Atwood Machines

An Atwood machine consists of two weights, of mass $m_1$ and $m_2$, connected by a light inextensible cord of length $l$, which passes over a pulley of radius $a\ll l$, and moment of inertia $I$. See Fig. 46.

Figure 46: An Atwood machine.

Referring to the diagram, we can see that this is a one degree of freedom system whose instantaneous configuration is specified by the coordinate $x$. Assuming that the cord does not slip with respect to the pulley, the angular velocity of pulley is $\dot{x}/a$. Hence, the kinetic energy of the system is given by

$$K = \frac{1}{2}\,m_1\,\dot{x}^{\,2} + \frac{1}{2}\,m_2\,\dot{x}^{\,2} + \frac{1}{2}\,I\, \frac {\dot{x}^{\,2}}{a^2}.$$ (694)

The potential energy of the system takes the form

$$U = -m_1\,g\,x - m_2\,g\,(l-x).$$ (695)

It follows that the Lagrangian is written

$$L=\frac{1}{2}\left(m_1+ m_2+\frac{I}{a^2}\right)\dot{x}^{\,2} + g\,(m_1-m_2)\,x + {\rm const}.$$ (696)

The equation of motion,

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{x}}\right) - \frac{\partial L}{\partial x} = 0,$$ (697)

thus yields

$$\left(m_1+ m_2+\frac{I}{a^2}\right)\ddot{x} - g\,(m_1-m_2) = 0,$$ (698)

or

$$\ddot{x} = \frac{g\,(m_1-m_2)}{m_1+m_2 + I/a^2},$$ (699)

which is the correct answer.

Figure 47: A double Atwood machine.

Consider the dynamical system drawn in Fig. 47. This is an Atwood machine in which one of the weights has been replaced by a second Atwood machine. The system now has two degrees of freedom, and its instantaneous position is specified by the two coordinates $x$ and $x'$, as shown.

For the sake of simplicity, let us neglect the masses of the two pulleys. Thus, the kinetic energy of the system is written

$$K = \frac{1}{2}\,m_1\,\dot{x}^{\,2} + \frac{1}{2}\,m_2\,(-\dot{x}+ \dot{x}')^2 + \frac{1}{2}\,m_3\,(-\dot{x} - \dot{x}')^2,$$ (700)

whereas the potential energy takes the form

$$U = - m_1\,g\,x - m_2\,g\,(l-x+x')- m_3\,g\,(l-x+l'-x').$$ (701)

It follows that the Lagrangian of the system is

$$L = \frac{1}{2}\,m_1\,\dot{x}^{\,2} + \frac{1}{2}\,m_2\,(-\dot{x}+ \dot{x}')^2 + \frac{1}{2}\,m_3\,(-\dot{x} - \dot{x}')^2$$

$$+ g\,(m_1-m_2-m_3)\,x+g\,(m_2-m_3)\,x' + {\rm constant}.$$ (702)

Hence, the equations of motion,

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{x}}\right) - \frac{\partial L}{\partial x} = 0,$$ (703)

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{x}'}\right) - \frac{\partial L}{\partial x'} = 0,$$ (704)

yield

$$m_1\,\ddot{x} + m_2\,(\ddot{x}- \ddot{x}') + m_3\,(\ddot{x}+\ddot{x}') - g\,(m_1-m_2-m_3) = 0,$$ (705)

$$m_2\,(-\ddot{x} + \ddot{x}')+m_3\,(\ddot{x}+\ddot{x}') - g\,(m_2-m_3) = 0.$$ (706)

The accelerations $\ddot{x}$ and $\ddot{x}'$ can be obtained from the above two equations via simple algebra.