Lagrangian Dynamics

Degrees of freedom (DOF) is the minimum number of independent coordinates required to completely specify the configuration of a system at any instant. For $N$ particles in space with $c$ independent constraint equations, $n = 3N - c$.

Generalised coordinates $q_1, q_2, \dots, q_n$ are any set of $n$ independent quantities (lengths, angles, etc.) that fix the configuration of a system with $n$ degrees of freedom; they need not have the dimension of length, and they automatically satisfy the constraints.

Constraints

• Holonomic: can be expressed as an equation connecting coordinates (and time) only, $f(x_1, x_2, \dots, t) = 0$; for example a rigid link of fixed length $x^2 + y^2 = l^2$. They reduce the number of coordinates.
• Non-holonomic: cannot be integrated into such an equation, being inequalities or non-integrable velocity relations; for example a particle moving on the outside of a sphere (it can leave, $r \ge R$), or a rolling sphere on a rough plane.

Further classes: scleronomic (not explicitly time dependent) and rheonomic (time dependent).

Examples

For the pendulum in (a) the particle has 2 coordinates in the plane $(x, y)$ and one constraint $x^2 + y^2 = l^2$, so $2 - 1 = 1$ DOF. Using generalised coordinates removes the constraint reaction (rod tension) from the equations of motion.

Consider $N$ particles with positions $\mathbf{r}_i(q_1,\dots,q_n,t)$ in terms of $n$ independent generalised coordinates $q_j$ (holonomic constraints). Let $\mathbf{F}_i$ be the applied force on particle $i$ (constraint forces do no virtual work on smooth constraints).

Step 1: D'Alembert-Lagrange principle

D'Alembert's principle with virtual displacements: the virtual work of the real forces plus the inertia forces is zero:

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

Step 2: Express in generalised coordinates

Since $\delta\mathbf{r}_i = \sum_j \dfrac{\partial\mathbf{r}_i}{\partial q_j}\delta q_j$, the applied-force term becomes $\sum_j Q_j\,\delta q_j$, where the generalised force is

$$Q_j = \sum_i \mathbf{F}_i\cdot\frac{\partial\mathbf{r}_i}{\partial q_j}$$

The inertia term is $\sum_j\left[\sum_i m_i\ddot{\mathbf{r}}_i\cdot\dfrac{\partial\mathbf{r}_i}{\partial q_j}\right]\delta q_j$.

Step 3: Convert the inertia term to kinetic energy

The velocity is $\dot{\mathbf{r}}_i = \sum_j \dfrac{\partial\mathbf{r}_i}{\partial q_j}\dot q_j + \dfrac{\partial\mathbf{r}_i}{\partial t}$. Two identities follow:

$$\frac{\partial\dot{\mathbf{r}}_i}{\partial\dot q_j} = \frac{\partial\mathbf{r}_i}{\partial q_j} \qquad \frac{d}{dt}\left(\frac{\partial\mathbf{r}_i}{\partial q_j}\right) = \frac{\partial\dot{\mathbf{r}}_i}{\partial q_j}$$

Then

$$\sum_i m_i\ddot{\mathbf{r}}_i\cdot\frac{\partial\mathbf{r}_i}{\partial q_j} = \frac{d}{dt}\left(\sum_i m_i\dot{\mathbf{r}}_i\cdot\frac{\partial\dot{\mathbf{r}}_i}{\partial\dot q_j}\right) - \sum_i m_i\dot{\mathbf{r}}_i\cdot\frac{\partial\dot{\mathbf{r}}_i}{\partial q_j}$$

With kinetic energy $T = \sum_i\tfrac12 m_i\dot{\mathbf{r}}_i^2$, this equals $\dfrac{d}{dt}\left(\dfrac{\partial T}{\partial\dot q_j}\right) - \dfrac{\partial T}{\partial q_j}$.

Step 4: Lagrange's equations

Because the $\delta q_j$ are independent, each coefficient must vanish:

$$\frac{d}{dt}\left(\frac{\partial T}{\partial\dot q_j}\right) - \frac{\partial T}{\partial q_j} = Q_j \qquad (j = 1, \dots, n)$$

Conservative systems

For conservative forces $Q_j = -\dfrac{\partial V}{\partial q_j}$, where $V = V(q)$ does not depend on velocities. Define the Lagrangian $L = T - V$. Since $\dfrac{\partial V}{\partial\dot q_j} = 0$:

$$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot q_j}\right) - \frac{\partial L}{\partial q_j} = 0$$

If some forces are non-conservative, they appear as $Q_j^{nc}$ on the right-hand side. The number of equations equals the number of DOF, and constraint forces never appear.

(a) Conservation theorems

1. Cyclic (ignorable) coordinate theorem: if the Lagrangian does not contain a coordinate $q_k$ (so $\partial L/\partial q_k = 0$), then Lagrange's equation gives $\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial\dot q_k}\right) = 0$, i.e. the generalised momentum $p_k = \dfrac{\partial L}{\partial\dot q_k}$ is constant. Examples: if $L$ is independent of $x$, the linear momentum $p_x = m\dot x$ is conserved; if independent of $\theta$ (central force), the angular momentum $p_\theta = mr^2\dot\theta$ is conserved.
2. Energy conservation: if $L$ does not depend explicitly on time and the constraints are scleronomic (time independent), the Hamiltonian $H = \sum_j\dot q_j\dfrac{\partial L}{\partial\dot q_j} - L$ is constant. For such conservative systems $H = T + V$, so the total mechanical energy is conserved.

(b) Atwood machine

Given: $m_1 = 3$ kg, $m_2 = 5$ kg; string length $l$ is constant.

       ___
      /   \   smooth massless pulley
     |     |
    [m1]  [m2]
     ^      | x (downward for m2)

Coordinate: one DOF. Let $x$ be the downward displacement of $m_2$ from the pulley level; then $m_1$ is at $(l - x)$ (it rises when $m_2$ goes down), and both have speed $\dot x$.

$$T = \tfrac12(m_1 + m_2)\dot x^2 \qquad V = -m_2 g x - m_1 g(l - x)$$ $$L = T - V = \tfrac12(m_1 + m_2)\dot x^2 + m_2 g x + m_1 g(l - x)$$

Lagrange's equation $\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial\dot x}\right) - \dfrac{\partial L}{\partial x} = 0$:

$$(m_1 + m_2)\ddot x - (m_2 - m_1)g = 0$$ $$\ddot x = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(5 - 3)(9.81)}{8} = 2.4525\ \text{m/s}^2$$

$m_2$ accelerates downward and $m_1$ upward at this rate.

Tension (from Newton's law for $m_1$): $T - m_1 g = m_1\ddot x$

$$T = m_1(g + \ddot x) = 3(9.81 + 2.4525) = 36.79\ \text{N}$$

(Check using $m_2$: $m_2(g - \ddot x) = 5(9.81 - 2.4525) = 36.79$ N. It agrees. Also $T = \dfrac{2m_1m_2g}{m_1+m_2}$.)

Answer: Acceleration = 2.453 m/s² (5 kg down, 3 kg up); tension = 36.79 N.