We looked at the physics behind launching pumpkins with a pneumatic air cannon in the last issue of The SPS Observer, but now it’s time to go old-school—dropping pumpkins from the roof! For maximum splatter, we recommend filling the pumpkins with liquid nitrogen!

…Did you say liquid nitrogen?

Yes! Nitrogen has a boiling point of 77 K (brrr!), so its liquid form is called a cryogenic. For some points of reference, pure water freezes at around 273 K, and the lowest recorded natural temperature is around 184 K (at Vostok Station on the Antarctic Plateau).1 Many physics experiments use liquid nitrogen (LN2) as a coolant, but it’s also perfect for freezing a pumpkin into a solid block of shatterable organic matter, as seen in Fig.1.

(Obvious) Disclaimer: Do not use LN2 without proper protection, training, and the appropriate permission!

Pumpkin Physics

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Figure 1 - Pouring liquid nitrogen into pumpkins. Pretty cool! Photo courtesy of Ravn Jenkins.

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Figure 2 - Illustration4,5,6 of correlating cross-sectional pumpkin shapes, with cross-sectional area A and volume V.

For science, let us suppose we want to calculate the force a pumpkin will exert on the ground at the moment of impact after falling from some known height. Let’s start with considering a pumpkin in free fall. A pumpkin of mass m will experience a gravitational force downward, Fg = mg, and a drag force Fd in the opposite direction of motion:

F_net = F_g - F_d

If the pumpkin is assumed to be moving through a uniform fluid (air) at a velocity v and has a cross-sectional area A perpendicular to the air flow, we can approximate the net force on the pumpkin F_net as 
F_net = mg -½ ρv² C_d A

where ρ is the mass density of air (1.225 kg/m3) and Cd is the drag coefficient of the falling object.

While there are many ways of doing this, we can estimate the average velocity with some good old conservation of energy. Before the pumpkin is released from rest at height h, the pumpkin has energy stored as gravitational potential energy U = mgh. Let’s assume that all of the pumpkin’s potential energy is converted to kinetic energy right before it hits the ground, which is not a poor assumption:

mgh = ½ mv²

Then the final velocity of the pumpkin should be given by

v = √2gh

To find the cross-sectional area, we must first consider the pumpkin’s shape. Pumpkins have three common shapes: oblate spheroid, prolate spheroid, and sphere. Fig. 2 shows each pumpkin shape and the corresponding cross-sectional area A. For the sake of this discussion, let us assume that the pumpkin is falling stem-up and will not spin. If you want a challenge, look up how spinning affects drag.^2,3

Finally, we must decide on the drag coefficient for the pumpkin. While there are a lot of complexities, Fig. 3 shows a simplified list of measured drag coefficients for various shapes, with an approximate Reynolds number of 10⁴, as the size of the object affects the value.

The drag coefficient of a sphere, C_d = 0.47, is a close-enough estimate for the drag coefficient of all pumpkin shapes.

Putting it all together, we find the final value for the force on the pumpkin right before it hits the ground:

F_net = mg - ρghC_d A

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Figure 3 - Drag coefficients for various shapes7.

Figure 3 - Drag coefficients for various shapes7.

Now, for the Pumpkin Puzzler

A. Chucking the squash: Say we’re dropping the pumpkin pictured in Figure 1 off the roof of the physics building. If the height and equatorial circumference of the pumpkin are 30 and 65 cm, respectively, the height of the roof is 9.1 m, the hollowed pumpkin has a mass of 6.2 kg, and we fill the pumpkin halfway with liquid nitrogen, what will the acceleration of the pumpkin be right before it hits the ground? More importantly, how does this differ with an uncarved pumpkin? Does mass even matter (pun intended)?

Hints: What shape is the pumpkin? The density of liquid nitrogen is 808.4 kg/m³ at 25℃ at 1 atm.⁸

B. Some devilish differentials: How would we calculate the velocity without using conservation of gravitational potential energy to estimate?
Hint: Use the identity ∫₀x 1/(a - bx’² ) dx’ = (tanh^-1 (√a/b x)/√ab).
Check your answers against our solutions.⁹ 

References:

1. World: Lowest Temperature, World Meteorological Organization, archived from the original June 16, 2010, retrieved December 10, 2013, from https://web.archive.org/web/20100616025722/
http://wmo.asu.edu/world-lowest-temperature.
2. Nathan, A., The Effect of Spin on the Flight of a Baseball, Am. J. Phys., vol. 76, no. 2, 2008.
3. Kagan, D. and Nathan, A., Statcast and the Baseball Trajectory Calculator, Physics Teacher, vol. 55, no. 3, pp. 134–136, 2017.
4. Spheroid diagrams by Ag2gaeh [CC BY-SA 4.0
(https://creativecommons.org/licenses/by-sa/4.0)], from Wikimedia Commons.
5. By Marina Vladivostok [Public domain], from Wikimedia Commons.
6. By Petr Kratochvil [CC0 Public domain], from Public Domain Pictures.
7. By TheOtherJesse [Public domain], from Wikimedia Commons.
8. Nitrogen—Thermophysical Properties, Engineering ToolBox, accessed November 1, 2018, from https://www.engineeringtoolbox.com/nitrogen-d_1421.html, 2008.
9. https://www.spsnational.org/sites/default/files/files/publications/obser....

More Halloween Fun

Bookmark these activities for your chapter’s fall calendar

Ghoulish Galileo: Bring your pumpkin drop to the next level by replicating Galileo’s experiment! Will two pumpkins dropped at the same time hit the ground simultaneously? Bonus points if someone dresses up as Galileo!

Spooky Physics Conference: Once a year, William and Mary’s chapter of SPS transforms into the SSP, the Society of Spooky Physics! Undergraduates present satirical talks of physical explanations for the paranormal to a panel of esteemed spooky science experts (aka graduate students and professors in costumes). The graphs are ghoulish, the tables are terrifying, and the equations will haunt your very soul!