Atwood machine and conservation theorems

(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.