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Moment of inertia of a cube

  1. Jul 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the moment of inertia of a cube of mass M and edges d round an axiz, z, that passes in the middle.
    Then calculate around an axis parallel to the z axis and passing on one of the edges

    2. Relevant equations
    Moment of inertia: ##I=\int r^2dm##

    3. The attempt at a solution
    I slice the cube into rectangular slices of thickness dz. i sum infinitesimal masses dm along the radius till it reaches the edge. this way i get I for a triangle. there are 8 like these.
    The radius's length, from the center to the edge is ##r=\frac{d}{2\cos\theta}##
    The infinitesimal mass is ##dm=\frac{M}{d^3}dr^2 dz##
    $$I=\frac{8M}{d^3} \int_{z=-\frac{d}{2}}^{z=\frac{d}{2}} dz \int_{\theta=0}^{\theta=\frac{\pi}{4}} d\theta \int_{r=0}^{r=\frac{d}{2\cos\theta}} r^2 dr =\frac{8M}{3d^3} \int_{z=-\frac{d}{2}}^{z=\frac{d}{2}} dz \int_{\theta=0}^{\theta=\frac{\pi}{4}} d\theta (r^3)\vert_0^{\frac{d}{2\cos\theta}}$$
    $$I=\frac{M}{3}\int _{z=-\frac{d}{2}}^{z=\frac{d}{2}} dz \int_{\theta=0}^{\theta=\frac{\pi}{4}} \frac{1}{\cos^3\theta} d\theta$$
    $$\int \sec^3 x=\frac{1}{2} (\sec x \tan x+\ln |\sec x +\tan x|)$$
    $$\int_{\theta=0}^{\theta=\frac{\pi}{4}} \frac{1}{\cos^3\theta} d\theta=\frac{1}{2} (\sec \theta \tan \theta+\ln |\sec \theta +\tan \theta|)\vert_0^{\frac{\pi}{4}}=\frac{1}{2} (\sqrt{2}+\ln(\sqrt{2}+1)-\ln 1)=2.29$$
    It's incorrect
     

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  2. jcsd
  3. Jul 9, 2015 #2

    TSny

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    This doesn't look right.

    Do you have enough factors of r in the integral. Does your expression have the right dimensions for moment of inertia?
     
  4. Jul 9, 2015 #3

    SteamKing

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    You started with the right equation for the MOI, then you went into the weeds with it.

    Remember, r2 = x2 + y2. Use this relation in the integral expression for the MOI and you can save a trip thru TrigLand.
     
  5. Jul 9, 2015 #4
    $$dm=\frac{M}{d^3}dr d\theta dz$$
    $$I=\frac{8M}{d^3}\int_v dm\; r^2=\frac{8M}{d^3}\int_v dz d\theta dr\; r^2=\frac{8M}{d^3}\int_{-d/2}^{d/2} dz \int_0^{\pi/4} d\theta \int_0^{\frac{d}{2\cos\theta}} r^2 dr$$
    $$I=\frac{8M}{d^2}\int_0^{\pi/4} \left( \frac{1}{3} r^3 \right)_0^\frac{d}{2\cos\theta}=\frac{Md}{3}\int_0^{\pi/4} \frac{1}{\cos^3\theta} d\theta$$
    It doesn't help
     
  6. Jul 9, 2015 #5

    SteamKing

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    You've just repeated your original calculation.

    You didn't even substitute x2 + y2 for r2 in the definition of the MOI.

    How do you expect things to work if you don't follow hints and suggestions?
     
  7. Jul 9, 2015 #6

    Nathanael

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    It's better, but it's still not quite right. (Check the dimensions, dθ is dimensionless.)


    Your method is the way I would've done it, but Karol's method gives the right answer, too. (Karol is just making one small mistake.)
     
  8. Jul 9, 2015 #7

    TSny

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    You can tell this is incorrect by dimensional analysis. The right side does not have the dimensions of mass. The volume element in cylindrical coordinates is not ##dr d\theta dz##. [EDIT: oops, I now see that Nathanael already pointed this out.]
     
