Lagrangian for a particle in a bowl with parabolic curvature

[vbrasic]

Homework Statement

A particle of mass $m$ moves without slipping inside a bowl generated by the paraboloid of revolution $z=b\rho^2,$ where $b$ is a positive constant. Write the Lagrangian and Euler-Lagrange equation for this system.

Homework Equations

$$\frac{\partial\mathcal{L}}{\partial q_i}-\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{q_i}}$$

The Attempt at a Solution

Potential energy is given by, $$V=mgz=mgb\rho^2.$$ Kinetic energy in cylindrical coordinates is given by, $$T=\dot{\rho}^2+\rho^2\dot{\theta}^2+\dot{z}^2=\dot{\rho}^2+\rho^2\dot{\theta}^2+4b^2\rho^2\dot{\rho}^2.$$ Then the Lagrangian is, $$\mathcal{L}=\dot{\rho}^2+\rho^2\dot{\theta}^2+4b^2\rho^2\dot{\rho}^2-mgb\rho^2.$$ Then, we can write the Euler-Lagrange equations for both $\theta$ and $\rho$. The $\theta$ equation is very simple. We have, $$\frac{\partial\mathcal{L}}{\partial\theta}=0,$$ and $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\theta}}=\frac{d}{dt}\bigg(2\rho^2\dot{\theta}\bigg)=4\rho\dot{\rho}\dot{\theta}+2\rho^2\ddot{\theta}.$$ So the Euler-Lagrange equation reads, $$4\rho\dot{\rho}\dot{\theta}+2\rho^2\ddot{\theta}=0.$$ I'm getting messy with my derivatives in the $\rho$ equations. Any help would be appreciated.

 

[tnich]

Homework Statement

A particle of mass $m$ moves without slipping inside a bowl generated by the paraboloid of revolution $z=b\rho^2,$ where $b$ is a positive constant. Write the Lagrangian and Euler-Lagrange equation for this system.

Homework Equations

$$\frac{\partial\mathcal{L}}{\partial q_i}-\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{q_i}}$$

The Attempt at a Solution

Potential energy is given by, $$V=mgz=mgb\rho^2.$$ Kinetic energy in cylindrical coordinates is given by, $$T=\dot{\rho}^2+\rho^2\dot{\theta}^2+\dot{z}^2=\dot{\rho}^2+\rho^2\dot{\theta}^2+4b^2\rho^2\dot{\rho}^2.$$ Then the Lagrangian is, $$\mathcal{L}=\dot{\rho}^2+\rho^2\dot{\theta}^2+4b^2\rho^2\dot{\rho}^2-mgb\rho^2.$$ Then, we can write the Euler-Lagrange equations for both $\theta$ and $\rho$. The $\theta$ equation is very simple. We have, $$\frac{\partial\mathcal{L}}{\partial\theta}=0,$$ and $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\theta}}=\frac{d}{dt}\bigg(2\rho^2\dot{\theta}\bigg)=4\rho\dot{\rho}\dot{\theta}+2\rho^2\ddot{\theta}.$$ So the Euler-Lagrange equation reads, $$4\rho\dot{\rho}\dot{\theta}+2\rho^2\ddot{\theta}=0.$$ I'm getting messy with my derivatives in the $\rho$ equations. Any help would be appreciated.

Your work looks correct so far except for a factor of $\frac 1 2 m$ missing from T and its partials. That doesn't matter in your partials with respect to $θ$ and $\dot θ$, but it will in the partials with respect to $ρ$ and $\dot ρ$. If you would post your results so far for partials with respect to $ρ$ and $\dot ρ$, maybe we could point out where you are going wrong.

 

[Dr Transport]

\rho is a constant...

 

[TSny]

A particle of mass $m$ moves without slipping inside a bowl

How can the particle move without slipping? I can't visualize that.

 

[tnich]

How can the particle move without slipping? I can't visualize that.

I think the idea is that a ball is rolling in a bowl without slipping.

 

[TSny]

If it's a ball, then it will be a really difficult problem. There would be the additional kinetic energy of rotation of the ball. Also, a (non-holonomic) constraint condition would need to be included that corresponds to the rolling without slipping.

Maybe the problem meant to state that the particle moves without friction rather than without slipping.

 

[vbrasic]

Maybe the problem meant to state that the particle moves without friction rather than without slipping.

Yes, I meant without friction.

 

[vbrasic]

The Lagrangian (with the m/2 factor added), is, $$\mathcal{L}=\frac{1}{2}m(\dot{\rho}^2+\rho^2\dot{\theta}^2+4b^2\rho^2\dot{\rho}^2)-mgb\rho^2.$$ So, $$\frac{\partial\mathcal{L}}{\partial\rho}=m\rho\dot{\theta}^2+4mb^2\rho\dot{\rho^2}-2mgb\rho,$$ and, $$\frac{\partial\mathcal{L}}{\partial\dot{\rho}}=m\dot{\rho}+4mb\rho^2\dot{\rho}.$$ Taking the time derivative, we have, $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\rho}}=m\ddot{\rho}+4mb(2\rho\dot{\rho}^2+\rho^2\ddot{\rho}).$$ Do my derivatives look alright?

 

[tnich]

Your $b$ should be $b^2$ in these two equations:

$$\frac{\partial\mathcal{L}}{\partial\dot{\rho}}=m\dot{\rho}+4mb\rho^2\dot{\rho}.$$
$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\rho}}=m\ddot{\rho}+4mb(2\rho\dot{\rho}^2+\rho^2\ddot{\rho}).$$

Otherwise, they look right.

 

[vbrasic]

Your $b$ should be $b^2$ in these two equations:

Otherwise, they look right.

As a final question. Supposing that $\rho$ is constant, we have that $\dot{\rho}=\ddot{\rho}=0,$ so the Euler-Lagrange equation for $\rho$ reads $$0\frac{mp\dot{\theta}^2-2mgb\rho}{(m+4mb^2\rho^2)}\to\dot{\theta}=\sqrt{2gb}.$$ Does it physically make sense that change in angular velocity has no dependence in radius if the path is circular with constant radius?

 

[tnich]

As a final question. Supposing that $\rho$ is constant, we have that $\dot{\rho}=\ddot{\rho}=0,$ so the Euler-Lagrange equation for $\rho$ reads $$0\frac{mp\dot{\theta}^2-2mgb\rho}{(m+4mb^2\rho^2)}\to\dot{\theta}=\sqrt{2gb}.$$ Does it physically make sense that change in angular velocity has no dependence in radius if the path is circular with constant radius?

I think it does. Assuming that $\dot ρ = 0$, the components of radial acceleration and gravitational acceleration tangent to the paraboloid surface must sum to zero. Try setting up that equation and you will get the result $\dot{\theta}=\sqrt{2gb}$.