This is purely a "for-fun" question. I was minding my own business in the washroom this morning when I began to unroll some toilet paper from the roll, and in typical Breaking Bad fashion (sorry if you don't understand the reference) I had a serious moment where I wondered how the rate of change of the volume of the roll of paper changes as it is unrolled. Obviously the circumference of the cylinder keeps decreasing and as a result the ROC changes as well. I thought about it a bit but got stumped when I started writing things out (my math skills are a bit rusty).
I believe the beginning part of this solution Accounting for changing radius of a paper roll to always unroll the same amount of paper is the correct starting point for this question.
For simplicity, assume that the paper is rolled onto itself – no cardboard piece in the centre and the innermost layer's circumference approaches 0.
Also, assume that the roll is being unravelled at 1 circumference per second ($2\Pi r$/s) of the outermost (initial) layer.
Lastly, assume that there are 100 layers (100 circles as defined in the mentioned question) and each is 1 unit-of-your-liking.
Feel free to adjust any of these parameters to your liking!
Best Answer
If the roll is rotating at a constant angular rate, you will remove one layer per second all the way down. As you started with $100$ layers, it will take $100$ seconds to unroll. The remaining volume is proportional to the square of the remaining time. If the starting volume is $V$, at time $t$ (seconds) there is $V\frac {(100-t)^2}{100^2}$ of the initial volume remaining.
Added: if you want the paper removed at a constant rate, the angular velocity goes up as $\frac 1r$. This is because the linear velocity is constant and $v=r\omega$. Then we have $\frac {dr}{dt}=-k\frac 1r, \frac 12r^2=C-kt, r=\sqrt{2(C-kt)}$. With $r(0)=r_0, r(100)=0$ we have $C-100k=0, C=\frac 12r_0^2,\\ r=r_0\sqrt{1-\frac t{100}}$