Maxwell's Wheel: Understanding Conservation of Energy and Derivatives

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Well, I have to do experiment with Maxwell's wheel, but i struggle with theoretical background.
If you don't know what Maxwell's wheel is,
http://www.nikhef.nl/~h73/kn1c/praktikum/phywe/LEP/Experim/1_3_18.pdf

I understand conservation of energy and how we made that equation, but this part i don't get;

Clipboard01.jpg

Ok, i understand derivative, but how do i get s(t) and v(t).
Thanks!
 
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You just need a bit of algebra. Hint: first solve for the acceleration.
 
You have a very simple ODE. Solve it for v(t) using the given I.C. for v, then solve for s(t) = ∫v(t)dt using the I.C. for s.

EDIT: go with Doc Al, don't need formal ODE approach. Divide by v and solve for accel.
 
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Edit:
I see that v derivated is acceleration, so when i divide equation and kill v(t) i get acceleration,
a= mg/(m+I/r2)

should i just double integrate it now for s(t) and once for v(t), yes,i get it.
thanks!
 
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One more thing
http://www.officeplayground.com/Assets/ProductPreview/pi3600-3799/3653_maxwellswheel_1.jpg
In scheme of wheel, where does ds point to?
 
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diracdelta said:
Edit:
I see that v derivated is acceleration, so when i divide equation and kill v(t) i get acceleration,
a= mg/(m+I/r2)
Right!

diracdelta said:
should i just double integrate it now for s(t) and once for v(t), yes,i get it.
Sure. (Or, since the acceleration is constant, you can use the standard kinematic formulas.)

diracdelta said:
In scheme of wheel, where does ds point to?
Down.
 
Alright. I can't remember why, if ds=d(phi)xr, where x i vector cross multiply. Using right hand rule, it should be perpenicular towards angle and radiaii?