Lagrange equation particle on an inverted cone

[bobred]

Homework Statement

Derive the equations of motion and show that the equation of motion for $\phi$ implies that $r^2\dot{\phi}=K$ where $K$ is a constant

Homework Equations

Using cylindrical coordinates and $z=\alpha r$
The kinetic and potential energies are
$$T=\dfrac{m}{2}\left[\left(1+\alpha^{2}\right)\dot{r}^{2}+r^{2}\dot{\phi}^{2}\right]$$ and
$$V=mg\alpha r$$
The Lagrangian is
$$L\left(r,\dot{r},\dot{\phi}\right)=\dfrac{m}{2}\left[\left(1+\alpha^{2}\right)\dot{r}^{2}+r^{2}\dot{\phi}^{2}\right]-mg\alpha r$$
The equation of motion is
$$\dfrac{d}{d t}\left(\dfrac{\partial L}{\partial\dot{\mathbf{q}_{k}}}\right)-\dfrac{\partial L}{\partial\mathbf{q}_{k}}=0$$

The Attempt at a Solution

The equation of motion for $r$
$$\ddot{r}+\alpha^{2}\ddot{r}-r\dot{\phi}^{2}+g\alpha =0$$
The equation of motion for $\phi$
$$r^{2}\dot{\phi}-r^{2}\ddot{\phi} =0$$
Is this correct? If so how does it imply that $r^2\dot{\phi}=K$ is a constant?
Thanks

 

[Ray Vickson]

Homework Statement

Derive the equations of motion and show that the equation of motion for $\phi$ implies that $r^2\dot{\phi}=K$ where $K$ is a constant

Homework Equations

Using cylindrical coordinates and $z=\alpha r$
The kinetic and potential energies are
$$T=\dfrac{m}{2}\left[\left(1+\alpha^{2}\right)\dot{r}^{2}+r^{2}\dot{\phi}^{2}\right]$$ and
$$V=mg\alpha r$$
The Lagrangian is
$$L\left(r,\dot{r},\dot{\phi}\right)=\dfrac{m}{2}\left[\left(1+\alpha^{2}\right)\dot{r}^{2}+r^{2}\dot{\phi}^{2}\right]-mg\alpha r$$
The equation of motion is
$$\dfrac{d}{d t}\left(\dfrac{\partial L}{\partial\dot{\mathbf{q}_{k}}}\right)-\dfrac{\partial L}{\partial\mathbf{q}_{k}}=0$$

The Attempt at a Solution

The equation of motion for $r$
$$\ddot{r}+\alpha^{2}\ddot{r}-r\dot{\phi}^{2}+g\alpha =0$$
The equation of motion for $\phi$
$$r^{2}\dot{\phi}-r^{2}\ddot{\phi} =0$$
Is this correct? If so how does it imply that $r^2\dot{\phi}=K$ is a constant?
Thanks

Your $\phi$-equation is incorrect:
$$\frac{\partial L}{\partial \dot{\phi}} = m r^2 \dot{\phi},$$
so
$$\frac{d}{dt} (m r^2 \dot{\phi}) = \frac{\partial L}{\partial \phi} = 0 .$$

 

[bobred]

Hi, got it thanks. Integrating the left hand side gives
$$r^2\dot{\phi}=K$$
Meaning I can eliminate $\dot{\phi}$ from the equation of motion for $r$