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Work and Mechanical Energy & Moment of Inertia Derivation?

by danyalasdf
Tags: derivation, energy, inertia, mechanical, moment, work
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danyalasdf
#1
Mar29-09, 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))
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Dr.D
#2
Mar29-09, 02:20 PM
P: 619
So....what is your question?


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