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Math Help - Another center of mass problem

  1. #1
    Senior Member Twig's Avatar
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    Another center of mass problem

    Hi!

    Attached picture below.

    Problem: Calcuate center of mass of the slender rod.

    Solution attempt:

    Since the density varies, I did: With  dm = \rho \cdot A \cdot dx , because I guess the area is the same throughout the rod.

    \bar{x}=\frac{\int_{0}^{1}x\cdot \rho_{0}(1-\frac{x}{2})A\; dx}{\int_{0}^{1} \rho_{0}(1-\frac{x}{2})A\; dx} = \frac{4}{9}

    Thx for your help!
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Twig View Post
    Hi!

    Attached picture below.

    Problem: Calcuate center of mass of the slender rod.

    Solution attempt:

    Since the density varies, I did: With  dm = \rho \cdot A \cdot dx , because I guess the area is the same throughout the rod.

    \bar{x}=\frac{\int_{0}^{1}x\cdot \rho_{0}(1-\frac{x}{2})A\; dx}{\int_{0}^{1} \rho_{0}(1-\frac{x}{2})A\; dx} = \frac{4}{9}

    Thx for your help!
    I confirmed that \frac{\int_{0}^{1}x\cdot \rho_{0}(1-\frac{x}{2})A\; dx}{\int_{0}^{1} \rho_{0}(1-\frac{x}{2})A\; dx} = \frac{4}{9}

    but I really don't know if that's the correct formula for the centre of mass of a rod.
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  3. #3
    Super Member Random Variable's Avatar
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    It seems correct except for the constant A. The rod is a 1-dimensional object.
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  4. #4
    Senior Member Twig's Avatar
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    I was thinking about the cross sectional area of the rod.

    So you mean I should just take dm=\rho \cdot dx ?
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Twig View Post
    I was thinking about the cross sectional area of the rod.

    So you mean I should just take dm=\rho \cdot dx ?
    Yes. The word "slender" in this problem is telling you to treat the rod as 1-dimensional.
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  6. #6
    Senior Member Twig's Avatar
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    Quote Originally Posted by Random Variable View Post
    Yes. The word "slender" in this problem is telling you that can treat the rod as 1-dimensional.
    Ah, I see. Really didnīt know that word in english =)

    Note: But the constant A didnīt affect the result? \frac{4}{9} is correct?
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  7. #7
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Twig View Post
    Ah, I see. Really didnīt know that word in english =)

    Note: But the constant A didnīt affect the result? \frac{4}{9} is correct?
    your answer is correct.

    A doesn't affect the result because it is assumed to be constant.

    A two-dimensional figure is a as good as a one-dimensional as long as one dimension remains unchanged.

    For example, if we are dealing with all rectangles with the same same, then we have to consider only their relative length..hence problem comes in the realm of one-dimension.
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