You are given a pendulum with the distance $d$ from the anchor point to the center of the massive sphere, radius $r$ and mass $m$. Gravity is given by $g$.
I want to derive a formula for the angle $\theta$ between the vertical axis and the rope, given the time $t$ since the pendulum was dropped with the angle $\theta_0$. Let $\omega$ be the angular speed and $\alpha$ be the angular acceleration. At time $t=0$, let $\theta=\theta_0$ and $\omega=\omega_0$.
The torque applied to the pendulum is given by $\tau=d \bullet mg\sin\theta$.
The inertia of the pendulum is given by $I=\frac{2}{5}mr^2+md^2$.
Newton's 2nd law for torque gives: $\sum\tau=I\alpha$ which solved for $\alpha$ gives
$$
\alpha=\frac{dmg\sin\theta}{m(\frac{2}{5}r^2+d^2)}=\frac{dg\sin\theta}{\frac{2}{5}r^2+d^2}
$$
Due to $\alpha$ not being constant, we have to use the integral forms for the rotational kinematics formulas, which are
$$
\theta=\theta_0 + \int_0^t\omega\, dt
$$
$$
\omega=\omega_0 + \int_0^t\alpha\, dt
$$
Using the latter one, we get
$$
\omega=\omega_0 + \int_0^t \frac{dg\sin\theta}{\frac{2}{5}r^2+d^2} \, dt
$$
Then the first one
$$
\theta=\theta_0 + \int_0^t
\omega_0 + \int_0^t \frac{dg\sin\theta}{\frac{2}{5}r^2+d^2} \, dt
\, dt
$$
And this is how far i get with my knowledge. I can't think of any way to solve this integral, because of the dependence of $\sin\theta$. Might this be a (non-linear 2nd order) differential equation?
I hope the steps I've done are correct, please let me know if they are not. I hope someone here is able to solve this problem. Thanks in advance!
Best Answer
The time law of the physical pendulum is the solution of the following Cauchy problem:
$$ \begin{cases} -m\,g\,d\sin\theta(t) = I\,\ddot{\theta}(t) \\ \theta(0) = \theta_0 \\ \dot{\theta}(0) = \omega_0 \\ \end{cases} \quad \quad \overset{\alpha\,:=\,\sqrt{\frac{m\,g\,d}{I}}}{\Rightarrow} \quad \quad \begin{cases} \ddot{\theta}(t) + \alpha^2\sin\theta(t) = 0 \\ \theta(0) = \theta_0 \\ \dot{\theta}(0) = \omega_0 \\ \end{cases} $$
where, in the proposed case, we have $I = \frac{2}{5}\,m\,r^2+m\,d^2$.
Due to its heavy non-linearity it cannot be integrated elementarily, so two ways are presented.
We can reduce it to a system of first order ODEs:
$$ \begin{cases} \dot{\theta}(t) = \omega(t) \\ \dot{\omega}(t) = -\alpha^2\sin\theta(t) \\ \theta(0) = \theta_0 \\ \omega(0) = \omega_0 \\ \end{cases} $$
and consequently apply a numerical method such as, for example, Euler's method:
$$ \begin{cases} t_k = t_{k-1} + \Delta t \\ \theta_k = \theta_{k-1} + \omega_{k-1}\Delta t \\ \omega_k = \omega_{k-1} - \alpha^2\sin(\theta_{k-1})\Delta t \end{cases} \quad \quad \quad \text{with} \; k = 1,2,\dots,n $$
where $\Delta t := T/n$ with $T>0$ and $n \in \mathbb{N}$ are fixed parameters.
We can content ourselves with studying the physical pendulum with small angles ($\theta \ll 1\,\text{rad}$):
$$ \begin{cases} \ddot{\theta}(t) + \alpha^2\,\theta(t) = 0 \\ \theta(0) = \theta_0 \\ \dot{\theta}(0) = \omega_0 \\ \end{cases} \quad \Rightarrow \quad \theta(t) = \theta_0\cos(\alpha\,t) + \frac{\omega_0}{\alpha}\,\sin(\alpha\,t) $$
which, being a linear ODE of the second order, is elementary integrable.
As an example, assuming $\alpha=3$, $T=10$, $\theta_0=5°,10°,\dots,90°$ and $\omega_0=0$:
where the blue sine wave is exact, while the red one is approximate.