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:

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

Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} Now since F_r &= 0 we can ignore it and look only at F_\theta.\\ Since we know F_\theta &= maL and ma &= mLa". We can conclude \end{align}}

sin(\theta)*mg = mL\theta"

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

\theta" - sin(\theta)(g/L) = 0

now we take the laplace transform of this. Unfortunately the laplace transform of that is horrible. Long, complicated, and nearly impossible to solve. So we use the approximation sin(\theta) = \theta where \theta is small. (I tried to leave sin(\theta) in the equation. After 4 hours and many wolframalpha.com timeouts I gave up)

with this new equation we get:

g*\theta(t)/L +s^2*\theta(t) - s*\theta(0) - \theta'(0) = 0

we know that \theta(0) = \theta0 and \theta'(0) = 0

solving for \theta(t) we get

\theta(t) = s*\theta₀/((-g/L)+S^2)

now we take the inverse laplace transform of that which yields

\theta(t) = cosh(t*(g/L)^(1/2))