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.
Yes, you are right! Only when a force is applied purely through the center of mass it results in the body gaining a linear action with no rotational components.When any force is applied at a distance from the center of mass, it results in the body gaining the linear acceleration mentioned above plus an angular acceleration which depends on the moment arm (perpendicular distance to the point of application). In all, yes, the work done by a force does depend on its point of application. (and of course the time interval through which it acts.)
Best Answer
Although you haven't explicit said it, I get the impression that you are wondering if the force is somehow split between providing rotational vs. linear acceleration. That does not happen.
It's a both/and situation, not an either/or situation. A force can cause both accelerations simultaneously.
The linear will be (instantaneously) $$\vec{a}=\frac{F_{net}}{m}$$ and $$\vec{\alpha}=\frac{\Sigma (\vec{R}\times\vec{F})}{\mathcal{I}},$$ where $\vec{R}$ is the vector pointing from the point of rotational interest to the application point of each force in the sum. For your problem, you only have one force. The difficulty will become how does that force change over time, and how do you describe it mathmetically.
But the force isn't split into different roles. It plays both roles.