91Ó°ÊÓ

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

$$\begin{gathered} tan\mathrm{Î\pm }=\frac{r}{z_1} \\ {z}_1=\frac{r}{tan\mathrm{Î\pm }} \\ ≡\mathrm{β}r \end{gathered}$$

Define$\frac{1}{\mathrm{³Ù²¹²ÔÎ\pm }}≡\mathrm{β}$. Next, write a constraint that the string has a fixed length.

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

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

$$\begin{gathered} z_2=r\sqrt{1+{\mathrm{β}}^2}-1 \\ \stackrel{˙}{z_2}=\stackrel{˙}{r}\sqrt{1+{\mathrm{β}}^2} \end{gathered}$$

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:

$$\begin{gathered} V_1=m_{1}g{z}_1 \\ =\mathrm{β}{m}_{1}gr \end{gathered}$$

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

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

For particle $1$the potential energy reads

$$\begin{gathered} V_1=m_{1}g{z}_1 \\ =\mathrm{β}{m}_{1}gr \end{gathered}$$

And the potential energy for particle $2$is

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

Thus, finally the Lagrangian is

$$\begin{gathered} L=T_1+T_2-V_1-V_2 \\ =\frac{1}{2}{m}_1\left({\stackrel{˙}{r}}^2+r^2{\stackrel{˙}{\mathrm{θ}}}^2+{\mathrm{β}}^2{\stackrel{˙}{r}}^2\right)+\frac{1}{2}{m}_2{\stackrel{˙}{r}}^2\left(1+{\mathrm{β}}^2\right)-\mathrm{β}{m}_{1}gr-m_{2}g\left(r\sqrt{1+{\mathrm{β}}^2}-1\right) \end{gathered}$$

Calculate the value of $\mathrm{β}$for this given case when $\mathrm{Î\pm }=\frac{\mathrm{Ï€}}{3}$

$$\begin{gathered} \mathrm{β}=\frac{1}{\tan \mathrm{Î\pm }} \\ =\frac{1}{\mathrm{³Ù²¹²ÔÏ€}/3} \\ =\sqrt{3} \end{gathered}$$

The Lagrangian simplifies to:

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

Euler equations reads

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

Calculate the required derivatives.

Use all of the equations above,

$4\stackrel{¨}{r}\left(m_1+m_2\right)-m_{1}r{\stackrel{˙}{\mathrm{θ}}}^2+\sqrt{3}{m}_{1}g+2m_{2}g=0$

Next, move onto the degree of freedom.

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

Calculate the required derivatives.

$$\begin{gathered} \frac{∂L}{∂\stackrel{˙}{\mathrm{θ}}}=m_1{r}^2\stackrel{˙}{\mathrm{θ}} \\ \frac{∂L}{∂\mathrm{θ}}=0 \end{gathered}$$

from the Euler equation:

$$\begin{gathered} \frac{d}{dt}\left(m{r}^2\stackrel{˙}{\mathrm{θ}}\right)=0 \\ {m}_1{r}^2\stackrel{˙}{\mathrm{θ}}=constraint \end{gathered}$$

Therefore,the equation of the cone is $m_1{r}^2\stackrel{˙}{\mathrm{θ}}=constraint$.