91Ó°ÊÓ

The given value isthe potential energy of the spring is $\frac{1}{2}\mathrm{k}{\left(\mathrm{r}-{\mathrm{r}}_0\right)}^2$.

The Lagrange equations are used to construct the equations of motion of a solid mechanics issue in matrix form, including damping.

The kinetic energy in polar coordinates has the following form:

$T=\frac{1}{2}m\left({\stackrel{Ë™}{r}}^2+r^2{\stackrel{Ë™}{\mathrm{Ï•}}}^2\right)$

The potential energy has the form:

Therefore, the Lagrangian is:

Observe the Euler equation for degree of freedom. The Euler equations reads:

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

First, let's calculate the required derivatives.

Use all of the equations above, after diving by we obtain from the Euler equation:

The Euler equation for the $\mathrm{Ï•}$degree of freedom reads:

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

Calculate the required derivatives.

Therefore, combining these equations to obtain the final Euler equation of motion:

After diving by $\mathrm{m}$and $\mathrm{r}$,

Therefore,the Lagrangian is $\stackrel{¨}{\mathrm{r}}-r{\stackrel{˙}{\mathrm{ϕ}}}^2+\frac{k}{m}\left(r-r_0\right)-g\cos \mathrm{ϕ}=0$and the Lagrange equations for a simple pendulum is $2\stackrel{˙}{r}\stackrel{˙}{\mathrm{ϕ}}+r\stackrel{¨}{\mathrm{ϕ}}+g\sin \mathrm{ϕ}=0$.