How to Find Kinetic energy in terms of spherical co-ordinates

Question

How to Find Kinetic energy in terms of \((r,\theta ,\phi )\)

Solution

We have to find the kinetic energy in terms of \((r,\theta ,\phi )\) that is in terms of spherical c0-ordinates. Kinetic energy in terms of Cartesian co-ordinates is

\(T = \frac{1}{2}m\left( {{{\dot x}^2} + {{\dot y}^2} + {{\dot z}^2}} \right)\)                                                                (1)

Where \(\dot x,\dot y{\rm{ and }}\dot z\) are derivatives of z, y and z with respect to time.

Cartesian co-ordinates x, y, z in terms of \(r,\theta \) and \(\phi \) are

\(\begin{array}{l}x = r\sin \theta \cos \phi \\y = r\sin \theta \sin \phi \\z = r\cos \theta \end{array}\)

Now derivatives of x, y and z w.r.t. t are

\(\begin{array}{l}\dot x = \frac{{dx}}{{dt}} = \dot r\sin \theta \cos \phi  + r\cos \theta \cos \phi \dot \theta  - r\sin \theta \sin \phi \dot \phi \\\dot y = \frac{{dy}}{{dt}} = \dot r\sin \theta \sin \phi  + r\cos \theta \sin \phi \dot \theta  + r\sin \theta \cos \phi \dot \phi \\\dot z = \frac{{dz}}{{dt}} = \dot r\cos \theta  - r\sin \theta \dot \theta \end{array}\)  

Where

\(\dot r = \frac{{dr}}{{dt}},\dot \theta  = \frac{{d\theta }}{{dt}},\dot \phi  = \frac{{d\phi }}{{dt}}\)  means all \(r,\theta \) and \(\phi \) changes with time as the particle moves or changes its position with time.

Now calculate for \({\dot x^2},{\dot y^2},{\dot z^2}\) and add them. After adding them we get

\({(\dot x)^2} + {(\dot y)^2} + {(\dot z)^2} = {\dot r^2} + {r^2}{\dot \theta ^2} + {r^2}{\sin ^2}\theta {\dot \phi ^2}\)

Putting this value of \({(\dot x)^2} + {(\dot y)^2} + {(\dot z)^2}\)in equation 1 we get kinetic energy of particle or system in terms of \(r,\theta \) and \(\phi \).

Hence

\(T = \frac{1}{2}m({\dot r^2} + {r^2}{\dot \theta ^2} + {r^2}{\sin ^2}\theta {\dot \phi ^2})\)

Equations of motion of coupled pendulum using the lagrangian method

Question

Obtain the equations of motion of coupled pendulum using the lagrangian method.

Solution

Consider a system of coupled pendulums as shown below in the figure

The displacement of A is \({x_1}\) and B is\({x_2}\) , condition being \({x_1}\) < \({x_2}\). In such state the spring gets stretched. The lengths of the strings of both the pendulums are same (say l). The angular displacement of A is \({\theta _1}\) and that of B is \({\theta _2}({\theta _2} > {\theta _1})\).

Therefore

\(\begin{array}{l}{x_1} = l{\theta _1} \Rightarrow {\theta _1} = \frac{{{x_1}}}{l}{\rm{                      (1)}}\\{x_2} = l{\theta _2} \Rightarrow {\theta _2} = \frac{{{x_2}}}{l}{\rm{                     (2)}}\end{array}\)

As the spring gets stretched, it is clear from the figure that restoring force works along the direction of displacement \({\theta _1}\) and opposite to the direction of displacement\({\theta _2}\) . Now A and B at zero potential level, the total potential energy of the system is given as

\(V = mgl(1 - \cos {\theta _1}) + mgl(1 - \cos {\theta _2}) + \frac{1}{2}k{({x_2} - {x_1})^2}\)

Where m is the mass of each one of the bob and k is the spring constant.

Since \({\theta _1}\)and \({\theta _2}\)are small so,

\(\begin{array}{l}\cos {\theta _1} = 1 - \frac{{\theta _1^2}}{2} + \frac{{\theta _1^4}}{4} + ......\\\cos {\theta _2} = 1 - \frac{{\theta _2^2}}{2} + \frac{{\theta _2^4}}{4} + ......\end{array}\)

Neglecting the higher powers other than squares of \({\theta _1}\)and \({\theta _2}\)the expression of potential energy can be written as

\(\begin{array}{l}V = mgl\frac{{\theta _1^2}}{2} + mgl\frac{{\theta _2^2}}{2} + \frac{1}{2}k{({x_2} - {x_1})^2}\\{\rm{    = }}\frac{{mgx_1^2}}{{2l}} + \frac{{mgx_2^2}}{{2l}} + \frac{1}{2}k{({x_2} - {x_1})^2}\end{array}\)

Also the kinetic energy of whole system is

\(T = \frac{1}{2}m\dot x_1^2 + \frac{1}{2}m\dot x_2^2 = \frac{1}{2}m(\dot x_1^2 + \dot x_2^2)\)

Hence Lagrangian L would be

\(\begin{array}{l}L = T - V\\L = \frac{1}{2}m(\dot x_1^2 + \dot x_2^2) - \frac{{mgx_1^2}}{{2l}} - \frac{{mgx_2^2}}{{2l}} - \frac{1}{2}k{({x_2} - {x_1})^2}\end{array}\)

Now

 \(\begin{array}{l}  \frac{\partial L}{\partial {{x}_{1}}}=-\frac{mg{{x}_{1}}}{l}+k({{x}_{2}}-{{x}_{1}}) \\\end{array}\)

\(\begin{array}{l}\frac{\partial L}{\partial {{{\dot{x}}}_{1}}}=m{{{\dot{x}}}_{1}} \\\end{array}\)

\(\begin{array}{l}\therefore \frac{d}{dt}\left( \frac{\partial L}{\partial {{{\dot{x}}}_{1}}} \right)=\frac{d}{dt}(m{{{\dot{x}}}_{1}})=m{{{\ddot{x}}}_{1}} \\\end{array}\)

Hence Lagrangian equation in terms of \({x_1}\)is

\(\begin{array}{l}\frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial {{\dot x}_1}}}} \right) - \frac{{\partial L}}{{\partial {x_1}}} = 0\\or,\\m{{\ddot x}_1} + \frac{{mg{x_1}}}{l} - k({x_2} - {x_1}) = 0\\or,\\m{{\ddot x}_1} =  - \frac{{mg{x_1}}}{l} + k({x_2} - {x_1})\end{array}\)

Also,

\(\begin{array}{l}\frac{{\partial L}}{{\partial {x_2}}} =  - \frac{{mg{x_2}}}{l} - k({x_2} - {x_1})\\\frac{{\partial L}}{{\partial {{\dot x}_2}}} = m{{\dot x}_2}\\and\\\frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial {{\dot x}_2}}}} \right) = m{{\ddot x}_2}\end{array}\)

Hence Lagrangian equation in terms of \({x_2}\)is

\(\begin{array}{l}\frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial {{\dot x}_2}}}} \right) - \frac{{\partial L}}{{\partial {x_2}}} = 0\\or,\\m{{\ddot x}_2} =  - \frac{{mg{x_2}}}{l} - k({x_2} - {x_1})\end{array}\)

The equation of motion for given system are

\(\begin{array}{l}m{{\ddot x}_1} =  - \frac{{mg{x_1}}}{l} + k({x_2} - {x_1})\\m{{\ddot x}_2} =  - \frac{{mg{x_2}}}{l} - k({x_2} - {x_1})\end{array}\)