If you assume that the paper tape is zero mass and zero thickness, then the force applied to the left side of the spool can only be the result of tension in the tape. That force will be directly up and tangent to the surface of the spool.
We can deduce the tension on paper tape by noting that had there been a second tape attached to the right, the spool would be supported symmetrically between them - with each tape supporting exactly half the weight. The tension on the left tape can be expressed as a portion of the weight and as a function of time: L=f(t), where t=0 at the moment that the right tape is removed. So, for t<=0, L=f(t)=0.5.
For t>0, we need to look at what horizontal motion we should expect from the spool. And what we notice is that there are only two forces acting on it, both without a horizontal component. So, if that spool wasn't swaying at the start, nothing will make it sway once the drop starts. Of course, these two misaligned vertical forces on the spool will add angular momentum to the spool. But that continuous ramping up of spin will exactly match the continuous downward acceleration.
There's a very dynamic version of "steady state" here. If we put our origin {0,0} on the center of the falling spool, then we will see a steady pull in one direction from gravity, a steady pull of equal magnitude in the other from the tape, and a steady build up of spin.

So f(t)=0.5 for all t. And, except as part of spin, nothing moves horizontally at all.

What do you think: if you were a participant in the Physics Olympiad and you handed in such an explanation, would the examination board accept it as a solution?

wrobel said:

subj from an old Polish physics olympiad problem book

Can you tell which book is it from? Or maybe in which year it appeared on the olympiad (if it did).

Waldemar Gorzkowski

25 LAT
OLIMPIAD
FIZYCZNYCH

Warszawa 1979

wrobel said:

Yep, I have it Thank you.

.Scott said:

We can deduce the tension on paper tape by noting that had there been a second tape attached to the right, the spool would be supported symmetrically between them - with each tape supporting exactly half the weight.

I don't see how this helps. There is no reason to assume that the tension in the left tape would be be a continuous function throughout the cutting of the right tape, thus starting out at half the spool weight during the fall.

Consider the limit, where the moment of inertia of the spool ##I## goes to zero for a fixed mass ##M## (because the radius goes to zero, or the mass is concentrated at the center). Here, after cutting the right tape, the tension in the left tape would immediately fall to around zero, because only a tiny tension is needed to spin up the spool, so it can fall.

.Scott said:

The tension is obviously a function of ##I##, hence the problem specifies it's a "uniform cylinder".