Question

During a porn shoot involving a boy and a girl, a camera man notices that the girl is riding vertically on top of the boy. The up and down periodic motion of the girl stops as the boy releases an obscene amount of cum (of density $\rho$) in her pussy (assume cylindrical of radius $b$) filling it completely, the cum in pussy surrounds the boy's dick (also cylindrical of radius $a$) forming a coat of liquid film. The cum film slides under gravity and the cameraman observes that the speed of cum film at radius $r$ is given by: $v = \frac{\rho gb^2}{2\eta}ln(\frac{r}{a}) - \frac{\rho g}{4\eta}(r^2 - a^2)$

The cameraman knows coefficient of viscosity of cum is $\eta$ and he wants to find:

i) The force on unit length of the boy's dick due to the viscous cum.

ii) The integral to calculate the volume flow rate of the cum down the boy's dick. (He may not evaluate the integral as he is lazy)

Answer 1

i) The force on unit length of the boy's dick due to the viscous cum.

ii) The integral to calculate the volume flow rate of the cum down the boy's dick.

Answer 2

The problem describes the flow of a viscous fluid (cum) in an annular region between two concentric cylinders (boy's dick and pussy wall). The fluid slides under gravity, and its velocity profile is given. We need to calculate the viscous force on the inner cylinder (dick) and set up the integral for the volume flow rate.

i) The force on unit length of the boy's dick due to the viscous cum.

The force exerted by a viscous fluid on a surface is determined by the shear stress at that surface. According to Newton's law of viscosity, the shear stress ($\tau$) is given by: $\tau = \eta \frac{dv}{dr}$ where $\eta$ is the coefficient of viscosity and $\frac{dv}{dr}$ is the velocity gradient.

The given velocity profile is: $v = \frac{\rho gb^2}{2\eta}ln\left(\frac{r}{a}\right) - \frac{\rho g}{4\eta}(r^2 - a^2)$

First, we calculate the velocity gradient $\frac{dv}{dr}$: $\frac{dv}{dr} = \frac{d}{dr} \left[ \frac{\rho gb^2}{2\eta} (\ln r - \ln a) - \frac{\rho g}{4\eta}(r^2 - a^2) \right]$ $\frac{dv}{dr} = \frac{\rho gb^2}{2\eta} \left(\frac{1}{r}\right) - \frac{\rho g}{4\eta}(2r)$ $\frac{dv}{dr} = \frac{\rho gb^2}{2\eta r} - \frac{\rho g r}{2\eta}$ $\frac{dv}{dr} = \frac{\rho g}{2\eta} \left( \frac{b^2}{r} - r \right)$

The force on the boy's dick is due to the viscous cum at its surface, which is at radius $r=a$. We evaluate the velocity gradient at $r=a$: $\left(\frac{dv}{dr}\right)_{r=a} = \frac{\rho g}{2\eta} \left( \frac{b^2}{a} - a \right) = \frac{\rho g}{2\eta a} (b^2 - a^2)$

Now, calculate the shear stress at the dick's surface ($r=a$): $\tau_a = \eta \left(\frac{dv}{dr}\right)_{r=a} = \eta \left[ \frac{\rho g}{2\eta a} (b^2 - a^2) \right]$ $\tau_a = \frac{\rho g}{2a} (b^2 - a^2)$

The force on unit length of the boy's dick ($F'$) is the shear stress multiplied by the surface area per unit length. The surface area per unit length of the dick is its circumference, $2\pi a$. $F' = \tau_a \times (2\pi a)$ $F' = \left[ \frac{\rho g}{2a} (b^2 - a^2) \right] \times (2\pi a)$ $F' = \pi \rho g (b^2 - a^2)$ Since the cum is flowing downwards, it exerts a downward viscous force on the boy's dick.

ii) The integral to calculate the volume flow rate of the cum down the boy's dick.

The volume flow rate ($Q$) through a cross-section is found by integrating the velocity over the area of flow. The cum flows in an annular region between the inner radius $a$ (dick surface) and the outer radius $b$ (pussy wall).

Consider an elemental annular ring of radius $r$ and thickness $dr$. The area of this elemental ring is $dA = 2\pi r dr$. The volume flow rate through this elemental ring is $dQ = v \cdot dA = v (2\pi r) dr$.

To find the total volume flow rate, we integrate $dQ$ from the inner radius $a$ to the outer radius $b$: $Q = \int_{a}^{b} v (2\pi r) dr$ Substitute the given expression for $v$: $Q = \int_{a}^{b} \left[ \frac{\rho gb^2}{2\eta}ln\left(\frac{r}{a}\right) - \frac{\rho g}{4\eta}(r^2 - a^2) \right] (2\pi r) dr$

The final answer is $\boxed{\text{i) Force per unit length} = \pi \rho g (b^2 - a^2) \text{ (downwards)} \\ \text{ii) Volume flow rate} = \int_{a}^{b} \left[ \frac{\rho g b^2}{2\eta} \ln\left(\frac{r}{a}\right) - \frac{\rho g}{4\eta}(r^2 - a^2) \right] (2\pi r) dr}$.

Explanation of the solution:

i) Calculate the velocity gradient $\frac{dv}{dr}$ from the given velocity profile. Evaluate this gradient at the surface of the boy's dick ($r=a$). Use Newton's law of viscosity $\tau = \eta \frac{dv}{dr}$ to find the shear stress at $r=a$. The force per unit length is the shear stress multiplied by the circumference of the dick ($2\pi a$). ii) The volume flow rate $Q$ is the integral of velocity $v$ over the cross-sectional area of flow. For an annular flow, consider an elemental ring of radius $r$ and thickness $dr$ with area $dA = 2\pi r dr$. Integrate $v \cdot dA$ from the inner radius $a$ to the outer radius $b$.

Answer:

i) The force on unit length of the boy's dick due to the viscous cum is: $F' = \pi \rho g (b^2 - a^2)$ This force acts downwards, in the direction of the cum flow.

ii) The integral to calculate the volume flow rate of the cum down the boy's dick is: $Q = \int_{a}^{b} \left[ \frac{\rho gb^2}{2\eta}ln\left(\frac{r}{a}\right) - \frac{\rho g}{4\eta}(r^2 - a^2) \right] (2\pi r) dr$