Question

A mass $\mathbf{m}$moves without friction on the surface of the cone $\mathit{r}\mathbf{=}\mathit{z}$ under gravity acting in the negative $\mathbf{z}$ direction. Here localid="1664345422471" $\mathit{r}$ is the cylindrical coordinate $\mathit{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}$. Find the Lagrangianand Lagrange's equations in terms of $\mathit{r}$and $\mathbf{\theta }$ (that is, eliminate $\mathbf{z}$).

Short answer

TheLagrangian is $L=\frac{1}{2}m\left(2{\dot{r}}^2+r^2{\dot{\theta }}^2\right)=mgr$and the Lagrange equation is $m{r}^2\dot{\theta }=constrain$.

Step-by-step solution

Given Information.

The given cylindrical coordinate equation is $r=\sqrt{x^2+y^{2}t}$

Step 2: Meaning of the Lagrange equations.

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

Find the Lagrangian.

First, the kinetic energy becomes, after using the latter constraint

$$\begin{gathered} T=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2\right) \\ =\frac{1}{2}m\left(2{\dot{r}}^2+r^2{\dot{\theta }}^2\right) \end{gathered}$$

Use that fact

role="math" localid="1664347290750" $$\begin{gathered} r=z \\ \dot{r}=\dot{z} \end{gathered}$$

Now, the potential energy is due to gravity, so it's equal to

$$\begin{gathered} V=mgz \\ =mgr \end{gathered}$$

Combining the kinetic and potential energy, the Lagrangian is

$$\begin{gathered} L=T-V \\ =\frac{1}{2}m\left(2{\dot{r}}^2+r^2{\dot{\theta }}^2\right)-mgr \end{gathered}$$

Find the Lagrange's equations.

First observe the Euler equation for $\mathrm{r}$degree of freedom. The Euler equations reads

$\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}-\frac{\partial L}{\partial r}=0$

Calculate the required derivatives

$$\begin{gathered} \frac{\partial L}{\partial \dot{r}}=2m\dot{r} \\ \frac{d}{dt}\frac{\partial L}{\partial \dot{r}}=2m\ddot{r} \\ \frac{\partial L}{\partial r}=mr{\dot{\theta }}^2-mg \end{gathered}$$

Use all of the equations above and dividing by $\mathrm{m}$we obtain from the Euler equation:

$2\ddot{r}-r{\dot{\theta }}^2+g=0$

Next, let's move onto the $\mathrm{\theta }$degree of freedom. The Euler equations reads:

$\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta }=0$

First, let's calculate the required derivatives.

$$\begin{gathered} \frac{\partial L}{\partial \dot{\theta }}=m{r}^2\dot{\theta } \\ \frac{\partial L}{\partial \theta }=0 \end{gathered}$$

from the Euler equation:

$$\begin{gathered} \frac{d}{dt}\left(m{r}^2\dot{\theta }\right)=0 \\ m{r}^2\dot{\theta }=constraint \end{gathered}$$

Therefore,the Lagrangian is $L=\frac{1}{2}m\left(2{\dot{r}}^2+r^2{\dot{\theta }}^2\right)-mgr$and the Lagrange equation is $m{r}^2\dot{\theta }=constraint$.