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Newton's Second Law & Moment of Inertia Derivation?

by danyalasdf
Tags: derivation, inertia, moment, newton
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danyalasdf
#1
Mar29-09, 11:35 AM
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 Newton's Second to derive the expression for the experimentally determined moment of inertia


2. Relevant equations
Newton's Second Law is F=m*a

s=(1/2)*a*t^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

Torque= F*r= m*r*a

T= (mf + m)(g - a) = tension

3. The attempt at a solution

T= (mf + m)(g - (2*s/t^2))

T= (mf + m)((1/2)(a*t^2) - a)

T= (mf + m)(a((1/2)*t^2 - 1)

I am stuck write here
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Pythagorean
#2
Mar29-09, 11:59 AM
PF Gold
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P: 4,262
Is calculus is a prerequisite for this course?
danyalasdf
#3
Mar29-09, 12:04 PM
P: 6
Quote Quote by Pythagorean View Post
Is calculus is a prerequisite for this course?
yes it is Physics Calculus Based 2211K

Pythagorean
#4
Mar29-09, 12:23 PM
PF Gold
Pythagorean's Avatar
P: 4,262
Newton's Second Law & Moment of Inertia Derivation?

Calc may not be necessary.

So what's the physical system you're doing in this problem? Is it the same exact system as in the lab?
danyalasdf
#5
Mar29-09, 01:12 PM
P: 6
Quote Quote by Pythagorean View Post
Calc may not be necessary.

So what's the physical system you're doing in this problem? Is it the same exact system as in the lab?
yes it is the exact same system


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