The Physics Factbook

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The purpose of this activity is to determine the minimum speed needed to successfully complete the loop shown in the clip "The Mini Loop" from the 2006 movie, Jackass Number Two.

Famous from the MTV show Jackass, Johnny Knoxville, Bam Margera, Chris Pontius, Steve-O, Ryan Dunn, and Jason "Wee Man" Acuna, come back to the screen for an action packed theatrical movie. Jackass Number Two is a sequel of the original Jackass movie in which a group of brave men come together to make a movie solely based on crazy and dangerous stunts.

A minibike is a miniaturized version of a motorcycle. The four stroke motor that runs the bike is similar to one of a lawnmower. The only difference is that the shaft direction. They were first used as "pit bikes" so that drag racers could efficiently travel to the pit stop. A decent minibike can reach speeds of over 60 km/h and also attain a fast acceleration because of its light body frame.

In this clip Ehren McGhehey and Thor Drake try to fully complete their homemade loop by riding on a minibike. |

Play the movie in DivX Player and find the scene where Danger Erhen is standing right besides the loop.

Open WadRuler [exe] and take the coordinates from Danger Erhens head, toe, top of the loop, and bottom of the loop.

Knowing that Danger Erhen is 1.8542 meters tall we can find the diameter of the loop. The distance between Danger Erhen's foot and toe is 6.62 centimeters and the diameter of the loop is 8.73 centimeters. We can use ratios to figure out the diameter.

(1.8542 m) / (0.0662 m) = x / (0.0873 m)

(6.62 m) x = (0.16187166 m^{2})

x = 2.4452 m

The diameter of the loop was found to be 2.445 meters.

To find the velocity needed to clear the loop, we made the centripetal force equal to gravity. This way we could find the final velocity at the top of the loop.

F = F

mg = mv^{2}/r

g = v^{2}/r

v = √gr

v = √(9.8 m/s^{2})(1.2225 m)

v = 3.461 m/s

With the final velocity we could find the initial velocity by using the conservation of energy formula.

KE_{f} + PE_{f} = KE_{i} + PE_{i}

½ mv_{f}^{2} + mgh = ½ mv_{i}^{2} + mgh

v_{f}^{2} + 2gh = v_{i}^{2} + 0

v_{i}= √(v_{f}^{2} + 2gh)

v_{i} = √((3.461 m/s)^{2} + 2(9.8 m/s^{2})(2.445 m))

v_{i} = 7.74 m/s

Thor Drake completing the loop.

We concluded that the minimum speed needed to clear the loop was 7.74 m/s (17.3 mph).

One source of error was that there is air resistance on the rider and on the bike. Therefore, the velocity needed to clear the loop would be greater. Another source of error is that we calculated so that the bike could complete the loop in a straight path. However, the bike is going in a angle. Therefore, the bike needs more speed to clear the ramp.

Lorant Lee, Karmen Ho, and Jeanne Chau -- 2007

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