Generator without Prime Mover

Consider the given set of equations:

$V_t = -\frac{V_m}{\omega_o}\sin{\theta}\dot{\theta}$ $P_R = \frac{V_t^2}{R} = \frac{V_m^2}{\omega_o^2R}\sin^2(\theta)\dot{\theta}^2$

After rearrangement and simplification, the governing differential equation is:

$\ddot{\theta} = -A\sin^2(\theta)\dot{\theta}$ $A = \frac{V_m^2}{\omega_o^2IR}$

Now,

$\ddot{\theta} = \dot{\theta}\frac{d \dot{\theta}}{d \theta} = -A\sin^2(\theta)\dot{\theta}$

Separating the variables, integrating and applying the initial conditions gives:

$\dot{\theta} = -\frac{A}{2}\left(\theta - \frac{\sin{2\theta}}{2}\right) + \omega_o$

Now, again, Separating the variables and integrating from $t = 0$ to $t =2$ gives:

$\int_{0}^{\theta_f} \frac{d\theta}{-\frac{A}{2}\left(\theta - \frac{\sin{2\theta}}{2}\right) + \omega_o} = 2$

Now, the angular position $\theta_f$ is not known. It needs to be found by solving the above integral equation. Having found $\theta_f$ its value can be replaced in the equation:

$\dot{\theta}_f = -\frac{A}{2}\left(\theta_f - \frac{\sin{2\theta_f}}{2}\right) + \omega_o$

Solving for $\theta_f$ can be done iteratively. Using a Wolfram Alpha like tool, or a numerical integration scheme, one can incrementally increase the value of $\theta_f$ till the integral evaluates to $2$ .

The answer comes out to be: $\boxed{97.376}$ .

As for the bonus question:

One can see that as the kinetic energy of the rotor reduces, the voltage across the terminals undergoes damped oscillations with variable frequency. This makes sense as the energy of the rotor is completely dissipated in the resistor. So naturally, the voltage across the resistor also decreases with time. This explains why the magnitude of the voltage reduces with time. I do not have one for the decreasing frequency.