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danyalasdf
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Homework Statement
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.
Homework 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
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))
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