Lagrange equation particle on an inverted cone

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SUMMARY

The discussion focuses on deriving the equations of motion for a particle on an inverted cone using Lagrangian mechanics. The kinetic energy is defined as T = m/2[(1 + α²)ṙ² + r²φ̇²] and the potential energy as V = mgαr. The Lagrangian is L(r, ṙ, φ̇) = T - V. The equation of motion for φ is shown to imply that r²φ̇ = K, where K is a constant, confirming the conservation of angular momentum in this system.

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bobred
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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
 
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bobred said:

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
$$\frac{d}{dt} (m r^2 \dot{\phi}) = \frac{\partial L}{\partial \phi} = 0 .$$
 
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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
 

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