# Trouble simplifying the Lagrangian

• I
• p1ndol
In summary, Taking x and y as provided in the solution, I imagine we should have:\begin{align*}\dot x = -a\gamma\sin(\gamma t) + l\dot \phi \cos(\phi) \dot y = -l\dot \phi\sin(\phi) \end{align*}Calculating the kinetic energy:T = \frac {m} {2} (l^2\dot\phi^2\cos(\phi)^2 - 2l\dot\phi\cos(\phi)a\gamma\sin(\gamma t) + a^2\gamma^2\sin(\gamma t
p1ndol
Hello, I have posted a similar thread on this question before, but I'd like to get some help to simplify the answers I've got so far in order to match the solutions provided. If anyone could help me, I would really appreciate it. Since (c) is quite similar to (b), I'll leave here what I've done on (b).

Taking x and y as provided in the solution, I imagine we should have:

$$\dot x = -a\gamma\sin(\gamma t) + l\dot \phi \cos(\phi)$$ $$\dot y = -l\dot \phi\sin(\phi)$$

Calculating the kinetic energy:

$$T = \frac {m} {2} (l^2\dot\phi^2\cos(\phi)^2 - 2l\dot\phi\cos(\phi)a\gamma\sin(\gamma t) + a^2\gamma^2\sin(\gamma t)^2+l^2\dot \phi^2\sin(\phi)^2)$$
$$T = \frac {m} {2} (l^2\dot\phi^2 - 2l\dot\phi\cos(\phi)a\gamma\sin(\gamma t) + a^2\gamma^2\sin(\gamma t)^2$$

For the potential energy:

$$V = -mgy = -mgl\cos(\phi)$$

So, considering that the Lagrangian is the kinetic minus the potential energy, I am trying to find out a way to get to the answer provided in the book removing time derivatives.

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As you wrote, the Lagrangian for (b) is\begin{align*}
L = \frac{1}{2}ml^2 \dot{\phi}^2 - ma\gamma l \dot{\phi} \sin{(\gamma t)} \cos{(\phi)} + \frac{1}{2}ma^2 \gamma^2 \sin^2{(\gamma t)} + mgl\cos{(\phi)}
\end{align*}Next, notice that the second term can be re-written as\begin{align*}
-ma\gamma l \dot{\phi} \sin{(\gamma t)} \cos{(\phi)} = ma\gamma^2 l \cos{(\gamma t)} \sin{(\phi)} - \frac{d}{dt} \left( ma\gamma l \sin{(\gamma t)} \sin{(\phi)}\right)
\end{align*}The total time derivative can be omitted in the Lagrangian without changing the equations of motion. Furthermore, the third term ##\frac{1}{2}ma^2 \gamma^2 \sin^2{(\gamma t)}## in the Lagrangian depends only on time, so may also be omitted (i.e. it can be re-written as a total derivative). Therefore\begin{align*}
\tilde{L} = \frac{1}{2}ml^2 \dot{\phi}^2 + ma\gamma^2 l \cos{(\gamma t)} \sin{(\phi)} + mgl\cos{(\phi)}
\end{align*}is a Lagrangian for the system.

p1ndol and vanhees71
Thanks, it helped a lot!

berkeman

## 1. What is the Lagrangian and why is it important in physics?

The Lagrangian is a mathematical function that describes the dynamics of a physical system. It is important because it allows us to formulate the equations of motion for a system in a way that is independent of the coordinate system used to describe the system.

## 2. What makes simplifying the Lagrangian difficult?

Simplifying the Lagrangian can be difficult because it involves solving complex mathematical equations and manipulating multiple variables and parameters. It also requires a deep understanding of the underlying physical principles and concepts.

## 3. How can I simplify the Lagrangian?

There is no one-size-fits-all method for simplifying the Lagrangian, as it depends on the specific system being studied. However, some general strategies include using symmetries and conservation laws, making appropriate coordinate transformations, and using approximations or simplifying assumptions.

## 4. What are some common mistakes made when simplifying the Lagrangian?

One common mistake is neglecting certain terms in the Lagrangian that may seem small or insignificant, but can actually have a significant impact on the dynamics of the system. Another mistake is using incorrect mathematical techniques or not fully understanding the physical principles involved.

## 5. How can I check if I have correctly simplified the Lagrangian?

One way to check the correctness of a simplified Lagrangian is to use it to derive the equations of motion for the system and compare them to known solutions or experimental data. Additionally, double-checking all mathematical calculations and ensuring that all physical principles have been properly accounted for can help verify the accuracy of the simplified Lagrangian.

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