# Hamiltonian of a pendulum constrained to move on a parabola

• Wavefunction
In summary: I actually just had that thought and doing it that would also mean that I end up 2 canonical coordinates which is exactly what I want. Thank you for your help you are a life saver!...or at least a math saver :)
Wavefunction

## Homework Statement

The point of suspension of a simple pendulum of length $l$ and mass $m$ is constrained to move on
a parabola $z = ax^2$ in the vertical plane. Derive a Hamiltonian governing the motion of the
pendulum and its point of suspension. This is a two-dimensional problem.

## Homework Equations

eq(1) $H=T+U$

## The Attempt at a Solution

$T=\frac{m}{2}[\dot{x}^2+\dot{z}^2]$

$U=mgz$

eq(2) $H = \frac{m}{2}[\dot{x}^2+\dot{z}^2]+mgz$

Applying the constraint $z=ax^2$, $\dot{z} = 2ax\dot{x}$

$H = \frac{m}{2}[\dot{x}^2+4a^2x^2\dot{x}^2]+mgax^2$

The resulting Hamilton's equations I got were:

eq(3) $\dot{p_x} = -(4ma^2x\dot{x}^2+2mgax) = -\frac{∂H}{∂x}$

eq(4) $\dot{x} = \frac{p_x}{m(1+4a^2x^2)} = \frac{∂H}{∂p_x}$

and $p_x = \frac{∂L}{∂\dot{x}} = m(1+4a^2x^2)\dot{x}$

This problem seemed uncharacteristically easy, so I definitely appreciate anybody who checks it over (: Thank you in advance!

The constraint you implemented is for the point of suspension for the pendulum. The pendulum bob need not move on that parabola.

What you have there is the solution to a point mass constrained to move on a parabola, when instead you want to solve for a pendulum whose point of support is on a parabola.

1 person
Wavefunction said:

## Homework Statement

The point of suspension of a simple pendulum of length $l$ and mass $m$ is constrained to move on
a parabola $z = ax^2$ in the vertical plane. Derive a Hamiltonian governing the motion of the
pendulum and its point of suspension. This is a two-dimensional problem.

## Homework Equations

eq(1) $H=T+U$

## The Attempt at a Solution

$T=\frac{m}{2}[\dot{x}^2+\dot{z}^2]$

$U=mgz$

let $x= lsin(θ)$ and $z = lcos(θ)$

eq(2) $H = \frac{m}{2}[\dot{x}^2+\dot{z}^2]+mgz$

$H = \frac{m}{2}[l^2\dot{θ}^2]+mglcos(θ)$

Applying the constraint $z=ax^2 → lcos(θ) = al^2sin^2(θ)$

$H = \frac{m}{2}[l^2\dot{θ}^2]+mgal^2sin^2(θ)$

Okay how does it look now?

You've still just applied the constraint to the bob itself, and not the point of support...all you've done is make a change of variables.

You need a set of coordinates for the pendulum support, and then a set of coordinates for the bob. There should be two constraints, one for where the support may go, and one for where the bob may go since the pendulum is fixed to the support.

For example, I should have the point of support at coordinates:

$$(x_s,z_s)$$

The bob would then have coordinates:

$$(x_b,z_b)$$

I should express the bob coordinates in terms of the support coordinates. And then implement the constraint on my support coordinates!

Matterwave said:
You've still just applied the constraint to the bob itself, and not the point of support...all you've done is make a change of variables.

You need a set of coordinates for the pendulum support, and then a set of coordinates for the bob. There should be two constraints, one for where the support may go, and one for where the bob may go since the pendulum is fixed to the support.

For example, I should have the point of support at coordinates:

$$(x_s,z_s)$$

The bob would then have coordinates:

$$(x_b,z_b)$$

I should express the bob coordinates in terms of the support coordinates. And then implement the constraint on my support coordinates!

then constraint I should apply should be $(x_s-x_b)^2+(z_s-z_b)^2=l^2$ ?

Yes, but there's a much simpler way of applying that constraint, which you sort of did from your previous attempt, by using lcos(theta) and lsin(theta)

Matterwave said:
Yes, but there's a much simpler way of applying that constraint, which you sort of did from your previous attempt, by using lcos(theta) and lsin(theta)

Okay I'm going to try to resolve it using this new information. Thanks for pointing that out by the way.(:

Okay so far I think I've managed to express the bob coordinates in terms of the support coordinates

$x_b=x_s+x$ and $z_b = z_s+z$ such that $x^2+z^2 = l^2$ but I'm not exactly sure whether this is helpful.

That would work except the algebra will get very annoying very quickly. Try this:

$$x_b=x_s+l\sin (\theta)$$

and

$$z_b=z_s-l\cos (\theta)$$

Draw a picture to see how these coordinates work.

Notice that your constraint is automatically satisfied with these coordinates.

1 person
Matterwave said:
That would work except the algebra will get very annoying very quickly. Try this:

$$x_b=x_s+l\sin (\theta)$$

and

$$z_b=z_s-l\cos (\theta)$$

Draw a picture to see how these coordinates work.

Notice that your constraint is automatically satisfied with these coordinates.

I actually just had that thought and doing it that would also mean that I end up 2 canonical coordinates which is exactly what I want. Thank you for your help you are a life saver! (:

## 1. What is the Hamiltonian of a pendulum constrained to move on a parabola?

The Hamiltonian of a pendulum constrained to move on a parabola is a mathematical function that describes the energy of the system, taking into account both the kinetic and potential energy. It is given by the equation H = T + V, where T represents the kinetic energy and V represents the potential energy.

## 2. How is the Hamiltonian of a pendulum on a parabola different from a regular pendulum?

The Hamiltonian of a pendulum on a parabola takes into account the restriction on the pendulum's movement, which is constrained to the shape of a parabola. This means that the potential energy component of the Hamiltonian is different, as it is dependent on the parabolic shape rather than just the height of the pendulum.

## 3. Can the Hamiltonian of a pendulum on a parabola be derived from the Lagrangian?

Yes, the Hamiltonian of a pendulum on a parabola can be derived from the Lagrangian using the Hamiltonian formalism. This involves taking the partial derivatives of the Lagrangian with respect to the position and momentum variables, and using them to construct the Hamiltonian function.

## 4. How does the Hamiltonian of a pendulum on a parabola affect the motion of the pendulum?

The Hamiltonian of a pendulum on a parabola determines the equations of motion for the system, which describe how the pendulum will move over time. The specific shape of the parabola will affect the potential energy component of the Hamiltonian, which in turn will affect the trajectory of the pendulum.

## 5. Can the Hamiltonian of a pendulum on a parabola be used to predict the behavior of the system?

Yes, the Hamiltonian of a pendulum on a parabola can be used to predict the behavior of the system. By solving the equations of motion derived from the Hamiltonian, we can determine the position and velocity of the pendulum at any point in time. This allows us to make predictions about the motion of the pendulum on the parabolic surface.

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