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 $\theta ₀$. Find a function to determine the angle at any time t. The summation of forces yields $\begin{array}{rl}{F}_x& =T\sin (\theta )\\ F_y& =T\cos (\theta )-mg=0\end{array}$ Polar coordinates may be easier to use, lets try that. now: $\begin{array}{rl}{F}_r& =T-mg\cos (\theta )=0\\ F_{\theta }& =\sin (\theta )mg=maL\end{array}$

$\begin{array}{rl}\text{Now since }{F}_r& =0\text{ we can ignore it and look only at }{F}_{\theta }.\\ \text{ Since we know }{F}_{\theta }& =maL\text{ and }ma=mLa.\text{ We can conclude}\end{array}$

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

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

$\begin{array}{r}\ddot{\theta }-(g/L)\sin (\theta )=0\\ \\ \text{Now we take the laplace transform of this.}\\ \text{Unfortunately the laplace transform of that is horrible. Long, and impossible to solve.}\\ \text{So we use the approximation }\\ \\ \sin (\theta )=\theta \\ \text{ where }\theta \text{ is small. }\\ \\ \text{(I tried to leave }sin(\theta )\text{ in the equation.}\\ \text{After 4 hours and many wolframalpha.com timeouts I gave up) }\\ \\ \text{with this new equation we get:}\\ \\ g*\frac{\theta (t)}{L+s^2}*\theta (t)-s\theta (0)-\dot{\theta }(0)=0\\ \\ \text{we know that }\theta (0)={\theta }_0\text{ and }\dot{\theta }(0)=0\\ \\ \\ \text{solving for }\theta (t)\text{ we get}\\ \\ \theta (t)=s*\frac{{\theta }_0}{(\frac{-g}{L}+S^2)}\\ \\ \\ \text{now we take the inverse laplace transform of that which yields }\\ \\ \\ \theta (t)=cosh(t\sqrt{(}\frac{g}{L}))\\ \end{array}$

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

$\begin{array}{rl}{F}_x& =T\sin (\theta )=0\\ \text{ and recognizing }T=mg\\ \end{array}$

you can arrive at the same answer