## Saturday, November 30, 2013

### Punkin Chunkin

By Laura Aseltine
In a desperate attempt to find something to do for my physics news I googled “Weird News” hoping to find something I could apply my limited physics knowledge to and came across the annual event that is a physics lover’s dream known as “Punkin Chunkin.” The competition occurred on November 1-3 and a television special covering this year’s event aired on Thanksgiving day (unfortunately I was too busy eating to remember to watch). The ultimate goal is simple: how far can you chuck a pumpkin, but there are many different categories and specific rules for how you can chuck this pumpkin. Some of the categories include air, centrifugal, catapult, and trebuchet.

For my physics news I decided to look at the air cannon, which uses compressed air to propel the pumpkin (pretty self explanatory). The official Punkin Chunkin website lists the rules as follows:

1.     Pumpkins must weigh between 8 & 10 pounds.
2.     “Compressed air only"
3.     Pumpkin must be loaded before Pressurizing vessels, and an Official must see you load it.
4.     All Air inlets on vessel must have a Check Valve Installed.
5.     Horn or sound device must sound when firing down range for safety of spotter on the field.
This years winner of the air cannon category was a team by the name of American Chunker Inc who shot their pumpkin a world record distance of 4,694.68 ft. I decided to calculate what the initial velocity of that pumpkin must have been when it left the cannon.

What I know/Assumptions

·      I found online that the cannon is 120 ft long
o   120 ft  x 0.3048 m/1 ft = 36.576 m
·      I assumed the cannon is at a 70° angle
·      Calculated the initial height of the pumpkin
o   Sin70 x 36.576 m = 28.30 m
·      Distance traveled = 4694.68 ft
o   4694.68 ft x 0.3048 m/1 ft = 1430.94 m

-       To find the initial velocity I need to break it up in into x and y components
o   V0x= V0 cos70
o   Voy= V0 sin70
-       First I calculated the time it took to reach the maximum height
o   Vf= V0 + at
o   0 = V0 sin 70 – 9.8t
o   t = 0.0793 V0
-       Then I calculated the maximum height it reaches
o   Vf2 = V02 + 2a(Δy)
o   02 = (V0 sin70)2 + 2 (-9.8) Δy
o   -(V02 0.599) = -19.6 Δy
o   0.031 V02 = Δy
o   max height = (V02 0.031) + 28.3
-       Then I calculated the time it would take for the pumpkin to fall from the maximum height to the ground
o   Δy = V0t + ½ at2
o   – ((V02 0.031) + 28.3) = 0 + ½ (-9.8)t2
o   √.0063V02 + √5.776 = t
o   t = .0793V0 + 2.403
-       So the total time the pumpkin was traveling is…
o   t = 0.0793V0  + .0793V0 + 2.403
o   t = 0.1586V0 + 2.403
-       I then used this time to calculate the initial velocity in the x direction
o   Δx = V0xt + ½ at2
o   1430.94 = V0 cos70 (0.1586V0 + 2.403) + 0
o   1430.94 = 0.100V02 + 1.522V0
o   V0 = 119.38 m/s
-       I then converted this velocity to miles per hour
o   119.38 m/s x 1 mi/1609.34 m x 3600s/hr = 267.05 mph

#### 1 comment:

1. Is that right?!? 267 mph? Wouldn't that completely smash the pump before it even came out of the cannon? I think the fluid part of the pump would undergo a drastic hydrostatic shock at that speed which would smash it up. Anybody want to back me up ?!?