There are a myriad of variables that go into hitting a springboard

There are a myriad of variables that go into hitting a springboard, so I am going to have to simplify it to a certain extent. For this project, I am going to assume that the gymnast is a cylinder and the gymnast and springboard act as one linear spring. This is illustrated in figure 2. The first thing that I am going to do is find the stiffness coefficient of this new linear spring. This can be done using Hooke’s Law, F=Kx. To find the force that is applied on the spring, I will use Newton’s second law F=ma=66.7*9.8≈654N and to find the spring deflection, I will use pictures taken during vaults. This ends up being around .07 m from when the spring is loaded into board. Hence, we can calculate the spring coefficient of one spring by setting up the equation 654=.07K and then proceed to say K≈9343 N/m. Now this number needs to be multiplied by the number of springs to get the full coefficient. 9343*7= 65401 N/m.

Now that there is a model of what this should look like, it can be applied to vaulting. What I am trying to find is the force perpendicular or Rt as I would like to find that is propelling me forward. And as Rtas an acceleration velocity, we can use D.G. Medley’s method of finding acceleration of a rotating polar coordinate (r,Ө) where r = L+x. This equation can be expressed as at= 2drdt(dӨdt)+r(d2Өdt2). But because we know that r=L+x, and L is a constant, it can be said that drdt=dxdtand d2rdt2=d2xdt2Now we can rewrite the equation for acceleration as at=2dxdt(dӨdt)+(L+x)(d2Өdt2).

Now applying Newton’s Second Law (F=ma) we get the equation

Rt-mgcosӨ=m(2dxdt(dӨdt)+(L+x)(d2Өdt2)

But to find d2Өdt2we need to utilize an equation for torque given in “Mass-Spring Modelling of Vault Springboard Contact” (1999 page 75): T=ddt(IdӨdt). This can then be rewritten as Rt(L+x)=I(d2Өdt2)+dIdt(dӨdt) as Torque can be written as the force perpendicular/transversal times r which equals (L+x) and from the product rule of derivatives. This can then be simplified to Rt=I(d2Өdt2)+dIdt(dӨdt)L+x

Now we can use substitution to get the equation I(d2Өdt2)+dIdt(dӨdt)L+x-mgcosӨ=m(2dxdt(dӨdt)+(L+x)(d2Өdt2)

I(d2Өdt2)+dIdt(dӨdt)L+x=m(2dxdt(dӨdt)+(L+x)(d2Өdt2)+gcosӨ)

I(d2Өdt2)+dIdt(dӨdt)=m(L+x)(2dxdt(dӨdt)+(L+x)(d2Өdt2)+gcosӨ)

I(d2Өdt2)=m(L+x)(2dxdt(dӨdt)+(L+x)(d2Өdt2)+gcosӨ)-dIdt(dӨdt)

I(d2Өdt2)+m(L+x)2(d2Өdt2)=m(L+x)(2dxdt(dӨdt)+gcosӨ)-dIdt(dӨdt)

(d2Өdt2)=m(L+x)(2dxdt(dӨdt)+gcosӨ)-dIdt(dӨdt)m(L+x)2+I

Finally, I would like to know the force perpendicular when I am leaving the board to find the acceleration during the preflight. These conditions will prevail when when the velocity is changed from downward to upwards or when the Vertical component equals 0. This can be expressed by the equation:

Rv=RtsinӨ+RrcosӨ.

Then, because Rr=Kx, the equation can become

0=RtsinӨ+KxcosӨ.

With a little rearranging, this turns into

Rt=KxcosӨ-sinӨ

From the figure below, Өcan be plugged in as 80°, K as 65401 N/m, and x as 1.6-1.93=-.36 to get Rt=65401(-.36)cos80-sin80=4151.5N,

But the vertical and horizontal components are wanted so with some trigonometry, it can be found that the vertical force is 4088.4N and the horizontal force is 720.9N.

And from these forces, the vertical and horizontal accelerations can be found from F=ma and my mass being 66.68 kg. From this we get the vertical acceleration to be 4088.4=66.68a⟶a=61 m/s^2 upwards. The horizontal component is then 720.4=66.68a⟶a=10.8 m/s^2.

Now the components of the preflight can be calculated. The upwards acceleration is about 61, but gravity has to be taken into account, so 9.8 will be subtracted from it to get 51.2 m/s^2. This seems very fast, but when “SUVAT” equations are used, it can be seen that this will only propel me. It takes 11 frames for me to get from the board to the vault table which is approximately .183 seconds

s=ut+12at2

s=0+1261(1160)21.03m

As the vault table is 1.35 meters tall, this acceleration is even a little slow to reach the full height of the vault table. But when the height of the springboard and body position are accounted for, I am sure that this number would be correct.

For the horizontal component, the same equation and method for tracking time was used, but in this case, the initial velocity that was calculated in the first chapter must also be put into consideration.

s=ut+12at2

s=8.7(1160)+12(10.8)(1160)21.7m5.6 ft

This is about the distance from where I hit the springboard to the center of the vault table. The front of the springboard is 2 feet 4 inches away from the front of the base of the table, and I hit about ⅓ of the way down the board. As the board is 120 cm or 3.9 ft, I will have to travel an extra 1 foot 4 inches. Finally, the vault table is 46 inches wide, so to hit in the middle, I would have to travel 23. All of this distances combines to a total of 5 feet and 7 inches.

Source: Essay UK - https://www.essay.uk.com/essays/science/there-are-a-myriad-of-variables-that-go-into-hitting-a-springboard/


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