Question 1.18: Find a Lagrangian and the equations of motion for the mechan......

Let us employ the coordinates shown in Fig. 1.8 and assume that the string remains taut, which requires initial conditions carefully chosen (see Problem 1.2). The constraint x_{1} + x_{2} = l_{0}, where l_{0} is a constant determined by the string’s length and the radius of the pulley, shows that out of the three coordinates, only x_{3} and either x_{1} or x_{2} can be taken as generalised coordinates (the system has only two degrees of freedom). Let us choose x_{2} and x_{3} as generalised coordinates. The kinetic energy of the system is

T=\frac{m_{1}}{2}\dot{ x}^{2}_{1}+\frac{m_{2}}{2}\dot{ x}^{2}_{2}+\frac{m_{3}}{2}\dot{ x}^{2}_{3}=\frac{m_{1}+m_{2}}{2}\dot{ x}^{2}_{2}+\frac{m_{3}}{2}\dot{ x}^{2}_{3},  (1.114)

since from x_{1} = l_{0}−x_{2} one derives \dot{x}_{1} = −\dot{ x}_{2}. Setting the zero level of the gravitational potential energy on the horizontal plane through the centre of the pulley, we have

V=-m_{1}g x_{1}-m_{2}g x_{2}-m_{3}g x_{3}+\frac{k}{2}\left({x}_{3}-{x}_{2}-l\right)^{2}

=-\left(m_{2}-m_{1}\right)g x_{2}-m_{1}g l_{0} -m_{3}g x_{3}+\frac{k}{2}\left({x}_{3}-{x}_{2}-l\right)^{2} , (1.115)

since points below the zero level have negative gravitational potential energy. Dropping an immaterial additive constant, the Lagrangian is

L = T −V = \frac{m_{1}+m_{2}}{2}\dot{ x}^{2}_{2}+\frac{m_{3}}{2}\dot{ x}^{2}_{3}+\left(m_{2}-m_{1}\right)g x_{2} +m_{3}g x_{3}- \frac{k}{2}\left({x}_{3}-{x}_{2}-l\right)^{2} . (1.116)

We have

\frac{\partial L}{\partial \dot{ x}_{2}}=\left(m_{1}+m_{2}\right)\dot{ x}_{2} , \frac{\partial L}{\partial x_{2}}=\left(m_{2}-m_{1}\right)g + k \left({x}_{3}-{x}_{2}-l\right), (1.117)

\frac{\partial L}{\partial \dot{ x}_{3}}=m_{3}\dot{ x}_{3} , \frac{\partial L}{\partial x_{3}}=m_{3}g – k \left({x}_{3}-{x}_{2}-l\right), (1.118)

and Lagrange’s equations are

\frac{d}{dt} \left(\frac{\partial L}{\partial\dot{ x}_{2}}\right) – \frac{\partial L}{\partial  x_{2}}=0 ⇒ \left(m_{1}+m_{2}\right)\ddot{ x}_{2} – \left(m_{2}-m_{1}\right)g – k \left({x}_{3}-{x}_{2}-l\right)=0 ,   (1.119a)

\frac{d}{dt} \left(\frac{\partial L}{\partial\dot{ x}_{3}}\right) – \frac{\partial L}{\partial  x_{3}}=0 ⇒ m_{3}\ddot{ x}_{3} – m_{3}g + k \left({x}_{3}-{x}_{2}-l\right)=0 .   (1.119b)

If k = 0 there is no interaction between m_{2} and m_{3}. In this limiting case Lagrange’s equations predict correctly that m_{3} falls freely (\ddot{x}_{3} = g) and that the accceleration of m_{2} is \ddot{x}_{2} = (m_{2} − m_{1})g/(m_{1} + m_{2}), in agreement with the result obtained in the treatment of Atwood’s machine by d’Alembert’s principle in Example 1.12.