Help finding mass and center of mass?

Finding the center of mass?

  • i'm helping a friend with calculus, and we're working on a center of mass question. it says to find the x coordinate of the center of mass, x bar, we take M_y / M, where M_y is the moment about the y axis and M is the mass of the entire object. density is consistent M will be ∫ρ*f(x) dx from a to b M_y will be ∫ρ*x*f(x) from a to b this is simple, the density will cancel out from top and bottom and everything works out easily. i understand this perfectly. however, after thinking about what the moment about the y axis really MEANS, i thought this up. what if instead we take this integral instead: ∫ρ*(x - k)*f(x) = 0 from a to b where k is the x value of the center of mass, x bar. we don't know k, but we solve for it since we know it must equal zero. we can easily figure out the rest of the integral. ρ just becomes zero, so it doesn't affect the value of k so long as density is constant. i tried it for a few simple functions and i get the same result. However, i haven't read anything about this in the book. maybe it's because i haven't taken a physics class. anyway, my question(s) is/are this: does this formula work for all "find the center of mass questions" so long as density is constant? or are there exceptions? was my thinking completely wrong? is there any way this could work for objects with a density function?

  • Answer:

    Mathematically speaking, what you have written is valid and would also work for a solid with a density function; just replace ρ with ρ(x). However, it has a drawback in that what was a direct calculation now becomes an algebra problem with this unknown k. In essence, all it does is introduce an extra step. However, it could be useful as an identity or in a proof.

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Mathematically speaking, what you have written is valid and would also work for a solid with a density function; just replace ρ with ρ(x). However, it has a drawback in that what was a direct calculation now becomes an algebra problem with this unknown k. In essence, all it does is introduce an extra step. However, it could be useful as an identity or in a proof.

DavidK93

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