What are some Canadian moments in the 1970's?

Can you explain this method for finding the centre of mass using moments?

• I’ve been studying some introduction mechanics and the textbook I’m using has a method that seems odd to me. It concerns finding the centre of mass of a system of point masses. It explains that given the mass, [math] m_i [/math] and the coordinates [math] (x_i,y_i) [/math] we simply assume that this system of points masses is equivalent to a single mass located at the systems centre of mass, that is at the coordinates [math] (\bar{x},\bar{y}) [/math] with mass: [math]\sum\limits_{i=1}^n {m_i }=M [/math]. Up to this is fine, it’s what follows when it goes about finding [math] (\bar{x},\bar{y}) [/math] that I’m confused about. It says we do it by ‘equating the moments of the two systems about the y-axis’ where it does the following computation: [math] \sum\limits_{i=1}^n {m_ig x_i}=Mg\bar{x} [/math] and it then repeats the above reasoning for finding the y-coordinate for the centre of mass with moments about the x-axis: [math] \sum\limits_{i=1}^n {m_ig y_i}=Mg\bar{y} [/math] For the case with the y-axis that sort of makes sense to me since we can define the moment of a force to be the distance to the line of action of the force (the x-coordinate) multiplied by the magnitude of the force (in this case the weight [math] m_i g[/math]) but how do we take moments about an entire axis? I am reasoning that we are actually taking moments about the point [math] (0,0)[/math], but then how does this apply to the y-coordinate for the centre of mass since seemingly it shouldn’t be included in the equations (since the only thing that is relevant to the ‘distance to the lines of action of the force’ is the x coordinate). I’m sure there must be something I’m not understanding about moments and that there is something I’m missing. A possible solution I came across was when reading the Feynman lectures on physics he says the entire system of particles could be rotated and then we could be proceed to find the y-coordinate of the centre of mass; but my textbook makes no mention of anything like this and Feynman isn’t specifically talking about this (he was talking about the total torques adding to zero) so I’m assuming the reasoning is wrong. To me it seems more intuitively obvious to define the centre of mass as the mass weight average of the coordinates which explains the equation normally given: [math] \frac{\sum\limits_{i=1}^n {m_i x_i}}{\sum\limits_{i=1}^n {m_i }}=\bar{x}  [/math] but this textbook doesn’t define it as such and goes about using this moment technique. So if anyone could help me understand how this textbook is defining it and hopefully gain some insight and a deeper understanding of moments would be great. Thanks in advance!

• Answer:

  Well, solve for , and you get exactly the weighted average of the coordinates (since the g's cancel). However, I don't think that answer would satisfy your curiosity, so I'll elaborate. how do we take moments about an entire axis? Well, moments about a point don't make sense, because there are an infinite number of axes that go through a point. However, in the case of plane figures, what you think of as the moment about a point is just the moment about an axis perpendicular to the plane that passes through that point. Imagine a disk sitting in the xy-plane in 3D space centered at the origin. It has 3 components of rotational freedom (about the x, y, and z axes). Rotation about the z-axis would be equivalent to what you think of as a moment about the origin; however, since the disk can also rotate about the x and y axes (think of taking a disk and just flipping it over), rotation about a point isn't quite the same thing when we leave 2 space, because there are 3 degrees of rotational freedom about a point in 3-space (unlike 2-space where there is only one). the only thing that is relevant to the ‘distance to the lines of action of the force’ is the x coordinate That's true, but that only gives information regarding the moment about the y-axis (and therefore ). To find , we do need to find the weighted average of the distances to the x-axis. The idea is that this planar figure is sitting in 3-space, and the force due to gravity acts downwards; therefore we can ignore rotation about the z-axis, because gravity is acting parallel to it. However, if it was heavier on one side of the x (or y) axis, and if if it were to rest on that axis, it would have a tendency to rotate around it unless it was moved around so that the axis passed through its center of mass.