How to choose the perpendicular axis?

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This site https://en.m.wikipedia.org/wiki/Perpendicular_axis_theorem says: Define perpendicular axes $x$, $y$, and $z$ (which meet at origin $O$) so that the body lies in the $xy$-plane, and the $z$-axis is perpendicular to the plane of the body. Let $I_x$, $I_y$, and $I_z$ be moments of inertia about axes $x$, $y$, and $z$ respectively



The perpendicular axis theorem states $I_z = I_x + I_y $. Now if we have a rigid three dimensional body then how to choose perpendicular axis as it lies on $xy$, $yz$, and $xz$—all planes. Even I find it messy when I see the cylinder's axis to be the perpendicular or $z$-axis. I could assign axis perpendicular to it and say that's $z$. How can I determine what's the perpendicular axis?










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    up vote
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    down vote

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    This site https://en.m.wikipedia.org/wiki/Perpendicular_axis_theorem says: Define perpendicular axes $x$, $y$, and $z$ (which meet at origin $O$) so that the body lies in the $xy$-plane, and the $z$-axis is perpendicular to the plane of the body. Let $I_x$, $I_y$, and $I_z$ be moments of inertia about axes $x$, $y$, and $z$ respectively



    The perpendicular axis theorem states $I_z = I_x + I_y $. Now if we have a rigid three dimensional body then how to choose perpendicular axis as it lies on $xy$, $yz$, and $xz$—all planes. Even I find it messy when I see the cylinder's axis to be the perpendicular or $z$-axis. I could assign axis perpendicular to it and say that's $z$. How can I determine what's the perpendicular axis?










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      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      This site https://en.m.wikipedia.org/wiki/Perpendicular_axis_theorem says: Define perpendicular axes $x$, $y$, and $z$ (which meet at origin $O$) so that the body lies in the $xy$-plane, and the $z$-axis is perpendicular to the plane of the body. Let $I_x$, $I_y$, and $I_z$ be moments of inertia about axes $x$, $y$, and $z$ respectively



      The perpendicular axis theorem states $I_z = I_x + I_y $. Now if we have a rigid three dimensional body then how to choose perpendicular axis as it lies on $xy$, $yz$, and $xz$—all planes. Even I find it messy when I see the cylinder's axis to be the perpendicular or $z$-axis. I could assign axis perpendicular to it and say that's $z$. How can I determine what's the perpendicular axis?










      share|cite|improve this question















      This site https://en.m.wikipedia.org/wiki/Perpendicular_axis_theorem says: Define perpendicular axes $x$, $y$, and $z$ (which meet at origin $O$) so that the body lies in the $xy$-plane, and the $z$-axis is perpendicular to the plane of the body. Let $I_x$, $I_y$, and $I_z$ be moments of inertia about axes $x$, $y$, and $z$ respectively



      The perpendicular axis theorem states $I_z = I_x + I_y $. Now if we have a rigid three dimensional body then how to choose perpendicular axis as it lies on $xy$, $yz$, and $xz$—all planes. Even I find it messy when I see the cylinder's axis to be the perpendicular or $z$-axis. I could assign axis perpendicular to it and say that's $z$. How can I determine what's the perpendicular axis?







      moment-of-inertia rigid-body-dynamics






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          2 Answers
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          If you have a 3 dimensional object, then it doesn't lie entirely in any plane. You can't use the theorem directly. You would instead be limited to looking at the moment of inertia of a 2D "slice" from the object. You could then sum all the slices together.



          Other than the fact that the $z$ axis must be perpendicular to the plane of the object, the choice of axes is yours. Normally you would choose axes that make the calculation simpler (or possible). It depends completely on the problem you're trying to solve.



