Aaron Boyd's Assignment 8

I decided to use laplace transforms to solve a pendulum equation. A pendulum with a weight of mass m and a massless rod length L is released from an initial angle \theta₀. Find a function to determine the angle at any time t. The summation of forces yields ${\begin{aligned}F_{x}&=T\sin(\theta )\\F_{y}&=T\cos(\theta )-mg=0\end{aligned}}$

Polar coordinates may be easier to use, lets try that.

now:

${\begin{aligned}F_{r}&=T-mg\cos(\theta )=0\\F_{\theta }&=\sin(\theta )mg=maL\end{aligned}}$

${\begin{aligned}{\text{Now since }}F_{r}&=0{\text{ we can ignore it and look only at }}F_{\theta }.\\{\text{ Since we know }}F_{\theta }&=maL{\text{ and }}ma=mLa.{\text{ We can conclude}}\end{aligned}}$

${\begin{aligned}\sin(\theta )*mg=mL{\ddot {\theta }}\end{aligned}}$

canceling the common mass term and rearranging a bit we get.

${\begin{aligned}{\ddot {\theta }}-(g/L)\sin(\theta )=0\\\\{\text{Now we take the laplace transform of this.}}\\{\text{Unfortunately the laplace transform of that is horrible. Long, and impossible to solve.}}\\{\text{So we use the approximation }}\\\\\sin(\theta )=\theta \\{\text{ where }}\theta {\text{ is small. }}\\\\{\text{(I tried to leave }}sin(\theta ){\text{ in the equation.}}\\{\text{After 4 hours and many wolframalpha.com timeouts I gave up) }}\\\\{\text{with this new equation we get:}}\\\\g*{\frac {\theta (t)}{L+s^{2}}}*\theta (t)-s\theta (0)-{\dot {\theta }}(0)=0\\\\{\text{we know that }}\theta (0)=\theta _{0}{\text{ and }}{\dot {\theta }}(0)=0\\\\\\{\text{solving for }}\theta (t){\text{ we get}}\\\\\theta (t)=s*{\frac {\theta _{0}}{({\frac {-g}{L}}+S^{2})}}\\\\\\{\text{now we take the inverse laplace transform of that which yields }}\\\\\\\theta (t)=cosh(t{\sqrt {(}}{\frac {g}{L}}))\\\end{aligned}}$

You can solve for the same thing from the cartesian coordinates. Taking:

${\begin{aligned}F_{x}&=T\sin(\theta )=0\\{\text{ and recognizing }}T=mg\\\end{aligned}}$

you can arrive at the same answer