Do Example $\mathbf{3}$above, using cylindrical coordinates for ${\mathbf{m}}_{\mathbf{1}}$. Hint: Use ${\mathbf{z}}_{\mathbf{1}}$ and ${\mathbf{z}}_{\mathbf{2}}$ for the $\mathbf{z}$ coordinates of ${\mathbf{m}}_{\mathbf{1}}$ and ${\mathbf{m}}_{\mathbf{2}}$. What is the equation of the conein terms of $\mathbf{r}$ and ${\mathbf{z}}_{\mathbf{1}}$ ? Note that $\mathbf{r}\mathbf{\ne }\mathbf{p}$, and $\mathbf{\theta }$ in cylindrical coordinates is not the same as in spherical coordinates (see Chapter $\mathbf{5}$ , Figures 4.4 and4.5).

The given cylindrical coordinates are${\mathrm{z}}_1,{\mathrm{z}}_2{\mathrm{m}}_1$and${\mathrm{m}}_2$.

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

From trigonometry we find

$$\begin{gathered} tan\alpha =\frac{r}{z_1} \\ {z}_1=\frac{r}{tan\alpha } \\ \equiv \beta r \end{gathered}$$

Define$\frac{1}{\mathrm{tan\alpha }}\equiv \beta$. Next, write a constraint that the string has a fixed length.

$$\begin{gathered} l=\left|z_2\right|+\sqrt{z_1^2}+r^2 \\ =\left|z_2\right|+\sqrt{r^2+{\beta }^2{r}^2} \\ =-z_2+r\sqrt{1+{\beta }^2} \end{gathered}$$

Since the particle of mass${\mathrm{m}}_2$cannot get above the point$\mathrm{z}=0$

$$\begin{gathered} z_2=r\sqrt{1+{\beta }^2}-1 \\ \dot{z_2}=\dot{r}\sqrt{1+{\beta }^2} \end{gathered}$$

Now we are ready to write the kinetic energy for particles$1$and $2$. The kinetic energy for particle$1$is:

The kinetic energy for particle$2$is simply given by:

For particle $1$the potential energy reads:

$$\begin{gathered} V_1=m_{1}g{z}_1 \\ =\beta m_{1}gr \end{gathered}$$

And the potential energy for particle ${\mathrm{m}}_2$is

$$\begin{gathered} V_2=m_{2}g{z}_2 \\ =m_{2}g\left(r\sqrt{1+{\beta }^2}-1\right) \end{gathered}$$

For particle $1$the potential energy reads

$$\begin{gathered} V_1=m_{1}g{z}_1 \\ =\beta m_{1}gr \end{gathered}$$

And the potential energy for particle $2$is

$$\begin{gathered} V_2=m_{2}g{z}_2 \\ =m_{2}g\left(r\sqrt{1+{\beta }^2}-1\right) \end{gathered}$$

Thus, finally the Lagrangian is

$$\begin{gathered} L=T_1+T_2-V_1-V_2 \\ =\frac{1}{2}{m}_1\left({\dot{r}}^2+r^2{\dot{\theta }}^2+{\beta }^2{\dot{r}}^2\right)+\frac{1}{2}{m}_2{\dot{r}}^2\left(1+{\beta }^2\right)-\beta m_{1}gr-m_{2}g\left(r\sqrt{1+{\beta }^2}-1\right) \end{gathered}$$

Calculate the value of $\mathrm{\beta }$for this given case when $\alpha =\frac{\mathrm{\pi }}{3}$

$$\begin{gathered} \beta =\frac{1}{\tan \alpha } \\ =\frac{1}{\mathrm{tan\pi }/3} \\ =\sqrt{3} \end{gathered}$$

The Lagrangian simplifies to:

$L=\frac{1}{2}{m}_1\left(4{\dot{r}}^2+r^2{\dot{\theta }}^2\right)+2m_2{\dot{r}}^2-\sqrt{3}{m}_{1}gr-m_{2}g\left(2r-1\right)$

Euler equations reads

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

Calculate the required derivatives.

Use all of the equations above,

$4\ddot{r}\left(m_1+m_2\right)-m_{1}r{\dot{\theta }}^2+\sqrt{3}{m}_{1}g+2m_{2}g=0$

Next, move onto the degree of freedom.

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

Calculate the required derivatives.

$$\begin{gathered} \frac{\partial L}{\partial \dot{\theta }}=m_1{r}^2\dot{\theta } \\ \frac{\partial L}{\partial \theta }=0 \end{gathered}$$

from the Euler equation:

$$\begin{gathered} \frac{d}{dt}\left(m{r}^2\dot{\theta }\right)=0 \\ {m}_1{r}^2\dot{\theta }=constraint \end{gathered}$$

Therefore,the equation of the cone is $m_1{r}^2\dot{\theta }=constraint$.