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 A₀. Find a function to determine the angle at any time t. The summation of forces yields

F_x = T*sin(A)

F_y = T*cos(A) - mg = 0

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

now:

F_r = T- cos(A)*mg = 0

F_A = sin(A)*mg = maL

now since F_r = 0 we can ignore it and look only at FA.

Since we know F_A = maL and ma = mLa". We can conclude

sin(A)*mg = mLA"

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

A" - sin(A)(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(A) = A where A is small. (I tried to leave sin(A) in the equation. After 4 hours and many wolframalpha.com timeouts I gave up)

with this new equation we get:

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

we know that A(0) = A0 and A'(0) = 0

solving for A(t) we get

A(t) = s*A0/((-g/L)+S^2)

now we take the inverse laplace transform of that which yields

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