Right, so then my formula will end up like
##0 = (PE + K_{trans} + K_{rot} + E_{loss}) - (PE_0 + K_{0trans} + K_{0rot} + E_{0loss})##
##0 = \frac {1}{2}Mv^2 + \frac {1}{2}Iw^2 + E_{loss} - Mgh##
##E_{loss} = Mgh - \frac {1}{2}Mv^2 - \frac {1}{2}I\left(\frac{v}{r}\right)^2##
and then I put in...
Right, I should be using the shaft radius of 0.01005m because that is the part of the flywheel that is in contact with the surface.
Amazing graphic of the problem, thank you for that. I will have a read of that
Yes I can see that now, thank you for making it clear
Potential Energy should = KE...
Yes it did. It was a very gentle slope and also it was the shaft of the flywheel that was in contact with the surface. So most of the energy would have gone into rotating the central cylinder of the flywheel, if that makes sense.
Sure, I will show you what I have worked out so far. I am stuck because the I value I worked out of 1.0273 kg/m^2 is very different to when I use I = 1/2 m R^2, where I got 1/2 * 4.01 * 0.08025^2 = 0.0129 kg/m2
I am stuck on what to do to calculate the inertia of a flywheel using the method described.
I am supposed to use conservation of energy equations to calculate the inertia.
I have a picture of the experiment and also the measurements I have taken.
It seems each method I try I get a different...