# Lagrangian for a particle in a bowl with parabolic curvature

## 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.

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tnich
Homework Helper

## 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
Gold Member
$\rho$ is a constant....

TSny
Homework Helper
Gold Member
A particle of mass ##m## moves without slipping inside a bowl
How can the particle move without slipping? I can't visualize that.

tnich
Homework Helper
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
Homework Helper
Gold Member
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.

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

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
Homework Helper
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.

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
Homework Helper
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}##.