Thursday, November 15, 2012

The Physics of Olympic Trampolining

By Clara Slight
This summer, I loved to watch the Olympic Games. One sport that I previously didn’t know existed but found very interesting was Trampolining. I realized that this sport is full of physics and decided to take a look at it from two perspectives: conservation of energy and rotational motion. Take a look at a video from the 2004 Olympic Games (most clear example I could find):

First, I took a look at the conservation of energy when a person jumps on this trampoline.  I analyzed the physics from when the person gained their maximum height and fell from this height. The athlete jumps approximately 7 m in the air at the highest point, and weighs approximately 80 kg. So their potential energy at the highest point is PEgrav = mgh = (7 m) (9.8 m/s2) (80 kg) = 5488 J. When they fall from the highest point, all of this potential energy is transferred into kinetic energy and then stopped by the trampoline. You can find the velocity at the fastest point right before they hit the mat by transferring all of the PE into KE. PEgrav= mgh = 5488 J = KEbottom = 1/2mv2. So 5488 = 1/2 mv2 and solving for the final velocity gives you 11.7 m/s.

Then you can find the spring coefficient of the trampoline. All of the potential energy is transferred into kinetic energy and then into spring kinetic energy, which then bounces the athlete into the air again. So KEspring = 5488 J = 1/2kx2, and the jumper stretches the trampoline about .5 m, so the spring coefficient can be found here. (Note-the effect of gravity once the athlete hits the trampoline is not being considered since he or she is so close to the ground – only in the air does this matter, spring does the significant amount of work to stop the jumper). When you solve for x in the equation 5488 J = 1/2kx2, you get the spring coefficient of 43904 or almost 44000 Nm.

Next, I examined the athlete from an angular motion perspective. I observed that he makes about 3 revolutions per 2 seconds or 1.5 revolutoins per second, and when you multiply this by 2pi radians to get the angular speed you get 9.4 radians per second. Since it takes him approximately one second to get going at this angular speed, I estimated that his angular acceleration is 9.4 m/s2 (wf = wi + at). Then I was able to find the linear velocity when the athlete was flat by multiplying the angular speed by the radius, which I approximated to be .8 m. Their linear speed therefore came out to be (.8) (9.4) = 7.52 m/s, which is 27 km/hr or about 17 mph. That is pretty fast!!!

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