There is indeed a term involving the time derivative of the changing coupling between the masses.
First, let's derive the equation for a single mass.
$$L = \frac{1}{2} I\, \dot{\theta}^2 - V(\theta)$$
$$\frac{\partial L}{\partial \dot{\theta}} = I\, \dot{\theta}$$
$$\frac{\partial L}{\partial \theta} = -\frac{dV}{d\theta} = \tau$$
$$\tau = \frac{d}{dt} \left( I\, \dot{\theta} \right) = I\, \ddot{\theta}$$
This shows you that the angular acceleration is proportional to the torque.
Now, assume we have two masses. The driven mass has moment of inertia $I_1$ and angular velocity $\dot{\theta}$. The secondary mass has moment of inertia $I_2$ and angular velocity $\dot{\theta_2} = a(t)\, \dot{\theta}$, where $a(t)$ is the changing coupling (for example a changing belt position in a continuously variable transmission).
$$L = \frac{1}{2} I_1\, \dot{\theta}^2 + \frac{1}{2} I_2\, a(t)^2\, \dot{\theta}^2 - V(\theta)$$
$$\frac{\partial L}{\partial \dot{\theta}} = \left( I_1 + a(t)^2 I_2 \right) \dot{\theta}$$
$$\frac{\partial L}{\partial \theta} = -\frac{dV}{d\theta} = \tau$$
$$\tau = \frac{d}{dt} \left( \left( I_1 + a(t)^2 I_2 \right) \dot{\theta} \right)$$
$$\tau = \left( I_1 + a(t)^2 I_2 \right) \ddot{\theta} + 2 I_2\, a(t) \frac{da}{dt} \dot{\theta}$$
The last term, proportional to $a \dot{a} \dot{\theta}$, is the funny term you're looking for. It says that when the coupling is changing, you need to apply some torque just to keep the angular velocity $\dot{\theta}$ constant. Another way to think of it is that in the absence of external torque, $\dot{\theta}$ is no longer constant (as it was for the single mass), but instead $(I_1 + a(t)^2 I_2) \dot{\theta}$ is constant, because that's the real angular momentum.
I(ring) = Idisk(R2) - Idisk(R1).
The trick is figuring out the mass.
Mass of R2-sized disk would be MR2 = M*(pi*R2*R2)/((pi*R2*R2)-(pi*R1*R1))
Mass of R1-sized disk would be MR1 = MR2*(pi*R1*R1)/(pi*R2*R2)
So I(ring) = 1/2MR2*(R2*R2) - 1/2MR1*(R1*R1)
= 1/2( (M*(pi*R2*R2*R2*R2) - M*(pi*R1*R1*R1*R1))/((pi*R2*R2) - (pi*R1*R1)) ) )
I guess the pis can come out:
= 1/2*M*(R2^4-R1^4)/(R2^2 - R1^2)
Ugh, this means:
= 1/2*M*(R2^2+R1^2)*(R2^2-R1^2)/(R2^2 - R1^2)
= 1/2*M*(R2^2+R1^2)
So, yep.
Best Answer
In general, the moment of inertia is found by evaluating $$I = \iiint\limits_V{\rho r^2 dV},$$ where $\rho$ is the mass density and $r$ is the distance from the axis with respect to which you wish to calculate the moment of inertia (often the axis of symmetry). For complex objects, there usually will not be a simple formula for the moment of inertia and it may be necessary to evaluate this integral numerically, or approximate it with some simplifying assumptions.
If you think the gear can be approximated as a simple, uniform disk of radius $R$, its moment of inertia with respect to the axis of symmetry will be roughly $\frac{1}{2}MR^2$, where $M$ is the total mass. If the gear is uniform and has a constant thickness throughout, with a certain "inner radius" (not containing the teeth) and an "outer radius" (just containing the teeth), $\frac{1}{2}MR^2$ will still be valid, with $R$ having a value somewhere between these two radii.