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Work and Mechanical Energy & Moment of Inertia Derivation?

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data

    We did something very similar to this in lab

    http://webenhanced.lbcc.edu/physte/phys2ate/2A LAB HANDOUTS/Moment of Inertia.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


    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))
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 29, 2009 #2
    So....what is your question?
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