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Moment Inertia w/ density F(x)

  • Thread starter fisselt
  • Start date
  • #1
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Homework Statement


Calculate the moment of inertia of a uniform rigid rod of length L and mass M lying along the x-axis which rotates about an axis perpendicular to the rod (the y axis) and passing through it’s center of mass. The rod has a line density that is a function of location such that =3x.


Homework Equations


[itex]\int ρr^2 dm[/itex]


The Attempt at a Solution


I thought this would be done simply by replacing ρ with 3x and r with x. However, while integrating I just keep getting 0 as my answer.
[itex]\int ρr^2 dm[/itex] dm=ρdx=3xdx
=[itex]\int 3x(x)^2 dx[/itex]
=[itex]\frac{3x^4}{4}[/itex]

Now when I evaluate from L/2 and -L/2 I always get zero.

I assume I'm setting up my integral incorrectly as I should probably have an odd exponent that I totally forgot how to do calculate an integral over the summer. Appreciate the help.
 

Answers and Replies

  • #2
tiny-tim
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hi fisselt! :smile:
… Now when I evaluate from L/2 and -L/2 I always get zero.
hardly surprising …

half of your rod has negative density! :biggrin:
 
  • #3
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Should I be evaluating from 0 to L then?

Meaning it would be (3L^4)/4
 
  • #4
tiny-tim
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Calculate the moment of inertia of a uniform rigid rod of length L and mass M lying along the x-axis which rotates about an axis perpendicular to the rod (the y axis) and passing through it’s center of mass. The rod has a line density that is a function of location such that =3x.
Should I be evaluating from 0 to L then?

Meaning it would be (3L^4)/4
(try using the X2 button just above the Reply box :wink:)

hmm … i don't really understand the question …

where is x being measured from?

and how can it be uniform if ρ = 3x ? :confused:
 
  • #5
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I think this may be part of my problem as well. I know typically with rods spinning about the perp. axis we evaluate from the center point (hence the L/2 AND -L/2). For all the other problems we have done this has worked but I have never done one with density as a function. I expected an answer that would be at least similar to the general form but I can't find it. I'm stuck doing it the same way over and over.

Home internet is down so no TEX on this phone, too time consuming.
 
  • #6
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Because the line density is ρ=3x, wouldn't the center of mass not be in the center? And so when evaluating the integral of inertia wouldn't the bound not be from -L/2 to L/2? You'd first have to find the location of the actual center of mass, and then evaluate from the endpoints when the center of mass is zero.
 
  • #7
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When making the substitution from the original form of the inertia integral you substitute M*R^2/L dr. But M/L is simply the linear density (for any uniform object). So when substituting in dm=ρ*r^2 dr you can use the 3x even though you have a negative lower bound because definite integrals allow for that (just keep in mind that you have to evaluate where the center of mass is placed at x=0).

Sorry I don't know how to use LaTeX lol.
 
  • #8
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Isn't this effectively what I've done originally? Perhaps the range is incorrect and I should figure center of mass of a 1 meter rod of this varying density? Something like -(L2)/3 to L/3.
 
  • #9
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Something like that.

You know how to perform a center of mass integral, right?

integrate 3*x^2 dx from 0 to L and divide by M, giving L^3/M.

So that's the center of mass. It seems like it should be from -2L/3 to L/3, but it's from -L^3/M to (L-L^3/M) because the actual calculated center of mass is at L^3/M.
 
  • #10
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The length could still be L, but you're right on with the -L/3 to 2L/3 if the mass of the rod is 1kg.
 
  • #11
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Also, the M in the center of mass is not M given, but mass obtained when integrating the linear density across the rod. Working through the calculations, we find the center of mass at 2L/3, which verifies your guess.
 
  • #12
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While integrating with the 2/3 and 1/3, however I'm getting a negative answer. Should I switch the signs in this case?
 
  • #13
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What are your bounds?
 
  • #14
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I had -(2L)/3 to L/3.
 
  • #15
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I got - (5L^4)/36
 
  • #16
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If you switch the bounds, you'd get the same answer, just positive. That should be right...

integral(3*x^3 dx) from -L/3 to 2L/3

?
 
  • #17
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Yeah if you switch the bounds you get +5L^4/36
 
  • #18
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Yeah, thats what I was figuring. Thanks a lot for the help. This discussion helped me out a lot!
 
  • #19
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No problem! Good luck with everything else.
 

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