# Can popping a balloon demonstrate that air has mass?

Introduction

My 7 year old daughter has been learning about states of matter in school and had to do an experiment to see if gasses have mass. The experiment involved balancing 2 inflated balloons on either end of a stick and then popping one of them. The idea being that, because all the air escapes from the popped balloon, the un-popped balloon will be heavier (because of the weight of air inside the balloon. The video of her experiment is linked here.

I posted this in my junior level fluid mechanics class discussion board and asked my students if this was a valid experiment to show that gasses have mass. My daughter was convinced (and excited to get to pop a balloon for class). I actually asked my students a more directed question which was:

Watch the video carefully and then give a reason why the experiment does not show what it claims to show. That is, what is the main reason that the un-popped balloon falls? Explain why, even if the experiment  was run perfectly, the effect of the release of air [on the balance] would be tiny.

Next time I will ask a more open ended question. I got a few responses but, given that I did this near a test in the fall COVD semester, not as many as I had hoped. The two main answers were:

1. The balloon fragments and so there is less latex on the popped side of the balance compared to the un-popped side so it falls toward the un-popped side
2. There is an impulse generated by the balloon popping. This was not symmetric because of how the balloon was popped and ruptured. This generated an impulse and a temporary imbalance which was followed by the balance arm (the stick) sliding through the ribbon that was holding it up.

However, there is a third reason that this experiment would barely register in imbalance even if run perfectly. Below is a quick analysis of why.

Analysis

Consider a perfectly balanced pair of balloons as shown in the diagram below. We assume that the balance mechanism is massless and symmetric as shown below

The balloons have the same mass and volume. A free body diagram of the green balloon before and after popping is shown in below.

The five forces acting on the balloon are the mass of the latex (balloon fabric), the mass of air in the balloon, the force due to the support, and the buoyancy forces. The buoyancy forces are due to the hydrostatic pressure acting on both the balloon fabric and the air in the balloon. Before the balloon is popped the force balance is given by

Therefore, the force due to the support is

Assuming that the balloon stays intact and that the change in volume of the latex as a result of deflation is negligible then the force balance after the deflation is

The support force is, therefore

As the changes in the weight and volume of the balloon fabric from the popping are negligible, we can calculate that the change in the magnitude of the force due to the support is

The weight of air in the balloon is

Where the subscript ‘air.B’ indicates the air in the balloon. The buoyancy force is

Where the subscript ‘air.a’ indicates the ambient air. Therefore, the change in the support force is

So there will be a change in the force not equal to the weight of air leaving the balloon being taken out of the force balance, but only due to the change in density in the air due to decompression. This is likely quite small.

To quantify this number we need to know the pressure inside the balloon. Fortunately there are research papers on this topic (https://www.researchgate.net/publication/239815213_Balloons_revisited). The answer to what is the pressure in the balloon is, obviously, that it depends on how full the balloon is. The paper linked above shows that there is an initial rapid increase in internal pressure followed as the balloon inflates by a slow decrease until some critical radius is reached beyond which it again increases rapidly with radius. We used 9” balloons (23 cm) and we did not overinflate them. We also didn’t test the elastic properties of the balloon fabric so the pressure we assume is only an estimate. But, if we inflated it to a radius of 10 cm then the internal pressure would be around 1.5 kPa. Assuming a room temperature of 22oC and atmospheric pressure (101.3 kPa) the density of dry air is 1.19565 kg/m3. At a gauge pressure of 1.5 kPa the density increases to  1.21336 kg/m3. I used https://www.omnicalculator.com/physics/air-density to get the densities. Plugging these densities and the volume of a 11 cm sphere into the last equation leads to

This is obviously very small and certainly undetectable by any domestic scale. It is equivalent to the weight of 0.06 ml of air.

So, ultimately, if you ran the perfect experiment with a perfect balance you would see an effect due to the balloon popping. But it is not as simple as removing the weight of air in the balloon from the force balance.

https://www.scientificamerican.com/article/stretchy-balloon-science/

https://aapt.scitation.org/doi/10.1119/1.11486

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.

Can popping a balloon demonstrate that air has mass?

# Rotational lift – Magnus effect on golf balls

Many science museums have demonstrations in which a ball is levitated by an air jet. Typically the ball is quite light and the air jet quite broad. These setups can produce reasonably stable results with the ball staying supported by the air jet. They even work when the air jet is inclined. There are occasionally explanations associated with the demonstration talking about how the drag from the air jet supports the weight of the ball and as it moves off-center then Bernoulli means that there is a low pressure on the jet side of the ball that draws it back in. This is often unsatisfying as an explanation. It also ignores the fact that, when the air jet is inclined, the ball typically rotates. To get at this in a little more detail I tried it with a golf ball.

Equipment

1. Shop-vac that can blow air
2. golf ball
3. a steady hand or some sort of mount for the hose outlet. I had a spare trolley that I could have a mount built on. See figure 1 below for the setup. Figure 1. image of the shop-vac and adjustable mount for the hose outlet

The mount shown is adjustable so that one can change the outlet flow angle. The shop-vac I used has an outlet nozzle diameter of 2.5 cm and the air speed, measured 25 cm downstream of the outlet, was 36 m/s. The main downsides of this are that it is quite noisy and also it can be tricky getting the ball to sit in the air stream.

Demonstration

1. turn on the shop-vac and place the ball in the air stream
2. release the ball such that it remains supported by the flow (This is a lot easier said than done).
3. observe the behavior
4. slowly adjust the angle of the air jet (I started at vertical and adjusted from there).

Animated gifs of this demonstration for three different jet angles are shown in figure 2.   Figure 2. Animated gifs of the demonstration for three different jet angles showing the different behavior for each angle. the uninterrupted outlet velocity is the same for each case.

