Lagrangian for a particle in a bowl with parabolic curvature

  • #1
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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.
 

Answers and Replies

  • #2
tnich
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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.
 
  • #3
Dr Transport
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[itex] \rho [/itex] is a constant....
 
  • #4
TSny
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A particle of mass ##m## moves without slipping inside a bowl
How can the particle move without slipping? I can't visualize that.
 
  • #5
tnich
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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.
 
  • #6
TSny
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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.
 
  • #7
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Maybe the problem meant to state that the particle moves without friction rather than without slipping.
Yes, I meant without friction.
 
  • #8
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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?
 
  • #9
tnich
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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.
 
  • #10
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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?
 
  • #11
tnich
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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}##.
 

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