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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?
moment-of-inertia rigid-body-dynamics
edited 2 hours ago
Buzz
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up 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?
moment-of-inertia rigid-body-dynamics
edited 2 hours ago
Buzz
2,83321321
asked 3 hours ago
Nobody recognizeable
516516
add a comment |Â
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?
moment-of-inertia rigid-body-dynamics
edited 2 hours ago
Buzz
2,83321321
asked 3 hours ago
Nobody recognizeable
516516
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
moment-of-inertia rigid-body-dynamics
edited 2 hours ago
Buzz
2,83321321
asked 3 hours ago
Nobody recognizeable
516516
edited 2 hours ago
Buzz
2,83321321
asked 3 hours ago
Nobody recognizeable
516516
edited 2 hours ago
Buzz
2,83321321
edited 2 hours ago
Buzz
2,83321321
edited 2 hours ago
Buzz
2,83321321
2,83321321
asked 3 hours ago
Nobody recognizeable
516516
asked 3 hours ago
Nobody recognizeable
516516
asked 3 hours ago
Nobody recognizeable
516516
516516
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add a comment |Â
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.
answered 3 hours ago
BowlOfRed
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
2 hours ago
1
Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
– BowlOfRed
2 hours ago
add a comment |Â
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.
answered 2 hours ago
kamran
90639
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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.
answered 3 hours ago
BowlOfRed
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
2 hours ago
1
Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
– BowlOfRed
2 hours ago
add a comment |Â
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.
answered 3 hours ago
BowlOfRed
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
2 hours ago
1
Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
– BowlOfRed
2 hours ago
add a comment |Â
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.
answered 3 hours ago
BowlOfRed
15.1k22339
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.
answered 3 hours ago
BowlOfRed
15.1k22339
answered 3 hours ago
BowlOfRed
15.1k22339
answered 3 hours ago
BowlOfRed
15.1k22339
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
2 hours ago
1
Yes, theorem is only valid when z is chosen perpendicular to the plane containing the slice.
– BowlOfRed
2 hours ago
add a comment |Â
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
2 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
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
2 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
add a comment |Â
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.
answered 2 hours ago
kamran
90639
add a comment |Â
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.
answered 2 hours ago
kamran
90639
add a comment |Â
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.
answered 2 hours ago
kamran
90639
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.
answered 2 hours ago
kamran
90639
answered 2 hours ago
kamran
90639
answered 2 hours ago
kamran
90639
answered 2 hours ago
kamran
90639
90639
add a comment |Â
add a comment |Â
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