Total energy of earth in its circular orbit around the sun

Question :
Find out the total energy of earth in its circular orbit around the sun in terms of gravitational constant
Answer:
Let R be the total distance between the earth and the sun. If \({M_e}\) and \({M_s}\) are the mass of earth and sun respectively, the gravitational force of motion of earth and sun is given by

\(F =  - \frac{{G{M_e}{M_s}}}{{{R^2}}}\)

where G is the gravitational constant. Since the centripetal force balances the gravitational force of attraction, we have

\({F_c} = |{F_G}|\),

where

\({F_c} = \frac{{{M_e}{v^2}}}{R}\)

v being the velocity with which earth is moving. Hence we have

\(\frac{{{M_e}{v^2}}}{R} = \frac{{G{M_e}{M_s}}}{{{R^2}}}\)

or

\({M_e}{v^2} = \frac{{G{M_e}{M_s}}}{R}\)

Therefore kinetic energy of earth in motion is

\(T = \frac{1}{2}{M_e}{v^2} = \frac{1}{2}\frac{{G{M_e}{M_s}}}{R}\)

As we know that , force in terms of potential energy is

\({F_G} = - \frac{{\partial V}}{{\partial R}}\)
\(V = - \int {{F_G}dR} = \int {\frac{{G{M_e}{M_s}}}{{{R^2}}}dR = - } \frac{{G{M_e}{M_s}}}{R}\)

Now total energy of the earth in the orbit around the sun is

\(E = T + V\)
\(E = \frac{1}{2}\frac{{G{M_e}{M_s}}}{R} - \frac{{G{M_e}{M_s}}}{R}\)

\(E = - \frac{1}{2}\frac{{G{M_e}{M_s}}}{R}\)

This is the required expression.

Brachistochrone Problem

Question

If a particle falls from rest under the influence of gravity from higher to lower point in the minimum time, what is the curve that the particle will follow?

Solution

Suppose v is the speed of the particle along the curve, then in traversing ds portion of the curve time spent would be \(\frac{{ds}}{v}\) so that total time taken by the particle in moving from highest point 1 to lowest point 2 will be

\[{t_{12}} = \int\limits_1^2 {\frac{{ds}}{v}} \]

Suppose vertical distance of fall upto point 2 be x, then from the principle of conservation of energy of the particle we find that

\(\begin{array}{l}\frac{1}{2}m{v^2} = mgx\\or\\v = \sqrt {2gx} \\then\\{t_{12}} = \int\limits_1^2 {\frac{{\sqrt {d{x^2} + d{y^2}} }}{{\sqrt {2gx} }}} dx\\{\rm{      = }}\int\limits_1^2 {\frac{{\sqrt {1 + {{\dot y}^2}} }}{{\sqrt {2gx} }}} dx\\{\rm{      = }}\int\limits_1^2 {fdx} \end{array}\)

Where,

\( f=\left ( \frac{1+\dot{y}^{2}}{2gx} \right )^{\frac{1}{2}} \)

For \({t_{12}}\) to be minimum equation

\(\frac{d}{{dx}}\left( {\frac{{\partial f}}{{\partial \dot y}}} \right) - \frac{{\partial f}}{{\partial y}} = 0\)

must be satisfied. From expression for \(f\) we find that

\(\begin{array}{l}\frac{{\partial f}}{{\partial y}} = 0\\\frac{{\partial f}}{{\partial \dot y}} = \frac{{\dot y}}{{\sqrt {2gx} \sqrt {1 + {{\dot y}^2}} }}\\\frac{d}{{dx}}\left( {\frac{{\dot y}}{{\sqrt {2gx} \sqrt {1 + {{\dot y}^2}} }}} \right) = 0\\or\\\frac{{{{\dot y}^2}}}{{2gx(1 + {{\dot y}^2})}} = c'\end{array}\)

Where c’ is the constant of integration.

Since c’ is a constant we can also write

\(\frac{{{{\dot y}^2}}}{{x(1 + {{\dot y}^2})}} = c\)

where c is also a constant. On integrating above equation we find

\(\begin{array}{l}\frac{{{{\dot y}^2}}}{c} = x(1 + {{\dot y}^2})\\or,\\{{\dot y}^2}\left( {\frac{1}{c} - x} \right) = x\\{{\dot y}^2}\left( {\frac{x}{c} - {x^2}} \right) = {x^2}\\\dot y = \frac{x}{{\sqrt {\frac{x}{c} - {x^2}} }}\end{array}\)

Putting \(\frac{1}{c} = 2a\) , and on integration we get

\(\int dy=\int \frac{x}{\sqrt{2ax-x^{2}}}dx\)

\( y=acos^{-1}(1-\frac{x}{a})-(2ax-x^{2})^{1/2}+c''\)

where c'' is new constant of integration.
In case c'' is zero then y will be zero for x=0.

As such the equation

\( y=acos^{-1}(1-\frac{x}{a})-(2ax-x^{2})^{1/2}\)

And this y represents an inverted cycloid with its base along y axis and cusp at the origin and is the curve that particle will follow.

Double pendulum

Question 

In case of a double pendulum find the expression for the kinetic energy of the system

Solution

We take a simple case where lengths and masses are same.see below in the figure

Here on being displaced the co-ordinates of two pendulums are

\(\begin{array}{l}{x_1} = l\sin {\theta _1}\\{y_1} = l\cos {\theta _1}\end{array}\)

For the first pendulum where \({\theta _1}\) is the angle through which the first pendulum have been displaced.

For second pendulum

\(\begin{array}{l}{x_2} = l\sin {\theta _1} + l\sin {\theta _2}\\{y_2} = l\cos {\theta _1} + l\cos {\theta _2}\end{array}\)

Where \({\theta _2}\) is the angle through which second pendulum has been displaced.

The total kinetic energy of the system is given by the expression

\(T = \frac{1}{2}m(\dot x_1^2 + \dot y_1^2) + \frac{1}{2}m(\dot x_2^2 + \dot y_2^2)\)

Now

And

Which is the required result

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

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