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Applying Rotational Dynamics to a Childrens Show About Colorful Ponies.


Ac3xAssasin

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So I have a test on Circular Motion in physics coming up on Monday, and I thought "What better way to study then apply what I learned to MLP:FIM?"

 

Well, many of you may remember this scene from "wonderbolts academy"

 

http://www.youtube.com/watch?v=kpyDq0nIWuM

 

Well through some observations I calculated various values relating to the circular motion of the pony (Wildflower according to the internet).

 

Here is my work on paper

 

Front:

 

 

 

Back:

 

 

 

So without further ado here is my explanation:

 

Based on observation and a educated guess that the length of an outstreched pony is about 4ft I found the distance from the dizzitron to Wildflowers COM to be about 1.5 pony lengths or about 1.83m.

 

Another educated guess, I assumed the average adult pegasi probalby weight about 60lb or 27.22 Kg.

 

At maximum angular velocity Wildflower makes 1 rotation about every 1 seccond about the dizzitron making her final angular velocity to be about 6.28 rad/s.

 

The time required for the dizzitron to reach maximum angular velocity came out to be about 6 secconds.

 

Finally, Wildflowers moment of inertia, without counting the dizzitron itself plugged into the formula I=mr^2 came out be about 91.16 Kg * m^2

 

Based on this I was able to find angular acceleration, Tangential Velocity, Radial Acceleration, Centripetal Force, Torque, Angular Momentum, and finally Rotational Kinetic Energy.

 

Now, the calculations.

 

Angular Acceleration

α = ωf - ωi / t

 

since intitial angular velocity is zero we get

 

α = 6.28rad/s / 6s

 

α = 1.047 rad/s^2

 

Tangential Velocity:

 

VT = ωf * r

 

VT = 6.28 rad/s * 1.83

 

VT = 11.49 m/s

 

Radial Acceleration

 

aR = (VT)^2 / r

 

aR = (11.49m/s)^2 / 1.83m

 

aR = 72.14 m/s^2 = 7.36 g's

 

Centripetal Force

 

Fc = mwf * aR

 

Fc = (27.22 Kg)(72.14 m/s^2)

 

Fc = 1963.65 N

 

Torque on Wildflower

 

T = I * α

 

T = (91.16 Kg * m^2) * (1.047 rad/s^2)

 

T = 95.44 m * N

 

Angular Momentum of Wildflower

 

L = I * ω

 

L = (91.16 Kg * m^2) * (6.28 rad/s)

 

L = 572.76 (Kg * m^2)/ s

 

Rotational Kinetic Energy of Wildflower

 

KE = (1/2) I * ω^2

 

KE = (1/2)(91.16 Kg * m^2) * (6.28 rad/s)^2

 

KE = 1797.60 J

 

 

Studying with ponies is the only way to study. :D

  • Brohoof 5

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