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).

Mgh

_{1}= mgh_{2}+ ½ kx^{2}= mgh_{3}+ ½ mv^{2}
(220 kg)(9.8
m/s

^{2})(2.8m) = (220 kg)(9.8 m/s^{2})(2.1m) + ½ k(.7m)^{2}
k = 6200 N/m

_{1}=

_{2}+ ½

^{2}

(9.8 m/s

^{2})(2.8m) = (9.8 m/s^{2})(3.2m) + ½ v^{2}
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.

Mgh

_{1}+ ½ mv^{2}= mgh_{2}
(220 kg)(9.8
m/s

^{2})(0m) + ½ (220 kg)v^{2}= (220 kg)(9.8 m/s^{2})(50m)
V = 31.3 m/s

This can
also be calculated with kinematics:

V

_{F}^{2}= v_{o}^{2}+ 2aΔx
0 = v

^{2}+ 2(-9.8 m/s^{2})(50m)
V = 31.3 m/s

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

Δx
= v

_{o}t + ½ at^{2}
50m = (31.3
m/s)t + ½ (-9.8)t

^{2}
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 10

^{9}Pa, we can determine the force on Tigger’s femur during the jump.
F = [(EA)/L

_{o}]ΔL
F = {[(15x10

^{9 }Pa)(0.017m)^{2}π]/.417m} (0.00005 m)
F = 1600 N

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