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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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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

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))
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Mar29-09, 02:20 PM
P: 619
So....what is your question?

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