Question 1.19: Apply the Lagrangian formalism to obtain the equations of mo......

Let (x_{1}, y_{1}) and (x_{2}, y_{2}) be the Cartesian coordinates of masses m_{1} and m_{2}, respectively. Taking the angles θ_{1} and θ_{2} as generalised coordinates (see Fig. 1.1 in Section 1.2), we have

x_{1} = l_{1} \sin \theta _{1} ,    y_{1} = l_{1} \cos \theta _{1} ,

x_{2} = l_{1} \sin \theta _{1}+l_{2} \sin \theta _{2} ,    y_{2} = l_{1} \cos \theta _{1} + l_{2} \cos \theta _{2},   (1.120)

whence

\dot{x}_{1}= l_{1}\dot{\theta}_{1} \cos \theta _{1} ,  \dot{x}_{2}= l_{1}\dot{\theta}_{1} \cos \theta _{1}+l_{2}\dot{\theta}_{2} \cos \theta _{2} , (1.121)

\dot{y}_{1}=- l_{1}\dot{\theta}_{1} \sin \theta _{1} ,  \dot{y}_{2}= -l_{1}\dot{\theta}_{1} \sin \theta _{1}-l_{2}\dot{\theta}_{2} \sin \theta _{2} . (1.122)

The kinetic energy relative to the inertial reference frame (x, y) is

T= \frac{m_{1}}{2} \left(\dot{x}^{2}_{1}+\dot{y}^{2}_{1}\right) + \frac{m_{2}}{2} \left(\dot{x}^{2}_{2}+\dot{y}^{2}_{2}\right) , (1.123)

which in terms of the generalised coordinates and velocities takes the form

T=\frac{m_{1}+m_{2}}{2} l^{2}_{1}\dot{\theta}^{2}_{1} + \frac{m_{2}}{2} l^{2}_{2}\dot{\theta}^{2}_{2} + m_{2} l_{1}l_{2}\dot{\theta}_{1}\dot{\theta}_{2} \cos \left(\theta _{1}-\theta _{2}\right). (1.124)

On the other hand, with the zero level of the gravitational potential energy on the horizontal plane through the point of suspension of m_{1}, we have

V = −m_{1}gy_{1} − m_{2}gy_{2} = −(m_{1} + m_{2})gl_{1} \cos \theta_{1} − m_{2}gl_{2} \cos \theta_{2} . (1.125)

Finally, the Lagrangian L = T − V is given by

L = \frac{m_{1}+m_{2}}{2}l^{2}_{1} \dot{\theta}^{2}_{1} +\frac{m_{2}l^{2}_{2}}{2}\dot{\theta}^{2}_{2}+ m_{2}l_{1}l_{2}\dot{\theta}_{1}\dot{\theta}_{2} \cos \left(\theta_{1}-\theta_{2}\right) + \left(m_{1}+m_{2}\right)g l_{1} \cos \theta_{1} + m_{2}g l_{2} \cos \theta_{2} . (1.126)

Making use of

\frac{\partial L}{\partial \dot{\theta }_{1} } =\left(m_{1}+m_{2} \right) l^{2}_{1} \dot{\theta }_{1}+m_{2}l_{1} l_{2} \dot{\theta }_{2}\cos \left(\theta _{1} -\theta _{2}\right) , (1.127)

\frac{\partial L}{\partial \theta_{1} } =-m_{2}l_{1} l_{2}\dot{\theta }_{1} \dot{\theta }_{2}\sin \left(\theta _{1} -\theta _{2}\right) -\left(m_{1}+m_{2} \right) g  l_{1}\sin \theta _{1} , (1.128)

the Lagrange equation

\frac{d}{dt} \left(\frac{\partial L }{\partial\dot{\theta }_{1}} \right) -\frac{\partial L }{\partial\theta_{1}}=0 (1.129)

becomes

\left(m_{1}+m_{2}\right)l^{2}_{1} \ddot{\theta }_{1} + m_{2} l_{1} l_{2}\ddot{\theta }_{2}\cos \left(\theta _{1} -\theta _{2}\right) +m_{2} l_{1} l_{2}\dot{\theta }^{2}_{2}\sin \left(\theta _{1} -\theta _{2}\right)+\left(m_{1}+m_{2}\right) g l_{1}\sin \theta _{1} =0.  (1.130)

In an entirely analogous fashion, one finds

m_{2} l^{2}_{2}\ddot{\theta }_{2} +m_{2} l_{1} l_{2}\ddot{\theta }_{1}\cos \left(\theta _{1} -\theta _{2}\right) – m_{2} l_{1} l_{2}\dot{\theta }^{2}_{1}\sin \left(\theta _{1} -\theta _{2}\right) + m_{2}gl_{2}\sin \theta _{2}=0   (1.131)

for the second of Lagrange’s equations.