Why does the rope fly up from the nail against gravity? 
Why does the rope fly up from the nail against gravity? As shown in the picture, the blue arrow is the nail fixed on the wall, the yellow arrow is the rope, and the green arrow is the iron block. The rope on the right side of the nail moves downward under the action of the iron block, and the rope on the left side of the nail moves upward. At some point, the rope goes up and away from the nail. Why? How to calculate the height away from the nail?
Video of rope rising into the air




 A: According to @weeeeliam's answers to the following questions:
What is the formula for calculating the tension of the rope section?
There is a circle of rope that rotates at a uniform angular velocity $ω$. What is the formula for calculating the tension of the rope section? Without gravity, the density of the rope is $ρ$, the radius of the rope circle is $R$, and the section radius of the rope is $r$.

@weeeeliam:
Consider an infinitesimally small section of the of the string $d\theta$. The following diagram illustrates this: 

The tension is of the same magnitude throughout the rope, and it acts perpendicular to the vector from the center of the string to the point of action.
From this diagram, you can tell that only the x-components to the left matter, since the y-components of the tensions cancel out. The x-components of the two tension vectors must be equal to the force required for centripetal acceleration.
$$2T\text{ sin}(d\theta/2) = Td\theta = (dm)\omega^2R$$
The small bit of mass can be found as follows:
$$dm = \rho dV = \rho A dx = \rho \pi r^2 (r d\theta) = \rho \pi r^3 d\theta$$
Then, cancelling out the $d\theta$ on both sides of the equation:
$$\boxed{T = \pi \rho r^3 \omega^2 R}$$
Note: This solution assumes $r << R$.
Here is my explanation: suppose the rotation is uniform, then

As shown in the figure, if
$T=πρr^3ω^2R＞ρHπr^2$
$⇒$
$rω^2R > H$
Then the rope will go up.
