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An Atwood machine couples two hanging masses by an inextensible mechanical string over a pulley. With ideal massless hardware, Newton's second law gives the downward acceleration of the heavier mass as $a=g(m_2-m_1)/(m_1+m_2)$. To prove it, write $m_{2}g-T=m_{2}a$ and $T-m_{1}g=m_{1}a$, then add. The apparatus associated with George Atwood turns a small difference of weights into slow, measurable acceleration.

A compound Atwood machine replaces one hanging mass by another pulley carrying two masses. Let the upper mechanical string carry mass $M$ opposite the moving pulley, and let the lower mechanical string carry masses $a,b$ at upward heights $x,y$. The length constraints on the mechanical strings give the height of $M$ as $-(x+y)/2$ up to a constant. Thus its Lagrangian isThis follows by adding the three kinetic energies and subtracting their gravitational potential energies. Constant coordinate shifts annihilating the linear potential energy give conserved generalized momentum components by Noether's theorem.

Use upward generalized coordinates. Let $h$ be the displacement of the lower pulley and $z$ that of the $4m$ mass. The two inextensible mechanical strings imposeFor ideal massless pulleys and mechanical strings, only the three masses contribute kinetic energy. HenceUp to a constant, their gravitational potential energy isThe Lagrangian of this compound Atwood machine is therefore

A massless fixed-axis pulley carries a rim point mass $M$ at radius $R$. A mechanical string gripping its rim without slipping passes over a fixed guide pulley to a hanging mass $m$. Choose the angle $\theta$ from downward vertical so the hanging mass descends by $R\theta$. The Lagrangian is $(M+m)R^2{\dot{\theta }}^2/2+MgR\cos \theta +mgR\theta$. The stable equilibrium satisfies $\sin {\theta }_e=m/M$ with $\cos {\theta }_e>0$, and exists for $0<m<M$. Its small-amplitude angular frequency is $[Mg\cos {\theta }_e/((M+m)R){]}^{1/2}$.