I'm trying to find equation of motion the following pendulum.
To do this, let
$$\vec r (t) = (l \sin (\theta (t)), -l \cos(\theta(t))).$$
then
$$m D^2 \vec r(t) = ( -l \sin(\theta(t))\cdot \dot{\theta}(t) + l\cos(\theta(t)) \cdot \ddot{\theta}(t) , -l\cos(\theta(t)) \cdot \dot{\theta}(t) – l \sin(\theta(t)) \ddot{\theta}(t) ) = (mg (-\hat y)) + \vec T.$$
By observing that the bob does not move in the radial direction, we can conclude that the net force on the radial direction has to be zero at any given moment, hence
$$T = mg\cos(\theta(t)).$$
Therefore, for $\theta = \theta(t)$, we get (x-axis)
$$ -l \sin(\theta)\cdot \dot{\theta} + l\cos(\theta) \cdot \ddot{\theta} = – mg\cos(\theta) \sin(\theta),$$
and (y-axis)
$$-l\cos(\theta) \cdot \dot{\theta} – l \sin(\theta) \ddot{\theta} = mg – mg\cos^2 (\theta).$$
so, the function $\theta(t)$ must satisfy both of these DEs at the same time.
However, anyone to took freshman year physics courses solves this problem by noting that $r = l \theta$, which is kind of weird considering the fact that what I have to solve in order to get the solution.
Question:
How to solve these above second order nonlinear ODEs ?
I have tried Laplace transform an got, for $L (\theta (t)) = F(s)$
$$F'(s) = (-1) \frac{(s^2 + 1 + g/l)}{s} F(s) +\frac{ \dot\theta(0)}{s} + \theta(0),$$
however, I do not know how transfer back the solution to the above first order ODE analytically, so leave that there.
Best Answer
There is a first error in your first equation $$ \ddot{\pmb r}(t)=\left(-l\sin\theta\;\dot\theta^2+l\cos\theta\;\ddot\theta,\;l\cos\theta\;\dot\theta^2+l\sin\theta\;\ddot\theta\right) $$ that is $$ m\ddot{\pmb r}(t)=-mg\hat y+\pmb T $$ and we have \begin{align} ml(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta)&=-T\sin\theta. & \tag {eq. 1} \\ ml(\ddot{\theta}\sin\theta+\dot{\theta}^2\cos\theta )&=T \cos\theta - mg & \tag {eq. 2} \end{align} Observe that $T\neq mg\cos\theta$ (this is your second error).
Taking the difference $\sin\theta\times (\text{eq. }1)-\cos\theta \times (\text{eq. }2)$ we find $$ ml\dot\theta^2=T-mg\cos\theta \tag 3$$ and taking the sum $\cos\theta\times (\text{eq. }1)+\sin\theta \times (\text{eq. }2)$ we find $$ ml\ddot\theta=-mg\sin\theta \tag 4$$ that is \begin{align} ml\dot\theta^2+mg\cos\theta &=T\tag 5\\ \ddot\theta+\frac{g}{l}\sin\theta &=0 \tag 6 \end{align}
I think you can solve these equations.