0$. Conversely, if the virtual work of applied forces vanishes for all admissible virtual displacements, the system must be in equilibrium (otherwise one could choose a virtual displacement in the direction of net force to get non-zero work). $\blacksquare$

Theorem 2.2 (D’Alembert’s Principle). For a system of $N$ particles:

$\sum_{i=1}^N (\mathbf{F}_i - m_i \ddot{\mathbf{r}}_i) \cdot \delta \mathbf{r}_i = 0$

Where $\mathbf{F}_i$ includes both applied and constraint forces. For ideal constraints, the Constraint forces do no virtual work, so only the applied forces contribute.

Proof. D’Alembert’s principle extends the principle of virtual work to dynamics by treating $-m_i \ddot{\mathbf{r}}_i$ as a “fictitious force” (the inertia force). Starting from Newton’s second law $\mathbf{F}_i + \mathbf{C}_i = m_i \ddot{\mathbf{r}}_i$:

$\sum_i (\mathbf{F}_i + \mathbf{C}_i - m_i\ddot{\mathbf{r}}_i) \cdot \delta\mathbf{r}_i = 0$

This is true. For ideal constraints $\sum_i \mathbf{C}_i \cdot \delta\mathbf{r}_i = 0$Giving:

$\sum_i (\mathbf{F}_i - m_i\ddot{\mathbf{r}}_i) \cdot \delta\mathbf{r}_i = 0$

$\blacksquare$