Undergrad Lagrangian for pendulum

[wnvl2]

I want to calculate the equation of motion of a planar pendulum of length l with a mass m at its end and a pivot point that moves uniformly along a vertical circular path (radius a) with a constant frequency ω.

The Lagrangian and the equation of motion for a planar pendulum with a moving pivot point can be derived using the Lagrange formulation.

- Length of the pendulum: $l$
- Mass of the pendulum bob: $m$
- Angle of the pendulum from the vertical: $\theta$
- The pivot point moves along a vertical circular path with a radius $a$ and a constant angular frequency $\omega$.
- Therefore, the coordinates of the moving pivot point can be described as:
$$
(x_0, y_0) = (a \sin(\omega t), -a \cos(\omega t))
$$

The total kinetic energy $T$ of the system is the sum of the kinetic energy of the pendulum bob and the motion of the moving pivot. The position of the bob in terms of the angle $\theta$ and the moving pivot is given by:
$$
x = x_0 + l \sin(\theta) = a \sin(\omega t) + l \sin(\theta)
$$
$$
y = y_0 - l \cos(\theta) = -a \cos(\omega t) - l \cos(\theta)
$$

The velocity of the pendulum bob is the time derivative of these coordinates:
$$
\dot{x} = \frac{d}{dt}(a \sin(\omega t) + l \sin(\theta)) = a \omega \cos(\omega t) + l \dot{\theta} \cos(\theta)
$$
$$
\dot{y} = \frac{d}{dt}(-a \cos(\omega t) - l \cos(\theta)) = a \omega \sin(\omega t) + l \dot{\theta} \sin(\theta)
$$

The kinetic energy of the pendulum bob is:
$$
T = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2)
$$

The potential energy $U$ of the system is due to gravity acting on the pendulum bob. The height $y$ of the bob (with the pivot at $y_0 = -a \cos(\omega t)$) is:
$$
y = -a \cos(\omega t) - l \cos(\theta)
$$
The gravitational potential energy is:
$$
U = m g y = -m g \left(a \cos(\omega t) + l \cos(\theta)\right)
$$

The Lagrangian $L$ is the difference between the kinetic and potential energies:
$$
L = T - U
$$

No problem, but to derive the equation of motion, I have to apply the Euler-Lagrange equation:
$$
\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = ???
$$

I don't know what I should put in place of the ??? I assume it is not zero as there is a driving force. How can I calculate that driving force?

 

[anuttarasammyak]

The Euler-Lagrange equation of constrainsts would work. The constraints would be circular motion of pivot with constant omega. But the constraint seems incorporated in your Lagrangian. How about trying RHS=0 and investigate the result whether it is reasonable or not ?

 

[PeroK]

I want to calculate the equation of motion of a planar pendulum of length l with a mass m at its end and a pivot point that moves uniformly along a vertical circular path (radius a) with a constant frequency ω.

The Lagrangian and the equation of motion for a planar pendulum with a moving pivot point can be derived using the Lagrange formulation.

- Length of the pendulum: $l$
- Mass of the pendulum bob: $m$
- Angle of the pendulum from the vertical: $\theta$
- The pivot point moves along a vertical circular path with a radius $a$ and a constant angular frequency $\omega$.
- Therefore, the coordinates of the moving pivot point can be described as:
$$
(x_0, y_0) = (a \sin(\omega t), -a \cos(\omega t))
$$

The total kinetic energy $T$ of the system is the sum of the kinetic energy of the pendulum bob and the motion of the moving pivot. The position of the bob in terms of the angle $\theta$ and the moving pivot is given by:
$$
x = x_0 + l \sin(\theta) = a \sin(\omega t) + l \sin(\theta)
$$
$$
y = y_0 - l \cos(\theta) = -a \cos(\omega t) - l \cos(\theta)
$$

The velocity of the pendulum bob is the time derivative of these coordinates:
$$
\dot{x} = \frac{d}{dt}(a \sin(\omega t) + l \sin(\theta)) = a \omega \cos(\omega t) + l \dot{\theta} \cos(\theta)
$$
$$
\dot{y} = \frac{d}{dt}(-a \cos(\omega t) - l \cos(\theta)) = a \omega \sin(\omega t) + l \dot{\theta} \sin(\theta)
$$

The kinetic energy of the pendulum bob is:
$$
T = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2)
$$

The potential energy $U$ of the system is due to gravity acting on the pendulum bob. The height $y$ of the bob (with the pivot at $y_0 = -a \cos(\omega t)$) is:
$$
y = -a \cos(\omega t) - l \cos(\theta)
$$
The gravitational potential energy is:
$$
U = m g y = -m g \left(a \cos(\omega t) + l \cos(\theta)\right)
$$

The Lagrangian $L$ is the difference between the kinetic and potential energies:
$$
L = T - U
$$

No problem, but to derive the equation of motion, I have to apply the Euler-Lagrange equation:
$$
\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = ???
$$

I don't know what I should put in place of the ??? I assume it is not zero as there is a driving force. How can I calculate that driving force?

