Hamiltonian Mechanics - The Pendulum

Classical Mechanics Level 4

This is a guided exercise on Hamiltonian mechanics. One does NOT need any extra or advanced knowledge of mechanics to solve this problem.

Consider a pendulum as shown in the figure. The rigid uniform rod which is hinged at the origin is free to rotate about the Z-axis due to ambient gravity. The mass of the rod is m m and its length is L L . Perform the following steps:

Step - 1:

At any general configuration, compute the Lagrangian of this system. The Lagrangian is defined as the difference between the system's kinetic energy ( T T ) and potential energy ( V V ).

L = T − V \mathcal{L} = T - V

If the angular velocity of the rod at a general instant is θ ˙ \dot{\theta} , then compute:

p = ∂ L ∂ θ ˙ p = \frac{\partial \mathcal{L}}{\partial \dot{\theta}}

The quantity p p is known as the generalised momentum, which will result to be a function of θ ˙ \dot{\theta} . What does this quantity physically mean for this system?

Step - 2:

After computing the generalised momentum, compute the following quantity which is henceforth referred to as the Hamiltonian ( H H ):

H = p θ ˙ − L H = p\dot{\theta} -\mathcal{L}

Express the Hamiltonian purely as a function of p p and θ \theta . What does the Hamiltonian physically mean?

Step - 3:

Calculate the equations of motion as such:

θ ˙ = ∂ H ∂ p \dot{\theta} = \frac{\partial H}{\partial p} p ˙ = − ∂ H ∂ θ \dot{p} = -\frac{\partial H}{\partial \theta}

One should obtain a set of two ODEs.

Step - 4:

Finally, use the pair of resulting ODEs to solve for p 2 p^2 as a function of θ \theta . Consider that the pendulum is released from rest from an angular position θ o \theta_o . The resulting expression will look as such:

p 2 = a m b L c g d ( cos ⁡ θ − cos ⁡ θ o ) \boxed{p^2 = \frac{am^bL^cg}{d}\left(\cos{\theta} - \cos{\theta_o}\right)}

Here, a a , b b , c c and d d are positive integers and a a and d d are co-prime. Compute a b c d \boxed{abcd} .

Inspiration

Bonus: Interpret the final expression. Is there an easier way to derive it?


The answer is 18.

Karan Chatrath by Karan Chatrath , 27, Netherlands

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1 solution

Steven Chase
Mar 21, 2020

Lagrangian:

T = 1 2 I θ ˙ 2 = 1 2 1 3 m L 2 θ ˙ 2 = 1 6 m L 2 θ ˙ 2 V = − m g L 2 cos ⁡ θ L = T − V = 1 6 m L 2 θ ˙ 2 + 1 2 m g L cos ⁡ θ T = \frac{1}{2} I \dot{\theta}^2 = \frac{1}{2} \frac{1}{3} m L^2 \dot{\theta}^2 = \frac{1}{6} m L^2 \dot{\theta}^2 \\ V = - m g \frac{L}{2} \cos \theta \\ \mathcal{L} = T - V = \frac{1}{6} m L^2 \dot{\theta}^2 + \frac{1}{2} m g L \cos \theta

Generalized momentum. This ends up being equal to I θ ˙ I \dot{\theta} , which is the rotational analog of linear momentum.

p = ∂ L ∂ θ ˙ = 1 3 m L 2 θ ˙ = I θ ˙ θ ˙ = 3 p m L 2 p = \frac{\partial{\mathcal{L}}}{\partial{\dot{\theta}}} = \frac{1}{3} m L^2 \dot{\theta} = I \dot{\theta} \\ \dot{\theta} = \frac{3 p }{m L^2}

Hamiltonian. This ends up being equal to the total energy of the system (kinetic plus potential):