    Last edited: Jul 9, 2015
  9. Jul 10, 2015 #8
    $$dm=\frac{M}{d^3}dz( r d\theta )dr,\; I=\frac{8M}{d^3}\int_0^d dz \int_0^{\pi/4} d\theta \int_0^{\frac{d}{2\cos\theta}}=\frac{1}{3}Md^2$$
    With the method ##r^2=x^2+y^2## i get the same answer:
    $$dm=dx\; dy\; dz,\; I=\frac{8M}{d^3}\int_0^d dz \int_0^{d/2} dy \int_0^{d/2} (x^2+y^2)dx=\frac{1}{3}Md^2$$
    But in the tables it should be ##\frac{1}{6}Md^2##
     
  10. Jul 10, 2015 #9

    SteamKing

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    You're still not setting up this problem correctly.

    You can express the mass of the cube as the volume, V, of the cube multiplied by the density of the cube, ρ, or

    M = ρ ⋅ V = ρ ⋅ d3

    From this relation, you can work out what dm must be.

    In your integrals, you integrate z from 0 to d but you integrate x and y both from 0 to d/2. Why?

    If you are going to calculate the MOI about the center of the cube, you must pay careful attention to using the proper limits of integration to account for the MOI of the entire cube.
     
  11. Jul 10, 2015 #10

    TSny

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    What is the integrand for the r integration? If you get that set up correctly, it should give you the correct answer.
     
  12. Jul 10, 2015 #11
    I thought ##\frac{M}{d^3}=\rho \left[ \frac{kg}{m^3} \right]##
    I integrate z from 0 to d because it's easier and doesn't matter if i start the count from the bottom or the middle.
    I integrate both x and y from 0 to d/2 since i calculate MOI only for an eighth, see drawing.
    $$I=8\int_v dm\cdot r^2=8\int_v\frac{M}{d^3}dz(rd\theta)dr\cdot r^2=\frac{8M}{d^3}\int_0^d dz \int_0^{\pi/4} d\theta \int_0^{\frac{d}{2\cos\theta}} r^3 dr$$
     

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  13. Jul 10, 2015 #12

    SteamKing

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    You're backsliding into Trig Land again.

    If you are going to use rectangular Cartesian coordinates, you've got to concentrate on those.

    When calculating the moment of inertia, there's less chance for error if you set up the integrals referenced to the center of the cube (which is also called the centroid) and evaluate them using the limits based on that location.

    You come to PF for help, but yet you stubbornly resist taking our advice to help simplify your calculations. Why is that?

    This is a cube. It's tailor made to use rectangular cartesian coordinates to set up the integrals, not cylindrical or any other coordinate system.
     
  14. Jul 10, 2015 #13

    Nathanael

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    This integral is correct. You must've made a mistake somewhere in evaluating it, because it is correct.
     
  15. Jul 10, 2015 #14
    In post #8 i used cartesian coordinates:
    $$dm=dx\; dy\; dz,\; I=\frac{8M}{d^3}\int_0^d dz \int_0^{d/2} dy \int_0^{d/2} (x^2+y^2)dx=\frac{1}{3}Md^2$$
    And yet the answer is wrong, it should be ##\frac{1}{6}Md^2##
    I am not trying only to solve, i want to understand and solve in other ways too. i am curious to know what's my error with cartesian coordinates as well, it will teach me.
    I have an error, i want to know why, if possible. and solve in cartesian also
     
  16. Jul 10, 2015 #15

    Nathanael

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    Why did you multiply this one by 8? You only broke it up into 4 sections, right?
     
  17. Jul 10, 2015 #16
    Each section is made up of 2 rectangles, see drawing again.
    I integrated θ only till π/4, not to π/2
     

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  18. Jul 10, 2015 #17

    Nathanael

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    No I'm talking about your cartesian integral. As I said in post #13, your cylindrical is correct.

    But for your cartesian integral you multiplied it by 8, even though you only broke it up into 4 pieces.
     
  19. Jul 10, 2015 #18
    Thanks for reminding me, i skipped post #13! and i found my mistake in the cylindrical coordinates, it gives ##\frac{1}{6}Md^2##
    About the cartesian coordinates you are right, it's for 4 quarters, thanks!

    And thank you all
     
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