Thursday, November 15, 2012

Bouncy, Trouncy,Flouncy, Pouncy…


By Emily Fennell

http://www.youtube.com/watch?v=dJFyz73MRcg
In “The Wonderful Thing About Tiggers” song, Tigger tells Winnie the Pooh that his “bottom is made out of springs.” If we look at Tigger’s tail bouncing, we can determine its spring coefficient.

If we model Tigger from a real tiger, his nose-to-tail length will be an average of 2.8 meters and his tail alone will be approximately 1 meter in length. His mass is about 220 kg. From Tigger video clips, we can determine that when he bounces on his tail, his tail rises 0.4 m from the ground. During the compression before the bounce, Tigger’s tail compresses to .3m long. Using conservation of energy, we can determine Tigger’s spring coefficient and the velocity of his bounce. Heights are measured from the top of Tigger’s head with the ground as the point of reference (h=0m).
Mgh1 = mgh2 + ½ kx2 = mgh3 + ½ mv2
(220 kg)(9.8 m/s2)(2.8m) = (220 kg)(9.8 m/s2)(2.1m) + ½ k(.7m)2
k = 6200 N/m

Mgh1 = mgh2 + ½ mv2
(9.8 m/s2)(2.8m) = (9.8 m/s2)(3.2m) + ½ v2
V = 2.8 m/s

From the spring coefficient, we can calculate the spring force.

F = -kx
F = -(6200 N/m)(.7m)
F = -4300 N

Tiger tails are used for balance and portraying emotion – not for structural support of the body. It is safe to say that a real tiger would not be able to withstand the compressive force and of its entire weight on its tail. Tigers have 30 bones in their tails, which are vertebrae that extend from their spine. Because the bones extend all the way to the base of the tail, it is impossible for vertebrae in the tail to compress to 1/3 the tail’s length in preparation for one of Tigger’s bounces.

In the “Tigger’s Shoes” episode, Tigger brags that he can jump to the top of a cliff in two hops. In one hop (from his feet – not his tail) I estimate that he makes it half way up the cliff, which I assume is 100m. Therefore he jumps 50 m. A regular tiger can leap at best 10 feet into the air, or ~3m. Tigger’s jumping abilities are again exaggerated in this cartoon, this time about 16x that of a real tiger. I used this information to calculate Tigger’s initial velocity for him to reach a height of 50m.
Mgh1 + ½ mv2 = mgh2
(220 kg)(9.8 m/s2)(0m) + ½ (220 kg)v2 = (220 kg)(9.8 m/s2)(50m)
V = 31.3 m/s

This can also be calculated with kinematics:

VF2 = vo2 + 2aΔx
0 = v2 + 2(-9.8 m/s2)(50m)
V = 31.3 m/s

With this information, we can determine how long it should take to reach 50m.

Δx = vot + ½ at2
50m = (31.3 m/s)t + ½ (-9.8)t2
t = 3.2s

In the “Tigger’s Shoes” clip, I timed Tigger’s actual jump to be around 3.2-3.5s, so it looks like my estimates were pretty close to what they were in the cartoon.

We can also determine the distance that Tigger’s femur will compress in this situation by looking at the stress and strain equation.  For this equation we assume that Tigger’s femur is of uniform size, .417 m in length and 0.034 m in diameter. In problem set #10, the human femur compresses .05 mm, so this compression distance is assumed to be the same for tiger femur. From this information and the estimation that Young’s Modulus for bone is 15 x 109 Pa, we can determine the force on Tigger’s femur during the jump.

F = [(EA)/Lo]ΔL
F = {[(15x109 Pa)(0.017m)2π]/.417m} (0.00005 m)
F = 1600 N

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