- #1
Moment of Inertia Homework: Parts C & D
- Thread starter athrun200
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Homework Help Overview
The discussion revolves around calculating the moment of inertia for a rod with a varying density, specifically focusing on parts C and D of the homework assignment. Participants are exploring how to approach the problem, particularly in relation to the density distribution along the rod and the implications of calculating the moment of inertia about different ends.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants discuss the implications of starting calculations from the heavy end versus the light end of the rod. There is an exploration of the density function and how it affects the moment of inertia calculation. Some participants suggest breaking the integral into two parts to simplify the process.
Discussion Status
There is ongoing dialogue about the correct setup for the integral and the interpretation of the density function. Some participants have expressed confusion about the approach and are seeking clarification on specific steps. While some have reported progress, others are still grappling with the complexity of the problem.
Contextual Notes
Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is a noted emphasis on ensuring that calculations are aligned with the problem's requirements regarding the moment of inertia about the heavy end of the rod.
- #2
cupid.callin
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the density you used is defined from the lighter end and you have to find I about heavier end.
you may take the dx element at distance x from heavy end and then use (10-x) in the equation of density
you may take the dx element at distance x from heavy end and then use (10-x) in the equation of density
- #3
gneill
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The parallel axis theorem works fine for objects with constant density. This object doesn't have constant density -- the density varies from end to end. Therefore there will be a light end and a heavy end. If you grab the rod by the light end and swing the heavy end around, you'll find a lot more angular inertia than if you grab the rod by the heavy end and swing the light end around.
So, see if you can set up your integral to directly find the moment of inertia about the heavy end.
So, see if you can set up your integral to directly find the moment of inertia about the heavy end.
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- #4
athrun200
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cupid.callin said:
Is it a rule that I must start from heavy end?
Why it is wrong to start from light end?
- #5
gneill
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athrun200 said:
It's not a rule, and it's not wrong to start at the light end. You may start anywhere you wish! But note that the problem is asking for the moment of inertia about the heavy end. Therefore, in the the calculation of the moment of inertia, the radial distances of the mass elements must be with respect to the heavy end. And be sure that the density at that distance from the heavy end is the correct density for that position along the rod.
- #6
athrun200
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gneill said:
Oh! Sorry for my carelessness, I didn't read the question carefully.
I have solved part d, but part c is still very complicated.
The result I obtained is so ugly that I don't want to write it out.
But I have check it and it is wrong.
So what's wrong with my work this time?
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- #7
gneill
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Suppose you break the integral into two. The first runs from x=0 to x= xcm, the second from xcm to 10. The density function remains as 2x + 4 for any value of x, so leave it alone. What is the distance from xcm to the mass element in each case?
- #8
athrun200
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gneill said:
I am so sorry that I don't understand.
Can you explain it more clearly?:shy:
- #9
gneill
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athrun200 said:
x is the position along the rod from the LHS. For the region to the left of the center of mass, the distance from the center of mass to x is xcm - x. To the right of the center of mass, the distance is x - xcm. The density at point x remains [itex]\rho (x) = 2 x + 4[/itex], since here x is always with respect to the light end of the rod.
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- #10
athrun200
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gneill said:
x is the position along the rod from the LHS. For the region to the left of the center of mass, the distance from the center of mass to x is xcm - x. To the right of the center of mass, the distance is x - xcm. The density at point x remains [itex]\rho (x) = 2 x + 4[/itex], since here x is always with respect to the light end of the rod.
Do you mean this?
I have just finished half of the integral.
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- #11
athrun200
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Oh! It works and I get the answer!
But can you explain what's wrong with my method?
Why I can't set the center of mass as the orgin?
But can you explain what's wrong with my method?
Why I can't set the center of mass as the orgin?
- #12
athrun200
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Work in #6 turns out to be correct, I just type it wrongly in my calculator.
By the way, how to generate the picture in #9?
What software are you using?
By the way, how to generate the picture in #9?
What software are you using?
- #13
gneill
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athrun200 said:
You can set the center of mass at the origin if you translate the density equation accordingly for each section of the rod. Personally I tend to find it easier to leave the density function alone and work out the required distances as functions of the same x.
- #14
gneill
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athrun200 said:
I use Visio to draw the pictures, then paste it into MS Paint to save it as a .gif type file.