Moments of Inertia Derivation, Please Help


by Ush
Tags: moment of inertia
Ush
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#1
Jun27-10, 11:50 AM
P: 98
1. The problem statement, all variables and given/known data

I have attached the problem in one file and I have attached my attempt in the second file.
I only need help deriving the moment of inertia for the first (1) and fourth (4) objects but I have attached my solutions to the other objects in case it helps jog someones memory onto how to do this =p


2. Relevant equations

I = ∑miri2

A = area
M = total mass
dm = change in mass
dA = change in area
dr = change in radius

3. The attempt at a solution

attempt is attached

--
Thank you for taking the time to read through my problem and helping me solve it, I appreciate your help
Attached Thumbnails
attempt.jpg   question.jpg  
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Doc Al
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#2
Jun27-10, 12:05 PM
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Number 1 is exactly the same problem as number 2, just with different limits of integration.

In number 4 note that all the mass is at the same distance from the axis.
Ush
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#3
Jun27-10, 12:21 PM
P: 98
I'm not sure how to integrate one so that I'll get 1/12ML2

I tried doing something similar

dm/M = dr/0.5 L because dr starts from the pivot point in the center and max dr will only cover half of the total length. After doing the integration I didn't get 1/12ML2

I still don't understand how to begin the fourth one =[

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#4
Jun27-10, 12:37 PM
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Moments of Inertia Derivation, Please Help


Quote Quote by Ush View Post
I'm not sure how to integrate one so that I'll get 1/12ML2
You've already done the integral (in #2)--the only change is the limits of integration.

I still don't understand how to begin the fourth one =[
I = ∫r2 dm. How does r vary as you move around the shell?
Ush
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#5
Jun27-10, 12:54 PM
P: 98
another attempt attached
Attached Thumbnails
attempt 2.jpg  
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#6
Jun27-10, 01:21 PM
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Quote Quote by Ush View Post
another attempt attached
For some reason, you are integrating from 0 to R/2. That's from the center of the rod to one end. But the rod goes from end to end.
Ush
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#7
Jun27-10, 01:26 PM
P: 98
oh wow =o I can't believe I missed that.
Thanks so much! i understand how to do the first one now =)

could you give me another hint onto how to do the fourth one?
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#8
Jun27-10, 01:34 PM
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Quote Quote by Ush View Post
could you give me another hint onto how to do the fourth one?
I thought I did:
Quote Quote by Doc Al View Post
How does r vary as you move around the shell?
I'll rephrase it. What's the distance from the axis of every element of mass dm as you go around the shell?
Ush
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#9
Jun27-10, 01:43 PM
P: 98
the distance from the axis of every element of mass, dm, is R ?
if R increases, the mass increases because you get a bigger shell

dr/R = dm/M ?? =S
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#10
Jun27-10, 01:50 PM
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Quote Quote by Ush View Post
the distance from the axis of every element of mass, dm, is R ?
Exactly. Is R a variable or a constant? (For a given shell.)
Ush
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#11
Jun27-10, 01:58 PM
P: 98
R is constant for a given shell
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#12
Jun27-10, 02:03 PM
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Quote Quote by Ush View Post
R is constant for a given shell
Right! So simplify and complete the integral: I = ∫R2 dm
Ush
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#13
Jun27-10, 02:17 PM
P: 98
if radius is constant. then mass is constant. there is no dr or dm =S
how do i sub dm for something?
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#14
Jun27-10, 02:35 PM
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Quote Quote by Ush View Post
if radius is constant. then mass is constant.
Not sure what you mean. Hint: How do you deal with constants within the integral sign?
Ush
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#15
Jun27-10, 02:39 PM
P: 98
if you have a constant then you take it out of the integral.
..oh my

I = ∫R2 dm
= R2∫dm
= R2 ∑m
= R2M


THANK YOU SO MUCH!


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