- #1

p1ndol

- 7

- 3

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:

[tex] \dot x = -a\gamma\sin(\gamma t) + l\dot \phi \cos(\phi) [/tex] [tex] \dot y = -l\dot \phi\sin(\phi) [/tex]

Calculating the kinetic energy:

[tex] 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)[/tex]

[tex] 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 [/tex]

For the potential energy:

[tex] V = -mgy = -mgl\cos(\phi) [/tex]

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.

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

[tex] \dot x = -a\gamma\sin(\gamma t) + l\dot \phi \cos(\phi) [/tex] [tex] \dot y = -l\dot \phi\sin(\phi) [/tex]

Calculating the kinetic energy:

[tex] 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)[/tex]

[tex] 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 [/tex]

For the potential energy:

[tex] V = -mgy = -mgl\cos(\phi) [/tex]

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.