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jolt527
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Hi, this is my first post here, so please forgive any errors in my post. :)
I'm recently thinking about switching from my major of computer science to physics, and have been brushing up on the first few semesters of physics I had taken a few years ago. I'm currently in the section on rotational motion and moments of inertia, and was looking at the parallel axis theorem and the proof they provided. I had a question about one of the parts of the proof, so I'll list what was written in my book (see attached image for reference):
https://www.physicsforums.com/attachments/19484
Suppose that an object rotates in the xy-plane about the z-axis. The coordinates of the center of mass are [tex] x_{CM}, y_{CM} [/tex]. Let the mass element dm have coordinates x, y. Because this element is a distance [tex]r = \sqrt{x^2 + y^2}[/tex] from the z-axis, the moment of inertia about the z-axis is:
[tex]I = \int r^2 dm = \int \left(x^2 + y^2\right) dm[/tex]
However, we can relate the coordinates x, y of the mass element dm to the coordinates of this same element located in a coordinate system having the object's center of mass as its origin. If the coordinates of the center of mass are [tex]x_{CM}, y_{CM}[/tex] in the original coordinate system centered on O, then from the attached figure we see that the relationships between the unprimed and primed coordinates are [tex]x = x' + x_{CM}[/tex] and [tex]y = y' + y_{CM}[/tex] Therefore,
[tex]I = \int \left[\left(x' + x_{CM}\right)^2 + \left(y' + y_{CM}\right)^2\right] dm[/tex]
[tex]I = \int \left[\left(x'\right)^2 + \left(y'\right)^2\right] \,dm + 2x_{CM} \int x' \,dm + 2y_{CM} \int y' \,dm + \left({x_{CM}}^2 + {y_{CM}}^2\right) \int \,dm[/tex]
The first integral is, by definition, the moment of inertia about an axis that is parallel to the z-axis and passes through the center of mass. The second two integrals are zero because, by definition of the center of mass,
[tex]\int x' dm = \int y' dm = 0[/tex]
The last integral is simply [tex]MD^2[/tex] because [tex]D^2 = {x_{CM}}^2 + {y_{CM}}^2[/tex] and
[tex]\int dm = M[/tex]
Therefore, we conclude that
[tex]I = I_{CM} + MD^2[/tex]
Now, I must admit that all of this proof totally makes sense, with the exception of the bold text "The second two integrals are zero because, by definition of the center of mass" followed by the integrals of [tex]x' dm[/tex] and [tex]y' dm[/tex] being equal to zero. This part stumps me - why are the equal to zero? I'm not sure what they mean by "by definition of the center of mass" and how it makes those integrals equal to zero. Could someone please explain why those two integrals are equal to zero? Thank you in advance! :)
-Keith
I'm recently thinking about switching from my major of computer science to physics, and have been brushing up on the first few semesters of physics I had taken a few years ago. I'm currently in the section on rotational motion and moments of inertia, and was looking at the parallel axis theorem and the proof they provided. I had a question about one of the parts of the proof, so I'll list what was written in my book (see attached image for reference):
https://www.physicsforums.com/attachments/19484
Suppose that an object rotates in the xy-plane about the z-axis. The coordinates of the center of mass are [tex] x_{CM}, y_{CM} [/tex]. Let the mass element dm have coordinates x, y. Because this element is a distance [tex]r = \sqrt{x^2 + y^2}[/tex] from the z-axis, the moment of inertia about the z-axis is:
[tex]I = \int r^2 dm = \int \left(x^2 + y^2\right) dm[/tex]
However, we can relate the coordinates x, y of the mass element dm to the coordinates of this same element located in a coordinate system having the object's center of mass as its origin. If the coordinates of the center of mass are [tex]x_{CM}, y_{CM}[/tex] in the original coordinate system centered on O, then from the attached figure we see that the relationships between the unprimed and primed coordinates are [tex]x = x' + x_{CM}[/tex] and [tex]y = y' + y_{CM}[/tex] Therefore,
[tex]I = \int \left[\left(x' + x_{CM}\right)^2 + \left(y' + y_{CM}\right)^2\right] dm[/tex]
[tex]I = \int \left[\left(x'\right)^2 + \left(y'\right)^2\right] \,dm + 2x_{CM} \int x' \,dm + 2y_{CM} \int y' \,dm + \left({x_{CM}}^2 + {y_{CM}}^2\right) \int \,dm[/tex]
The first integral is, by definition, the moment of inertia about an axis that is parallel to the z-axis and passes through the center of mass. The second two integrals are zero because, by definition of the center of mass,
[tex]\int x' dm = \int y' dm = 0[/tex]
The last integral is simply [tex]MD^2[/tex] because [tex]D^2 = {x_{CM}}^2 + {y_{CM}}^2[/tex] and
[tex]\int dm = M[/tex]
Therefore, we conclude that
[tex]I = I_{CM} + MD^2[/tex]
Now, I must admit that all of this proof totally makes sense, with the exception of the bold text "The second two integrals are zero because, by definition of the center of mass" followed by the integrals of [tex]x' dm[/tex] and [tex]y' dm[/tex] being equal to zero. This part stumps me - why are the equal to zero? I'm not sure what they mean by "by definition of the center of mass" and how it makes those integrals equal to zero. Could someone please explain why those two integrals are equal to zero? Thank you in advance! :)
-Keith