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 Figure 32.

Figure 32: 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

$$K = \frac{1}{2}\,m_1\,\dot{x}^{\,2} + \frac{1}{2}\,m_2\,\dot{x}^{\,2} + \frac{1}{2}\,I\, \frac {\dot{x}^{\,2}}{a^2}.$$ (624)

The potential energy of the system takes the form

$$U = -m_1\,g\,x - m_2\,g\,(l-x).$$ (625)

It follows that the Lagrangian is written

$$L=\frac{1}{2}\left(m_1+ m_2+\frac{I}{a^2}\right)\dot{x}^{\,2} + g\,(m_1-m_2)\,x + {\rm const}.$$ (626)

The equation of motion,

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{x}}\right) - \frac{\partial L}{\partial x} = 0,$$ (627)

thus yields

$$\left(m_1+ m_2+\frac{I}{a^2}\right)\ddot{x} - g\,(m_1-m_2) = 0,$$ (628)

or

$$\ddot{x} = \frac{g\,(m_1-m_2)}{m_1+m_2 + I/a^2},$$ (629)

which is the correct answer.

Figure 33: A double Atwood machine.

Consider the dynamical system drawn in Figure 33. This is an Atwood machine in which one of the weights has been replaced by a second Atwood machine with a cord of length $l'$. 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

$$K = \frac{1}{2}\,m_1\,\dot{x}^{\,2} + \frac{1}{2}\,m_2\,(-\dot{x}+ \dot{x}')^2 + \frac{1}{2}\,m_3\,(-\dot{x} - \dot{x}')^2,$$ (630)

whereas the potential energy takes the form

$$U = - m_1\,g\,x - m_2\,g\,(l-x+x')- m_3\,g\,(l-x+l'-x').$$ (631)

It follows that the Lagrangian of the system is

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

$$+ g\,(m_1-m_2-m_3)\,x+g\,(m_2-m_3)\,x' + {\rm const}.$$ (632)

Hence, the equations of motion,

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{x}}\right) - \frac{\partial L}{\partial x} = 0,$$ (633)

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{x}'}\right) - \frac{\partial L}{\partial x'} = 0,$$ (634)

yield

$$m_1\,\ddot{x} + m_2\,(\ddot{x}- \ddot{x}') + m_3\,(\ddot{x}+\ddot{x}') - g\,(m_1-m_2-m_3) = 0,$$ (635)

$$m_2\,(-\ddot{x} + \ddot{x}')+m_3\,(\ddot{x}+\ddot{x}') - g\,(m_2-m_3) = 0.$$ (636)

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