# How Do You Model a Variable Length Pendulum?

• Zoidberg
In summary, the conversation discusses the solution to a problem involving a simple pendulum with variable length. The Lagrangian, equations of motion, Hamiltonian, and total mechanical energy of the system are calculated and discussed. It is noted that the Hamiltonian is not conserved due to the time dependence of the length of the pendulum. Ultimately, the solution is considered correct and it is mentioned that the length of the pendulum may be considered as a constraint.
Zoidberg
[SOLVED] Variable Length Pendulum

Hi, everybody! I'm not sure about my solution to the problem.

## Homework Statement

Consider a simple pendulum with variable length r(t). The suspension point remains fixed.
(i)Write the Lagrangian for the pendulum and the equations of motion.
(ii)Write the Hamiltonian for the system.
(iii)Calculate the total mechanical energy of the system.

## Homework Equations

Lagrangian: L=T-V
Euler-Lagrange equation: $$\frac{d~}{dt} \ \left( \, \frac{\partial L}{\partial \dot{q}_i} \, \right) \ - \ \frac{\partial L}{\partial q_i} \ = \ 0$$
Legendre transform of the Lagrangian: $$H\left(q_j,p_j,t\right) = \sum_i \dot{q}_i p_i - L(q_j,\dot{q}_j,t)$$

## The Attempt at a Solution

(i) I think the system has only one degree of freedom, the angle $$\theta$$. The kinetic energy of the system is $$T=\frac{m}{2}\left( \dot{r}^2+r^2 \dot{\theta}^2\right)$$. The potential potential energy is $$V=-mgr \cos\theta$$.
$$\Rightarrow L=\frac{m}{2}\left( \dot{r}^2+r^2 \dot{\theta}^2\right)+mgr \cos\theta$$.
Using the Euler-Lagrange equation for $$q=\theta: 2mr\dot{r}\dot{\theta}+mr^2 \ddot{\theta}-mgr\sin\theta =0$$
(ii)With the generalized momenta $$p_{\theta}=\frac{\partial L}{\partial \dot{\theta}}=mr^2 \dot{\theta} \Leftrightarrow \dot{\theta}=\frac{p_{\theta}}{mr^2}$$ I can write the Hamiltonian $$H=p_\theta \dot{\theta} -L=\frac{p_\theta}{2mr^2}-\frac{m}{2}\dot{r}^2-mgr\cos\theta$$. The Hamiltonian is not conserved since r=r(t) is time dependent.
(iii) The total energy of the system is $$E=T+V=\frac{p_\theta}{2mr^2}+\frac{m}{2}\dot{r}^2-mgr\cos\theta$$

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Math seems fine, but $r$ and $\theta$ are both changing with time. How exactly are you defining the degree of freedom?

maverick280857 said:
Math seems fine, but $r$ and $\theta$ are both changing with time. How exactly are you defining the degree of freedom?

r is a fixed function of time r(t). You can't predict its values from the dynamics.

First I was confused about $$H(\theta,p_{\theta})\neq E$$, but after reading some books this seems to be a classic example of this type of problem. If I understand this correct r(t) is a holonomic constraint.
Anyway the solution seems to be correct. Thanks for your responses.

Zoidberg said:
First I was confused about $$H(\theta,p_{\theta})\neq E$$, but after reading some books this seems to be a classic example of this type of problem. If I understand this correct r(t) is a holonomic constraint.
Anyway the solution seems to be correct. Thanks for your responses.

What happens is that the system is not isolated, there is some extrenal force dictating how r changes with time. In that case H will not be E. The same thing happens, for example, when there is something rotating at a constant angular velocity, which means there is an extrenal force acting and forcing the rotation.

Dick said:
r is a fixed function of time r(t). You can't predict its values from the dynamics.

Ok, so its essentially given.

maverick280857 said:
Ok, so its essentially given.

Right. It could also as Zoidberg pointed out, could be considered as a constraint.

Zoidberg said:
I can write the Hamiltonian $$H=p_\theta \dot{\theta} -L=\frac{p_\theta}{2mr^2}-\frac{m}{2}\dot{r}^2-mgr\cos\theta$$.

Shouldn't pΘ appear on the second power here?

(off: why can't I write latex formulas? The preview shows a completely different formulas from that I write down. E.g. how $$p_\theta$$ looks like?)

edit: wow! now it displays good. Only the preview was wrong.

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## What is a variable length pendulum?

A variable length pendulum is a type of pendulum where the length of the string or rod can be adjusted. This allows for changes in the period and frequency of the pendulum's oscillations.

## How does the length of a variable length pendulum affect its motion?

The length of a pendulum directly affects its period and frequency of oscillation. A longer pendulum will have a longer period and lower frequency, while a shorter pendulum will have a shorter period and higher frequency.

## What are the real-world applications of a variable length pendulum?

Variable length pendulums have various applications in science, engineering, and technology. They are used in timekeeping devices, seismometers, and as a tool for measuring the acceleration due to gravity.

## What factors can affect the motion of a variable length pendulum?

The motion of a variable length pendulum can be affected by factors such as the length of the string or rod, the mass of the bob, air resistance, and the angle at which it is released.

## What are some limitations of a variable length pendulum?

Some limitations of a variable length pendulum include the influence of external factors such as air resistance and the accuracy of measuring the length of the string or rod. Additionally, the motion of a pendulum may be affected by the Earth's rotation and other factors in the environment.

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