91Ó°ÊÓ

An ordinary first-order differential equation that is linear in the independent variable and unknown function but not solved for the derivative is termed as Lagrange's equation.

A function that equals the difference between potential and kinetic energy and characterises the state of a dynamic system in terms of position coordinates and their time derivatives is termed as Lagrangian.

Given a simple pendulum

The equilibrium position is denoted by polar coordinates with $\mathrm{Ï•}=0$.

Now it is aconstraint that the length is fixed. This means that $r=l$where$l$is the length of the pendulum string and thus localid="1664362325356" $\stackrel{Ë™}{r}=0$

Then the kinetic energy will become:

And the potential energy will have the form:

$$\begin{gathered} V=mgh \\ =mgl\left(1-\cos \mathrm{Ï•}\right) \end{gathered}$$

According to the definition of the Lagrangian,

$$\begin{gathered} L=T-V \\ =\frac{1}{2}m{l}^2{\stackrel{Ë™}{\mathrm{Ï•}}}^2-mgl\left(1-\cos \mathrm{Ï•}\right) \end{gathered}$$

So, the lagrangian is$L=\frac{1}{2}m{l}^2{\stackrel{Ë™}{\mathrm{Ï•}}}^2-mgl\left(1-\cos \mathrm{Ï•}\right)$

Since the lagrangian has only one degrees of freedom

The $\mathrm{Ï•}$degree of freedom, the Euler equation will be given by:

$\frac{d}{dt}\frac{∂L}{∂\stackrel{˙}{\mathrm{ϕ}}}-\frac{∂L}{∂\mathrm{ϕ}}=0$

Then the required derivative will be:

Then the Euler equation will be:

Divide the above Euler equation by $\mathrm{l}$, the result will be

$\stackrel{¨}{\mathrm{ϕ}}+\frac{g}{l}\sin \mathrm{ϕ}=0$

So, the Lagrange’s equation is$\stackrel{¨}{\mathrm{ϕ}}+\frac{g}{l}\sin \mathrm{ϕ}=0.$