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

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