Undergrad Trouble simplifying the Lagrangian

[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.

 

[ergospherical]

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]

Thanks, it helped a lot!