Why can't I integrate the pendulum equation d²φ/dt² = -(g/R)sin φ twice?

[Moneer81]

Hello,

while working on a simple mechanics problem using polar coordinates, I got this equation

second derivative of phi = (-g/R) sin phi

now I need to solve this to get an equation for phi (t) but the books says that I cannot solve this using elementary functions and that the solution will be the more complex Jacobi elliptic function. My question is why can't I integrate twice to get the equation for phi (t) ?

thanks a lot

 

[MalleusScientiarum]

You can't just straight integrate that differential equation. I'm not even sure how one would attempt it. Here's one way to set up the elliptic integral:

Multiply through by $$\dot{\phi}$$ and get the differential equation
$$\ddot{\phi} \dot{\phi} = -g/R \dot{\phi} sin \phi$$
which we recognize as being the first time derivative of
$$\dot{\phi}^2 = g/R cos \phi + C$$
when then leads to the integral:
$$\int d\phi ~ 1/\sqrt{g/R cos \phi + C} = t - t_0$$
which is then the elliptic integral left behind. This is why we make things like the small angle approximation, where applicable. If you're working with exact numbers, you could pretty simply write a numerical algorithm

 

[Moneer81]

MalleusScientiarum ,

why can't we straight integrate it with respect to time ?

 

[Moneer81]

ok never mind...stupid question. i know why :)

 

[James R]

Yeah - because it's sine of phi, not sine of t.

 

[dextercioby]

In post #2, you missed a little "2" when going from $\dot{\phi}\ddot{\phi}$ to the square of the first derivative.

Daniel.

 

[psuchetic_edition]

i guess theylor expansion is the best choice

 

[MalleusScientiarum]

I advise numerical analysis.