Sharknado

The shark is kept afloat by a constant flux of momentum delivered at its surface by the air molecules in the wind stream.

$\displaystyle F = w_{shark}g = \frac{\Delta p}{\Delta t}$

If we slice the surface of the shark lengthwise into tiny rectangular strips, then each of these strips has a thin rectangular prism of wind hitting it at a constant rate.

Each prism has mass $\rho_{air}l_{length}l_{height}l_{width}$ where $\rho_{air}$ is the density of air and $l_{height}$ is the height of the rectangular prism. Each molecule in the air stream delivers an amount of momentum that depends on its radial position around the cylinder.

For instance, an air molecule hitting the center of the cylinder recoils with its initial velocity so that $\Delta v_y = 2 v_{wind}$ . At the edge of the cylinder, air hits the surface in a glancing collision and $\Delta v_y = 0$ .

In general, each particle hits the surface at and angle $-\theta$ relative to the radius at the point of contact and comes off at an angle $\theta$ , so that the change in vertical speed is given by $\Delta v(\theta) = v_{wind}\left(1+\cos 2\theta\right)$ .

In terms of the incoming angle, the width of each strip $l_{width}$ is given by the horizontal projection of the arc length $R\cos\theta d\theta$ .

Therefore, each prism of air delivers the momentum $\displaystyle \Delta p = \rho_{air}R\cos \theta \left(1+\cos 2\theta\right) v_{wind}l_{length}l_{height}d\theta$ . The last piece we need is the time scale $\Delta t$ over which the momentum is delivered.

At each point on the surface, the air molecules are arriving at speed $v_{wind}$ , if the prism has height $l_{height}$ then the time required is $\displaystyle \Delta t = v_{wind}/l_{height}$ .

Therefore, the momentum delivered per unit time to the piece of surface at angle $\theta$ is $\displaystyle d\frac{\Delta p}{\Delta t}\left(\theta\right) = \rho_{air}R\cos \theta \left(1+\cos 2\theta\right) v_{wind}^2l_{length}d\theta$ . The total rate of momentum delivery to the shark is then

$\displaystyle \begin{aligned}\frac{\Delta p}{\Delta t} &= \int\limits_{-\pi/2}^{+\pi/2} d\frac{\Delta p}{\Delta t}\left(\theta\right)d\theta \\ &= \rho_{air}R v_{wind}^2l_{length}\int\limits_{-\pi/2}^{+\pi/2} \cos \theta \left(1+\cos 2\theta\right) d\theta \\ &= \rho_{air}R v_{wind}^2l_{length}\frac83 \end{aligned}$

Therefore, we have $\boxed{\displaystyle v_{wind} = \sqrt{\frac{3w_{shark}g}{8\rho_{air}R l_{length}}}} =$ * 70.35 m/s *.

Relevant quantities

The weight of the shark is given by $w_{shark} = l_{length}\pi R^2 d_{shark}$ which is 4712.4 kg/m^3 using the numbers given.

The density of air is given by the molar weight of gas divided by the volume of one mole of gas which is $\frac{m_a}{RT/P} =$ 1.166 g/m^3 using the numbers given.

Consider a very small area dS in the surface of the shark and make an angle $\alpha$ with the horizontal plane.

After a period of t, the mass of air molecules bouncing off the shark is: $dm=\rho_a v t dS \cos \alpha$ . (with $\rho_a=\frac{P \mu}{RT}$ is the density of the air)

These molecules cause a pulse of: $dp=2dm v \cos \alpha$ .

Therefore, the vertical component of the force caused by these molecules is: $dF=\frac{dp \cos \alpha}{t}$

= $2v^2 \rho_a dS \cos^3 \alpha$

= $2v^2 \rho_a h r d\alpha \cos^3 \alpha$ .

The total force is: $F=\int dF=2v^2 \rho_a h r \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^3 \alpha d\alpha$

= $\frac{8}{3} v^2 \rho_a h r$ .

In order to keep the shark suspended, $F=mg=g \rho_w \pi r^2 h$ .

Therefore, $v=\sqrt{\frac{3g \rho_w \pi dRT}{16P \mu}} =70.3(m/s)$