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.
Best Answer
I think there is a little nuance to be addressed here.
On one hand, if we take the problem at face value then there is no restorative force to slow the swing down. Hence by the law of inertia, the angular speed will remain $v_i/R$ indefinitely, where R is the length of the swing and $v_i$ is the initial tangential speed.
If we take the swing to be under gravity then we have more to work with. I do not think this problem can be solved by taking a simple time derivative like most calc 1 related rates problems.
I will try and explain. Let's first draw a picture of the swing mid swing:
In this diagram $T$ is the tension in the (mass-less) rope, $W$ is the weight of the swing seat, $\phi$ is the angle of travel as measured from the vertical, and $g$ is the magnitude of the gravitational acceleration. If we define the gravitational potential to be vanish at the bottom of the swing, then the total mechanical energy of the system is given by:
$$E = \frac{1}{2} m R^2 \dot{\phi}^2 + mgR \left( 1 - \cos (\phi) \right)$$
Note that I am ignoring frictional forces for simplicity. If we take a time derivative we have:
$$0 = R \dot{\phi} \ddot{\phi} + g \sin (\phi) \dot{\phi} $$
We do not have enough data about $\phi$ or $\ddot{\phi}$ to solve for $\dot{\phi}$. I will now attempt to solve this problem with other ideas from calc 1 and basic physics. We can state the position, velocity, and acceleration:
$$\vec{d} = x\hat{x} + y\hat{y}$$ $$\vec{v} = \dot{x}\hat{x} + \dot{y}\hat{y}$$ $$\vec{a} = \ddot{x}\hat{x} + \ddot{y}\hat{y}$$
In the above equations $\hat{x}$ and $\hat{y}$ refer to vectors of length $1$ in the $x$ and $y$ directions, respectively. As shown in the diagram, the system benefits from a change in co-ordinates. We will now convert these equations from Cartesian to polar co-ordinates:
$$x = R \sin{\phi}$$ $$y = - R \cos{\phi}$$
With this change of co-ordinates we have:
$$\dot{x} = R \cos ( \phi ) \dot{\phi}$$ $$\ddot{x} = - R \sin ( \phi ) \dot{\phi}^2 + R \cos ( \phi ) \ddot{\phi}$$ $$\dot{y} = R \sin ( \phi ) \dot{\phi}$$ $$\ddot{y} = R \cos ( \phi ) \dot{\phi}^2 + R \sin ( \phi ) \ddot{\phi}$$
We also need to express the Cartesian unit vectors in terms of the polar unit vectors $\hat{r}$ and $\hat{\phi}$. Let's start by constructing the radial unit vector:
$$\hat{r} = \frac{\vec{r}}{||\vec{r}||} = \frac{\vec{d}}{R} = \sin ( \phi ) \hat{x} - \cos ( \phi ) \hat{y}$$
Now, we know that $\hat{r}$ and $\hat{\phi}$ are perpendicular. We can use this to draw a diagram:
We can now easily express $\hat{\phi}$ as $\cos ( \phi ) \hat{x} + \sin ( \phi ) \hat{y}$. If we reverse these equations for $\hat{x}$ and $\hat{y}$ we have:
$$\hat{x} = \sin ( \phi ) \hat{r} + \cos ( \phi ) \hat{\phi}$$ $$\hat{y} = - \cos ( \phi ) \hat{r} + \sin ( \phi ) \hat{\phi}$$
This allows us to completely express the kinematic quantities in terms of polar co-ordinates:
$$\vec{d} = R \hat{r}$$ $$\vec{v} = R \dot{\phi} \hat{\phi}$$ $$\vec{a} = - R \dot{\phi}^2 \hat{r} + R \ddot{\phi} \hat{\phi}$$
With that, we can now write down Newton's Law in both the radial and angular directions:
$$ - m R \dot{\phi}^2 = m g \cos ( \phi ) - T $$ $$ m R \ddot{\phi} = - m g \sin ( \phi ) $$
Let's focus on the angular equation since we don't need to worry about the tension. This expression can be rearranged to
$$\ddot{\phi} = - \frac{g}{R} \sin ( \phi )$$
which is the well known equation of motion for a pendulum. So if you had memorized this equation from a physics course then by all means, start here. This equation is very difficult to solve in general, so let's make a rather strong assumption. Suppose that the maximum angular displacement of the swing is small. We can now approximate the acceleration as
$$ \ddot{\phi} \approx - \frac{g}{R} \phi $$
which is the equation of motion for a harmonic oscillator. The general solution of this equation is given by:
$$ \phi(t) = A \sin \left( \sqrt{\frac{g}{R}} t \right) + B \cos \left( \sqrt{\frac{g}{R}} t \right)$$
Recall that our initial conditions for this system are
$$ \dot{\phi}(0) = \frac{v_i}{R}$$ $$ \phi(0) = 0$$
which will simplify $\phi(t)$:
$$ \phi(t) = \frac{v_i}{\sqrt{g R}} \sin \left( \sqrt{\frac{g}{R}} t \right) $$
If we want to find the angular velocity, 1 second after the swing is pushed then we simply have to differentiate and substitute in the required values:
$$ \dot{\phi}(t) = \frac{v_i}{R} \cos \left( \sqrt{\frac{g}{R}} t \right) \implies \vec{\dot{\phi}} = \dot{\phi}(1) \hat{\phi} = \frac{3}{8} \cos \left( \sqrt{\frac{9.81}{8}} \right) \approx 0.1676 \, \, \mathrm{Hz} \, \, \hat{\phi} $$