--- title: "Atwood machine and conservation theorems" subject: "Engineering Mechanics" chapter: 9 key: "semester_3/engineering_mechanics:atwood-machine-lagrange-conservation" source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/atwood-machine-lagrange-conservation site: BCT Engineering --- # Atwood machine and conservation theorems Engineering Mechanics · Chapter 9: Lagrangian Dynamics - Asked in: year not given, 3+5 marks - Video: [Physics 9 Conservation of Energy (7 of 11) Atwood Machine](https://www.youtube.com/watch?v=HUnaZYkRZes) — Michel van Biezen - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/atwood-machine-lagrange-conservation - Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-9 ## Question (a) State the conservation theorems of Lagrangian dynamics (cyclic coordinate and energy). (b) Using Lagrange's equation, find the acceleration of the masses of an Atwood machine having masses 3 kg and 5 kg connected by a light inextensible string over a smooth massless pulley. Also find the tension in the string. ## Short answer If Lagrangian does not contain a coordinate, its conjugate momentum is conserved; if L has no explicit time and constraints are scleronomic, total energy H = T + V is constant. For Atwood machine (masses 3 kg, 5 kg), using one coordinate x, Lagrange’s equation yields acceleration 2.45 m/s² (5 kg down, 3 kg up). Tension = 36.8 N. ## Answer #### (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. ```text ___ / \ 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. If L does not contain a coordinate q_k, what is conserved? A. Total energy B. The generalised momentum p_k = dL/dq_k-dot C. The coordinate q_k D. Kinetic energy Answer: B. dL/dq_k = 0 gives d/dt(dL/dq_k-dot) = 0. (Key point: Cyclic coordinate) 2. When is the Hamiltonian H (= T + V here) conserved? A. Whenever forces are non-zero B. When L does not depend explicitly on time and constraints are scleronomic C. Only for one particle D. When all coordinates are cyclic Answer: B. Time independence of L and the constraints makes H constant. (Key point: Energy conservation) 3. What is the equation of motion for the Atwood machine from Lagrange's equation? A. (m1 + m2) x'' = (m2 - m1) g B. (m1 - m2) x'' = (m1 + m2) g C. (m1 + m2) x'' = (m1 + m2) g D. m2 x'' = m1 g Answer: A. L = (1/2)(m1+m2)x-dot^2 + m2 g x + m1 g(l - x). (Key point: Atwood Lagrangian) 4. For 3 kg and 5 kg masses, what are the acceleration and tension? A. 2.453 m/s^2 and 36.79 N B. 1.96 m/s^2 and 29.4 N C. 2.453 m/s^2 and 49.05 N D. 4.9 m/s^2 and 36.79 N Answer: A. a = 2(9.81)/8 = 2.4525; T = 3(9.81 + 2.4525) = 36.79 N. 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