Actually you proved, with the third equation, that the translational speed does not depend on the radius $r$ of
the cylinder, or on its mass $m$, but only on $g$ and $\alpha$. Indeed it reads like a differential equation defining the motion of the cylinder.
If the translational speed of the cylinder axis does not depend on the radius, then the angular speed has to depend on it, since they are related by the relation $v=\omega r$.
I do not see how that fact can be overcome by any computation
including the total momentum of the system. This momentum can be
computed from the speed at distance $l$, but will depend on the radius
$r$, because it is squared in the moment of inertia, but not in the
angular speed. Getting the total momentum by other means, such as resolving the differential equation will not do any better. Imho (but I have not practiced this sport for a long time).
But having $r$ in the expression of the angular speed is not a crime.
BTW What are you supposed to do with the coefficient of friction $\mu$ ?
Do you have to check the cylinder does not slip?
The problem with attempting to do the analysis with the forward point of contact on the box when it is sliding is that the box is accelerating. This makes a non-inertial frame and there's more moving parts. Besides the force of gravity on the center of mass, there will be fictitious forces.
First, lets assume friction is zero. If so, we can calculate the acceleration of the box.
$$F_{x'} = mg\sin \theta$$
$$a_{x'} = g\sin \theta$$
In an accelerating frame, we get fictitious forces.
$$ F_{fict} = -ma_{frame}$$
$$ F_{fict} = -mg\sin \theta$$
The sign indicates the force is opposite the direction of the acceleration. The case where we expect the box to tip most likely is where it is tall and not wide. Let's assume the width is minimal (a rod sliding on its end). If so, the torque from gravity will be the COM at half the height, and leaning forward $\sin \theta$
$$\tau_g = \frac12 hmg \sin \theta$$
Meanwhile the fictitious force is located at half the height and points opposite the acceleration vector, so it has a lever arm of exactly half the height.
$$\tau_{fict} = F_{fict} d$$
$$\tau_{fict} = -\frac12 hmg\sin \theta$$
$$ \tau_{net} = \tau_{g} + \tau_{fict}$$
$$ \tau_{net} = \frac12 hmg \sin \theta - \frac12hmg \sin \theta$$
$$ \tau_{net} = 0$$
If you add friction, you will reduce the acceleration of the box, reducing the fictitious force in this frame. When that happens, the net torque would be sufficient to tip a rod, but would depend on the dimensions if a box would tip over.
I didn't do the math, but $\tau_{fict}$ should be pretty simple to calculate, and then $\tau_{g}$ becomes $\frac12h\sin \theta - \frac12w\cos \theta$ (I think)...
Sorry, I wanted to reply to this comment, and I couldn't do it in the comment space.
If we instead take the static plane to be the frame of reference (and
a minimal width box), wouldn't we get the same results anyways?
The problem is that in the frame of the plane, the axis you want to consider in the question (the front corner of the box) is accelerating. I don't think that's okay to do. Let's imagine a pendulum being accelerated via a force on the axis. And we let it reach steady state so the angle isn't changing.
The solid pendulum has a force accelerating it to the right, and it has gravity pulling it down. The force is applied directly at the axis and contributes no torque. Gravity pulls the weight down, so is applying a torque of $(mg \times L\sin \theta)$. Yet the pendulum does not rotate. Perhaps someone else has more information about this situation. My intuition is that this is simply the wrong technique for an accelerating axis, but I'm not sure what the correct analysis would be.
Best Answer
That is only true for rotational motion. As it stands there is nothing in your problem to suggest that any sort of rotational motion will happen. Since the plane is "FRICTIONLESS" (your words), it does not do the normal "rolling without slipping" motion but instead slips completely, sliding down without any rotation whatsoever.
This is why, as you say, the net torque about the center will be zero and angular acceleration will be zero.
Angular momentum will may or may not be conserved at the bottom of the incline; it depends on whether the new ground that it is rolling onto is frictionless (hence torquing the object) or not.