Atwood machine

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.

Two initially equal buckets of total mass $M_0$ share an ideal Atwood machine. If one loses sand at constant rate $\alpha$ and the sand leaves with zero initial velocity relative to that bucket, the leakage supplies no rocket thrust. The remaining masses satisfy $(M+M_0)\dot{v}=g(M_0-M)$ for upward velocity of the leaking side. Using $\dot{M}=-\alpha$ and integrating from rest gives $v=(g/\alpha )[M-M_0-2M_0\log ((M+M_0)/(2M_0))]$.

If the initial sand fraction is $ϵ$, emptying occurs at $M=(1-ϵ)M_0$. This gives the displayed speed, which tends to zero for $ϵ\to 0$ and to $(g{M}_0/\alpha )(2\log 2-1)$ for $ϵ\to 1^{-}$. The latter is a finite limiting result, not a physical claim about an exactly massless empty bucket. The derivation is an application of a variable-mass system, rather than of $d(Mv)/dt=F$ for a closed body.

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.