          The theorem just states the relationship. It doesn't mean that there is necessarily a unique choice, or that any choice is especially useful.






          share|cite|improve this answer




















          • So if i have any slice of cylinder which would be a shell like lying on xy plane then z is the perpendicular axis . Am i right?
            – Nobody recognizeable
            3 hours ago






          • 1




            Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
            – BowlOfRed
            2 hours ago

















          up vote
          1
          down vote













          I, by definition has to do with a surface not a volume and integrating sum of I of slices is meaningless. Hence, it is called second moment of area, not volume.



          lets use a cucumber as an example. If we cut an infinitely thin slice from the middle and look at the round cross section we can assign 3 axis to that surface, X,Y,Z.
          Then $I_z = I_x + I_y $ with X being in-plane of round surface horizontal axis and Y in-plane vertical. And Z an axis coming out of the surface vertically.



          But this Z should not be confused with deceptively similar sounding Z axis of the whole cucumber if we had positioned it at the same origin vertically.



          If we were to measure that Z we needed to cut the cucumber longitudinally like a long oval shape and then we could call that axis Z if we liked to, and the horizontal short axis X. But I of that axis would be $ int_-x^+x Z^2da $



          Which is obviously totally different.






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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote













            If you have a 3 dimensional object, then it doesn't lie entirely in any plane. You can't use the theorem directly. You would instead be limited to looking at the moment of inertia of a 2D "slice" from the object. You could then sum all the slices together.



            Other than the fact that the $z$ axis must be perpendicular to the plane of the object, the choice of axes is yours. Normally you would choose axes that make the calculation simpler (or possible). It depends completely on the problem you're trying to solve.



            The theorem just states the relationship. It doesn't mean that there is necessarily a unique choice, or that any choice is especially useful.






            share|cite|improve this answer




















            • So if i have any slice of cylinder which would be a shell like lying on xy plane then z is the perpendicular axis . Am i right?
              – Nobody recognizeable
              3 hours ago






            • 1




              Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
              – BowlOfRed
              2 hours ago














            up vote
            3
            down vote













            If you have a 3 dimensional object, then it doesn't lie entirely in any plane. You can't use the theorem directly. You would instead be limited to looking at the moment of inertia of a 2D "slice" from the object. You could then sum all the slices together.



            Other than the fact that the $z$ axis must be perpendicular to the plane of the object, the choice of axes is yours. Normally you would choose axes that make the calculation simpler (or possible). It depends completely on the problem you're trying to solve.



            The theorem just states the relationship. It doesn't mean that there is necessarily a unique choice, or that any choice is especially useful.






            share|cite|improve this answer




















            • So if i have any slice of cylinder which would be a shell like lying on xy plane then z is the perpendicular axis . Am i right?
              – Nobody recognizeable
              3 hours ago






            • 1




              Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
              – BowlOfRed
              2 hours ago












            up vote
            3
            down vote










            up vote
            3
            down vote









            If you have a 3 dimensional object, then it doesn't lie entirely in any plane. You can't use the theorem directly. You would instead be limited to looking at the moment of inertia of a 2D "slice" from the object. You could then sum all the slices together.



            Other than the fact that the $z$ axis must be perpendicular to the plane of the object, the choice of axes is yours. Normally you would choose axes that make the calculation simpler (or possible). It depends completely on the problem you're trying to solve.



            The theorem just states the relationship. It doesn't mean that there is necessarily a unique choice, or that any choice is especially useful.






            share|cite|improve this answer












            If you have a 3 dimensional object, then it doesn't lie entirely in any plane. You can't use the theorem directly. You would instead be limited to looking at the moment of inertia of a 2D "slice" from the object. You could then sum all the slices together.



            Other than the fact that the $z$ axis must be perpendicular to the plane of the object, the choice of axes is yours. Normally you would choose axes that make the calculation simpler (or possible). It depends completely on the problem you're trying to solve.



