Work and Mechanical Energy & Moment of Inertia Derivation?by danyalasdf Tags: derivation, energy, inertia, mechanical, moment, work 

#1
Mar2909, 12:11 PM

P: 6

1. The problem statement, all variables and given/known data
We did something very similar to this in lab http://webenhanced.lbcc.edu/physte/p...%20Inertia.pdf Use Work and Mechanical Energy to derive the expression for the experimentally determined moment of inertia. 2. Relevant equations Wf= work of friction = Delta E = Ef  Ei Wf= work of friction = Uf + Kf  Ui  Ki U= Potential Energy K = Kinetic Energy Kf = (1/2)(m + mf)(Vf)^2 + (1/2)(I)(omegaf)^2 I = Moment of Inertia Omegaf = angular acceleration Average Velocity = v = (Vf + Vi)/(2) If Neccessary s=(1/2)*a*t^2 Torque= F*r= m*r*a T= (mf + m)(g  a) = tension Ui= mgh Uf= mfgh K(rotate) = (1/2)(I)(omegaf)^2 I = Moment of Inertia K(linear) = (1/2)(m + mf)(Vf)^2 Experimentally Moment of Inertia I=r^2(m((gt^2/2s)t)  mf) Trying to get to this ^ mf= mass effective not much meaning just mass in kg If it confusing the gt^2 is divided by 2s then it is subtracted by t and multiplied by r^2 and then minus mf 3. The attempt at a solution Wf = work of friction = Uf + Kf  Ui  Ki The final potential energy and initial kinetic energy are both zero so this only leaves Wf = Kf  Ui Wf = ((1/2)(m + mf)(Vf)^2) + (1/2)(I)(omegaf)^2  mgh Wf = (1/2)(m + mf)(s/t)^2 + (1/2)(I)((s/t)*(1/r))^2  mgh Wf = (1/2)(m + mf)(s^2/t^2) + (1/2)(I)((.5*a*t^2)/(t) * (1/r))^2 Wf = ((1/2)(m + mf)((1/2)*(a*t^4)*(t^2)) + ((1/2)(I)(a*t) * (1/r))^2 Wf = ((1/8)(m + mf)(a^2 * t^2) + (1/8)(I)((a^2 * t^2)/(r^2)) Wf = (1/8)(a^2*t^2)((m + mf) + (I/r^2)) 



#2
Mar2909, 02:20 PM

P: 619

So....what is your question?



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