H = p θ ˙ − L = p θ ˙ − 1 6 m L 2 θ ˙ 2 − 1 2 m g L cos ⁡ θ = p 3 p m L 2 − 1 6 m L 2 9 p 2 m 2 L 4 − 1 2 m g L cos ⁡ θ = 3 p 2 m L 2 − 3 p 2 2 m L 2 − 1 2 m g L cos ⁡ θ = 3 p 2 2 m L 2 − 1 2 m g L cos ⁡ θ H = p \dot{\theta} - \mathcal{L} \\ = p \dot{\theta} - \frac{1}{6} m L^2 \dot{\theta}^2 - \frac{1}{2} m g L \cos \theta \\ = p \frac{3 p }{m L^2} - \frac{1}{6} m L^2 \frac{9 p^2 }{m^2 L^4} - \frac{1}{2} m g L \cos \theta \\ = \frac{3 p^2 }{m L^2} - \frac{3 p^2 }{2 m L^2} - \frac{1}{2} m g L \cos \theta \\ = \frac{3 p^2 }{2 m L^2} - \frac{1}{2} m g L \cos \theta

Equations of motion:

θ ˙ = ∂ H ∂ p = 3 p m L 2 p ˙ = − ∂ H ∂ θ = − 1 2 m g L sin ⁡ θ \dot{\theta} = \frac{\partial{H}}{\partial{p}} = \frac{3 p }{m L^2} \\ \dot{p} = -\frac{\partial{H}}{\partial{\theta}} = -\frac{1}{2} m g L \sin \theta

Re-arranging equations of motion:

d p d θ = p ˙ θ ˙ = − 1 2 m g L sin ⁡ θ m L 2 3 p p d p = − 1 6 m 2 L 3 g sin ⁡ θ d θ \frac{dp}{ d \theta} = \frac{\dot{p}}{\dot{\theta}} = -\frac{1}{2} m g L \sin \theta \frac{m L^2}{3 p} \\ p \ dp = -\frac{1}{6} m^2 L^3 g \sin \theta \, d \theta

Integrating both sides:

∫ 0 p p d p = − 1 6 m 2 L 3 g ∫ θ 0 θ sin ⁡ θ d θ p 2 2 = − 1 6 m 2 L 3 g ( − cos ⁡ θ + cos ⁡ θ 0 ) p 2 = 1 3 m 2 L 3 g ( cos ⁡ θ − cos ⁡ θ 0 ) = I 2 θ ˙ 2 \int_0^p p \, dp = -\frac{1}{6} m^2 L^3 g \int_{\theta_0}^{\theta} \sin \theta \, d\theta \\ \frac{p^2}{2} = -\frac{1}{6} m^2 L^3 g \Big( -\cos \theta + \cos \theta_0 \Big) \\ p^2 = \frac{1}{3} m^2 L^3 g \Big( \cos \theta - \cos \theta_0 \Big) = I^2 \dot{\theta}^2

The end result is essentially an energy conservation expression. The pendulum trades angle (and thus potential energy) for speed as it moves.

4 Helpful 3 Interesting 3 Brilliant 1 Confused

Thanks for the solution. The document you shared earlier is quite insightful. Also, an observation. In your solution, the Lagrangian and length have the same symbol. Perhaps you could could the symbol that I did '\mathcal{L}'.

Karan Chatrath - 1 year, 2 months ago

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Good catch. I have changed it accordingly. I like how the book actively explains things, and tries to address both questions and misconceptions from students before they arise.

Steven Chase - 1 year, 2 months ago

I have a question. Why when he derive the Generalized momentum don´t appear theta¨ (alpha)

Alejandro Guevara - 1 year, 2 months ago

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This is because the differentiation is not performed with respect to time. The notation is that of a partial derivative.

Karan Chatrath - 1 year, 2 months ago

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Oh, thank you. I see my error.

Alejandro Guevara - 1 year, 2 months ago

Why is the potential energy V equal to -mgLcos(θ)/2 instead of mg(L-Lcos(θ))?

Michael Wang - 1 year ago

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