Observations

1. when the air stream is vertical the ball is quite unstable and, after bouncing around for a while, it falls out of the stream. This differs from many museum exhibits where the ball is lighter and is able to change rotational direction more rapidly.
2. when the air jet is not vertical the ball is more stable. It rotates rapidly and oscillates backward and forward along the line of the air jet.

Qualitative explanation

The rotation of the ball is key here. The balls rotation deflects the air jet changing the momentum of the flow (see figure 3). To do this the ball must apply a force normal to the direction of the incoming air jet. The reaction to this is a lift force on the ball normal to the direction of the incoming air stream (see figure 4). This is the same process used to throw curve balls (see Lift, Boundary layer separation, and curve balls).

The vertical components of the drag and lift both act upward and together balance the weight of the ball. Figure 3. Schematic diagram showing the incoming air stream, the ball rotation, and the resulting wake deflection. Figure 4. (Left) free body diagram of the ball showing the drag, lift, and weight forces. (Right) force triangle showing the force balance when the ball is stable.

The rotation of the ball is established by the air stream. When the ball is placed in the air stream, if it is not rotating, it will fall out. As it does, the air flow across the top of the ball is substantially faster than across the bottom. This drives the ball to rotate which in turn deflects the wake downward and generates the lift force. This is also why it can take a few goes to get the demonstration to work as the rotational inertia of the golf ball limits the rotational acceleration.

Back of the envelope calculation

A standard golf ball has a mass of 46 grams and a diameter of 43 mm. I assumed the density of air to be 1.25 kg/mand the measured air velocity from my set up was 36 m/s and I took the jet angle to be 45 degrees. In this case the sum of the forces in the vertical direction becomes

mg=0.5ρu2A sin45 (CD+CL)                                                                        (1)

Substituting values into (1) leads to CD+CL=0.54. The Reynolds number is approximately 105 which is in the region of the drag crisis for a golf ball (Link) so the coefficient sum would appear to be reasonable.

Application

There are lots of circumstances in which the rotation of a compact object causes a lift force. The most obvious is the in-flight curve of a golf ball when not hit perfectly. One application I am working on is the lift-off and flight of compact debris in severe storms. In particular I am interested in the conditions under which loose-laid roof gravel is removed during hurricanes and tornadoes. On possible mechanism is that the wind shear on the gravel surface pushes a gravel piece to roll over its downwind neighbor. If it is deflected up during this process then the large velocity gradient (wind shear) near the surface could generate the rotation required to generate a lift force and launch the gravel up into the wind field. My student and I will be doing testing on this, and other potential mechanisms, at the Florida International University Wall of Wind Experimental Facility over the next few years.

The research and outreach project (including this post) are based upon work supported by the National Science Foundation under Grant No. 1760999. Any opinions, findings, and conclusions or recommendations expressed in the material are those of the author and do not necessarily reflect the views of the NSF.

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.

# Video of “Solid body rotation – measuring the rotational velocity of a potting wheel”

Here is a link to a video from the “Solid body rotation – measuring the rotational velocity of a potting wheel” demonstration. The video shows the parabolic cavity forming as the water spins up until a steady solid body rotation is achieved. In this video the wheel was slightly wobbly so the cavity was not perfectly parabolic.

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.

# Solid body rotation – measuring the rotational velocity of a potting wheel

This demonstration uses the pressure distribution in a fluid undergoing solid body rotation to estimate the rotational speed of the cylinder.

Pre-demonstration calculations

Under solid body rotation about a vertical axis the horizontal variation in pressure in the radial direction is

dP/dr=-ρa=ρω2r

The pressure in the vertical direction is also hydrostatic resulting in the free surface at the top of the water becoming curved concave up. Taking the origin to be at the free surface on the axis of rotation then the pressure along a horizontal radial line from that point will be

P=ρω2r2/2

such that the height to the free surface above that point is

z=P/ρg=ω2r2/2g.

For a cylinder of radius R and depth H when not rotating the water will rise such that the total distance from the new axial water surface to the surface at the edge of the cylinder is given by

h=ω2R2/2g

when fully spun up. The volume of the fluid is initially V=πR2H. This volume is conserved. The volume of the paraboloid of air is given by  V=πR2h/2. which is also the volume of the water above the height of the axial free surface. Denoting the distance that the water surface drops on the axis of rotation by δ we can write that the volume of water above this point must be the same regardless of whether the water is rotating or not. Therefore,

πR2δ=πR2h/2.

or

h=2δ.

That is, the water level at the side of the cylinder rises by the same height that the water level drops on the axis of rotation. Therefore, if the initial depth of water on the cylinder is marked prior to the experiment then one can measure the amount that the water level rises on the cylinder side (δ) using a marked scale and then back calculate the angular velocity as

ω=(4gδ)½/R

The figure below defines the dimensions used above. Equipment

You will need

1. a clear cylinder marked with a vertical scale that passes all the way around the cylinder (I bought a large cookie jar and used thin tape to make the scale and mark the initial depth of water I needed)
2. a method for rotating the cylinder at a constant rate (I was lucky enough to get a repaired potting wheel for the cost of a replacement switch and potentiometer).
3. A stopwatch to estimate the actual rotation rate

Demonstration

1. Fill the cylinder to the marked initial depth
2. start the cylinder rotating at a constant angular velocity such that an air cavity forms as described above. Ensure that the rotation speed is small enough that the cavity does not extend down to the base of the cylinder and that the top of the water surface does not reach the top of the container.
3. measure the height of the free surface above the marked initial depth line
4. measure the angular velocity by counting the number of revolutions over some short period of time.
5. calculate the angular velocity using the equation above and compare.