Collection Spotlight: Tornado Spheres


I do apologize, in advance for the horrible photograph of these Tornado Spheres. It was quite difficult to take a photo of the desired effect, so just let your imagination run wild!

Tornado Spheres

The Tornado Spheres consists of two steel ball bearings that have been welded together. They are then placed on a flat surface and set off spinning. Using a regular straw, air blown to speed up the spinning spheres to very high rpm’s (up to 12,000 rpm). Using an LED torch (the smear of purple in the right-hand picture), you can get a pretty nice looking effect (which can be seen in the left-hand picture).

The key to getting the spheres to spin so quickly with just a straw is to have the straw tapered at the end. This tapering creates a nozzle, which is a very common design in physics.

To explain exactly why this happens will require a little knowledge of fluid mechanics and the Bernoulli Equation. But to put it simply, just think of it in terms of conservation of mass. What air goes in must come out. Since you constrict the flow at the end, the air has to speed up to ensure that the same amount of air is leaving the nozzle as what is going in. This causes the air at the end of the straw to leave at a higher velocity than what was entering the straw. In fact, the air can leave the straw at speeds of more than 400 km/hr! Seems absurd, so let’s show this by doing a little physics!

We first need to make a few assumptions:

  • The flow of air through the straw is incompressible: In other words, the density of air through the straw remains constant. This assumption is valid because we are dealing with very low pressure (air coming out of your mouth)
  • The friction between the air and the inside of the straw is negligible: This will make our calculations very easy to do and is also a good assumption to make. The length of straw is very small, and the straw is perfectly straight, so it won’t have any impact on the air flow

The equation we will be using is the Bernoulli equation, but in order to do the calculations, we will need two pieces of information:

  1. The pressure differential between the inside of your mouth and the atmosphere: After a little bit of searching and some guesstimating, this comes up to around 1.3 psi or about 9000 Pascals
  2. The speed of the air leaving your mouth when blowing through the straw: Again, a little estimating will give you an answer of roughly 1.5 m/s
  3. The straw remains perfectly level: That means that there will be no change of potential energy of the air. While this is not always true (the straw can sometimes be held at an angle), the very small height difference will make a very small impact on the air velocity

Now, we can solve Bernoulli’s Equation below. You can solve it between two points in space, #1 and #2. We’ll take point #1 as the start of the straw, and point #2 as the end of the straw. We’ll assume that the temperature of air inside the straw is around 30°C. This is important because we need to know the density of air at that temperature.

\(
P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 =P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2\\
\text{where:}\\
P = \text{Pressure in Pascals}\\
\rho = \text{Density of air }= 1.164 \; \text{kgm}^{-3}\\
v= \text{Velocity of the air in ms}^{-1}\\
g = \text{Acceleration due to gravity }= 9.81 \;\text{ms}^{-1}\\
h = \text{Height in meters}\\
\)
\(
\text{Since the height is constant, } h_1 = h_2 \text{ and the equation simplifies to:}\\
P_1 + \frac{1}{2} \rho v_1^2  =P_2 + \frac{1}{2} \rho v_2^2\\
\text{We now re-arrange to get the outlet velocity, } v_2 \text{ :}\\
v_2 = \sqrt{\frac{2(P_1 – P_2) + \rho v_1^2}{\rho}}
P_1 – P_2 \text{ is the pressure differential, } \delta P\\
v_2 = \sqrt{\frac{2\Delta P + \rho v_1^2}{\rho}}\\
\text{Solving:}\\
v_2 = \sqrt{\frac{2(9000) + (1.164)(1.5)^2}{1.164}}\\
v_2 = 124.36 \text{ms}^{-1}\\
= 448 \text{kmhr}^{-1}
\)

The Bernoulli Equation is an extremely simple equation, yet incredibly powerful. It lies at the heart of fluid mechanics. It is simply an energy-balancing equation because energy must be conserved. But what you can see from the above is that the air exiting the straw travels at a speed of more than 400 km/hr, which is impressive. Catch is, it’s such a small amount of air and will quickly slow down as it exits the straw that you don’t really notice its speed.

This high speed of air that is exiting the straw is critical in getting the spheres spinning at a very high speed. I used the convex mirror that was provided with my Euler’s Disk. I found it helped keep the spinning spheres near the center, and also had a similar effect to when a disk was spun on it.

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