By including the motion of the pivot in your equations, the Euler-Lagrange equations should work normally.

You could double check by generating the equation of motion using forces in the accelerating reference frame of the pivot.

 

[wnvl2]

I think I understand it now better. I haven a system with 2 degrees of freedom $\phi$ en $\theta$. The equations of motion are

$$\frac{d}{dt}\frac{\delta L}{\delta \dot{\phi}}-\frac{\delta L}{\delta \phi} = -mg(a \sin \phi+l\sin\theta)$$
$$\frac{d}{dt}\frac{\delta L}{\delta \dot{\theta}}-\frac{\delta L}{\delta \theta} = 0$$

In the second equation of motion the RHS is zero, in the first not.

 

[BvU]

What is $\phi$ ? Haven't seen it thus far! Don't you just have $\theta$ ? A drawing, perhaps ?

$$\ $$

 

[anuttarasammyak]

I think $\phi=\omega t$ as a constraint. I don't think we need the first equation.

 

[PeroK]

I think I understand it now better. I haven a system with 2 degrees of freedom $\phi$ en $\theta$. The equations of motion are

$$\frac{d}{dt}\frac{\delta L}{\delta \dot{\phi}}-\frac{\delta L}{\delta \phi} = -mg(a \sin \phi+l\sin\theta)$$
$$\frac{d}{dt}\frac{\delta L}{\delta \dot{\theta}}-\frac{\delta L}{\delta \theta} = 0$$

In the second equation of motion the RHS is zero, in the first not.

There is only one Euler-Lagrange equation for this system. The Lagrangian is time dependent and $\omega$ is constant.

 

[PeroK]

PS if there is an Euler-Lagrange equation, there should be an equivalent Newtonian equation of motion. The equation of motion for the pivot point is fixed. That cannot be derived from Newton's laws without including the unspecified external forces. Likewise, there is no Euler-Lagrange equation for the pivot point.

 

[BvU]

A drawing, perhaps ?

No takers. Here's my attempt at stating the problem:

​

Suspension point A with coordinates $(x_0, y_0)$ is forced to describe a circle with radius $a$. So $$ \begin{align*}
\phi &= \omega t\\ x_0 &= a\cos\phi\\ y_0&=a\sin\phi\end{align*}

$$Mass $m$ swings and has coordinates $$
\begin{align*}
x&=x_0+l\cos\theta \\ y&= y_0+l\sin\theta\end{align*}

$$We want to set up a Lagragian $\mathcal L = T - V$ in terms of $\theta$ and solve the Euler-Lagrange equation$$\frac{\ d}{dt}\frac{\delta\mathcal L}{\delta \dot{\theta}}-\frac{\delta\mathcal L}{\delta \theta} = 0
$$The extreme case $\omega = 0$ should give us a simple pendulum and, at the other extreme, the case $\omega >> \sqrt{g/l}$ should give us $\theta = \phi$.

Without some further assumptions this seems to me a pretty difficult enterprise
Setting up $\mathcal L$ is already daunting. An attempt at simplifying with $\theta << 1$ spoils all the fun (forces $\omega \approx 0)$. I don't think I've ever seen this as an exercise in a textbook...

I wonder if someone can prove me wrong

$\$

 

[PeroK]

I don't see the problem deriving the equation of motion. The kinetic energy and potential energy can easily be written down in terms of the motion of the pivot plus the angle $\theta$.

Moreover, a Newtonian analysis in the accelerating reference frame of the pivot is equally straightforward.

Solving the equation of motion is another matter, as it often is.

 

[anuttarasammyak]

(post #9)

Though not an essential comment, in order that these formula are right, $\phi$ is measured from x axis as usual not from -y as shown in the diagram. In order that the diagram is right $\phi=\omega t - \frac{\pi}{2}$ instead of the first formula.

 

[BvU]

My mistake, thanks for for correcting!

$\$