Damped Pendulum

At a general time $t$ , let the coordinates of the ball be $(x,y) = (\cos{\theta},\sin{\theta})$ . The velocity and acceleration components are:

$\left[ \begin{matrix} x\\ y\end{matrix} \right] = \left[ \begin{matrix} \cos{\theta}\\ \sin{\theta} \end{matrix} \right] \implies \left[ \begin{matrix} \dot{x}\\ \dot{y} \end{matrix} \right] =\left[ \begin{matrix} -\dot{\theta}\sin{\theta}\\ \dot{\theta}\cos{\theta} \end{matrix} \right] \implies \left[ \begin{matrix} \ddot{x}\\ \ddot{y} \end{matrix} \right] =\left[ \begin{matrix}-\ddot{\theta}\sin{\theta}-\dot{\theta}^2\cos{\theta}\\ \ddot{\theta}\cos{\theta}-\dot{\theta}^2\sin{\theta} \end{matrix} \right]$

Let the magnitude of tension be $T$ and it is directed towards the origin along the length of the rod. Now, applying Newton's second law gives:

$m\left[ \begin{matrix} \ddot{x}\\ \ddot{y} \end{matrix} \right] = \left[ \begin{matrix}0\\ -mg \end{matrix} \right] -T\left[ \begin{matrix} \cos{\theta} \\ \sin{\theta} \end{matrix} \right] -D\left[ \begin{matrix} \dot{x}\\ \dot{y} \end{matrix} \right]$

This gives us a system of two equations in $\ddot{\theta}$ and $T$ . Eliminating $T$ from these set of equations leads to the final equation of motion which is the following. Simplifications left out.

$\ddot{\theta} = -\frac{\dot{\theta}}{10} -10\cos{\theta} \ ; \ \theta(0) = \frac{\pi}{3} \ ; \ \dot{\theta}(0) = 0$

We asked to compute the sum $\cos{\theta}+\sin{\theta}$ at time $t=30$ . This has been done numerically as follows:

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clear all, clc

% Initial conditions:
theta(1)    = pi/3; dtheta(1)   = 0;

% Time step, etc. :
dt          = 1e-5;  t  = 0:dt:30;

for k = 1:length(t)-1

    % Equation of motion:
    ddtheta     = -10*cos(theta(k)) - dtheta(k)/10;

    % Semi implicit Euler numerical integration:
    dtheta(k+1) = dtheta(k) + ddtheta*dt;  
    theta(k+1)  = theta(k) + dt*dtheta(k+1);    
end

ANSWER = cos(theta(end)) + sin(theta(end)) % ANSWER = -1.3438

I think it is interesting how @Karan Chatrath incorporated the rod tension into his solution. My approach was a bit more conventional.

Moment of Inertia:

$I = m L^2$

Position and velocity:

$x = L \cos \theta \\ y = L \sin \theta \\ \dot{x} = -L \sin \theta \, \dot{\theta} \\ \dot{y} = L \cos \theta \, \dot{\theta}$

Gravity, damping, and total force:

$\vec{F}_g = (0, -m g) \\ \vec{F}_D = (-D \dot{x}, -D \dot{y}) \\ \vec{F} = \vec{F}_g + \vec{F}_D$

Torque from vector cross product:

$\vec{r} = (x,y) \\ \vec{\tau} = \vec{r} \times \vec{F}$

The torque is one-dimensional, so it can be thought of as a scalar $\tau$ . The sign of $\tau$ changes appropriately to yield oscillation. Angular acceleration:

$\ddot{\theta} = \frac{\tau}{I}$

Numerical integration takes care of the rest