Simple question really.
$L (d^2/dt^2) \theta = - g \cdot \theta $
Or
$ I \theta \ddot (t) + m \cdot g \cdot l \cdot sin ( \theta \cdot t ) $
Assuming $ sin \theta = \theta $ for angles less than about 15 degrees.
Is it
$ L(s) \cdot s^2 = - g \cdot s $
Or
$ M(s) \cdot g \cdot l \cdot (s/(s^2 + \theta)) $
Or something else?!
I think I know what the Laplace transforms of trig functions are, and of normal time-based variables, but how do you deal with mgl?
I guess you can make theta become s if your variables are of the form $u(\theta)$.
The Laplace transform is from $\mathbb{R}_{\ge 0}$ to $\mathbb{C}$ so as we consider $t \ge 0$ we can proceed obtaining
$$ \mathcal{L}\left(\theta(t)\right) = \Theta(s) = \frac{I(s\theta_0+\theta'_0)}{I s^2+m g l} $$
and after anti transformation
$$ \theta(t) = \left(\cos\left(\sqrt{\frac {m g l}{I}}t\right)\theta_0 + \sqrt{\frac{I}{m g l}}\sin\left(\sqrt{\frac {m g l}{I}}t\right)\theta'_0\right)u(t) $$
with $u(t)$ the Heaviside step function.