If the empty tape shaft, that is, the drive shaft, rotates at a constant angular velocity, is the linear velocity constant It's a hypothesis

If the empty tape shaft, that is, the drive shaft, rotates at a constant angular velocity, is the linear velocity constant It's a hypothesis

The driving force for the tape to move forward is the leading shaft, not the empty belt shaft. The function of the empty belt shaft is to "take up the tape", that is, to pull the tape forward with an appropriate amount of force, so as to make the tape roll smoothly on the shaft. The shaft does not rotate at a constant angular speed. At the beginning, it turns fast, until there are more tapes on it, It turns slowly. Why? Because if it drives the tape, it can't move at a constant speed
The main shaft rotates at a constant speed. It and a rubber pulley clamp the tape and make the tape move forward at a constant speed

When the tape recorder works, the radius of the tape reel on the tape axis will be reduced by half after 10 minutes, and then it will take time to reduce by half?

2.5 min. you can calculate by the area reduced. 3 / 4 of the area is reduced in 10 min. at this rate, it will take 2.5 min to reduce 3 / 16 of the area

When listening to the tape recorder, I found that the radius of the tape reel on the tape axis was reduced by half after 30 minutes, and the axis was set

Analysis: if the thickness of the tape is D and the original radius is r, the total length is L = (π R2 - π R2) / D, so we can calculate the length of the tape according to the side area. If the speed of the driving wheel is fixed, the time is proportional to the length of the tape
Original tape area S1 = π * (R1 ^ 2-r2 ^ 2) (R1 = 2 * R2)
The area of the tape wound after S2 = π * (R2 ^ 2-r3 ^ 2) (R3 is the final radius)
(r2-r1) / T1 = (r3-r2) / (2 * T2) (considering the thickness increase in the radial direction)
T 2 = 5 minutes