            The theorem just states the relationship. It doesn't mean that there is necessarily a unique choice, or that any choice is especially useful.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            BowlOfRed

            15.1k22339




            15.1k22339











            • So if i have any slice of cylinder which would be a shell like lying on xy plane then z is the perpendicular axis . Am i right?
              – Nobody recognizeable
              3 hours ago






            • 1




              Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
              – BowlOfRed
              2 hours ago
















            • So if i have any slice of cylinder which would be a shell like lying on xy plane then z is the perpendicular axis . Am i right?
              – Nobody recognizeable
              3 hours ago






            • 1




              Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
              – BowlOfRed
              2 hours ago















            So if i have any slice of cylinder which would be a shell like lying on xy plane then z is the perpendicular axis . Am i right?
            – Nobody recognizeable
            3 hours ago




            So if i have any slice of cylinder which would be a shell like lying on xy plane then z is the perpendicular axis . Am i right?
            – Nobody recognizeable
            3 hours ago




            1




            1




            Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
            – BowlOfRed
            2 hours ago




            Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
            – BowlOfRed
            2 hours ago










            up vote
            1
            down vote













            I, by definition has to do with a surface not a volume and integrating sum of I of slices is meaningless. Hence, it is called second moment of area, not volume.



            lets use a cucumber as an example. If we cut an infinitely thin slice from the middle and look at the round cross section we can assign 3 axis to that surface, X,Y,Z.
            Then $I_z = I_x + I_y $ with X being in-plane of round surface horizontal axis and Y in-plane vertical. And Z an axis coming out of the surface vertically.



            But this Z should not be confused with deceptively similar sounding Z axis of the whole cucumber if we had positioned it at the same origin vertically.



            If we were to measure that Z we needed to cut the cucumber longitudinally like a long oval shape and then we could call that axis Z if we liked to, and the horizontal short axis X. But I of that axis would be $ int_-x^+x Z^2da $



            Which is obviously totally different.






            share|cite|improve this answer
























              up vote
              1
              down vote













              I, by definition has to do with a surface not a volume and integrating sum of I of slices is meaningless. Hence, it is called second moment of area, not volume.



              lets use a cucumber as an example. If we cut an infinitely thin slice from the middle and look at the round cross section we can assign 3 axis to that surface, X,Y,Z.
              Then $I_z = I_x + I_y $ with X being in-plane of round surface horizontal axis and Y in-plane vertical. And Z an axis coming out of the surface vertically.



              But this Z should not be confused with deceptively similar sounding Z axis of the whole cucumber if we had positioned it at the same origin vertically.



              If we were to measure that Z we needed to cut the cucumber longitudinally like a long oval shape and then we could call that axis Z if we liked to, and the horizontal short axis X. But I of that axis would be $ int_-x^+x Z^2da $



              Which is obviously totally different.






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                I, by definition has to do with a surface not a volume and integrating sum of I of slices is meaningless. Hence, it is called second moment of area, not volume.



                lets use a cucumber as an example. If we cut an infinitely thin slice from the middle and look at the round cross section we can assign 3 axis to that surface, X,Y,Z.
                Then $I_z = I_x + I_y $ with X being in-plane of round surface horizontal axis and Y in-plane vertical. And Z an axis coming out of the surface vertically.



                But this Z should not be confused with deceptively similar sounding Z axis of the whole cucumber if we had positioned it at the same origin vertically.



                If we were to measure that Z we needed to cut the cucumber longitudinally like a long oval shape and then we could call that axis Z if we liked to, and the horizontal short axis X. But I of that axis would be $ int_-x^+x Z^2da $



                Which is obviously totally different.






                share|cite|improve this answer












                I, by definition has to do with a surface not a volume and integrating sum of I of slices is meaningless. Hence, it is called second moment of area, not volume.



                lets use a cucumber as an example. If we cut an infinitely thin slice from the middle and look at the round cross section we can assign 3 axis to that surface, X,Y,Z.
                Then $I_z = I_x + I_y $ with X being in-plane of round surface horizontal axis and Y in-plane vertical. And Z an axis coming out of the surface vertically.



                But this Z should not be confused with deceptively similar sounding Z axis of the whole cucumber if we had positioned it at the same origin vertically.



                If we were to measure that Z we needed to cut the cucumber longitudinally like a long oval shape and then we could call that axis Z if we liked to, and the horizontal short axis X. But I of that axis would be $ int_-x^+x Z^2da $



                Which is obviously totally different.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                kamran

                90639




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