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Math Help - Find z-coordinate of center of mass

  1. #1
    Senior Member Twig's Avatar
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    Find z-coordinate of center of mass

    Hi

    I wasn´t quite sure where to post this, but it involves integrals so I chosed calculus.

    I have attached a picture.

    Problem:
    Find the z-coordinate of center of mass of the object.

    Not really sure how to start.

    But wont a differential slice of the cone be something like  \pi \cdot r_{0}^{2} \cdot dz ?

    But  r_{0}^{2}=x^{2}+y^{2} , for each different slice. But the radius  r_{0} is obviously changing with z.

    Should I try to express x and y as functions of z ?

    Thx
    Attached Thumbnails Attached Thumbnails Find z-coordinate of center of mass-pic_03.jpg  
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Twig View Post
    Hi

    I wasn´t quite sure where to post this, but it involves integrals so I chosed calculus.

    I have attached a picture.

    Problem:
    Find the z-coordinate of center of mass of the object.

    Not really sure how to start.

    But wont a differential slice of the cone be something like  \pi \cdot r_{0}^{2} \cdot dz ?

    But  r_{0}^{2}=x^{2}+y^{2} , for each different slice. But the radius  r_{0} is obviously changing with z.

    Should I try to express x and y as functions of z ?

    Thx
    Let R be the radius of circular base of the cone.

    and r be the radius of "slice" at distance z from the vertex.

    then, \frac{R}{h}=\frac{r}{z}

    There is no need of x and y in the equation, only z is required(since by symmetry, centroid lies on z-axis)
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  3. #3
    Senior Member Twig's Avatar
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    Hi

    Wow, nice thats what I did =)

    So got  r=\frac{zR}{h}

    Thus..

     \int_{0}^{h} z\cdot \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz = \frac{\pi R^{2}h^{2}}{4}
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Twig View Post
    Hi

    Wow, nice thats what I did =)

    So got  r=\frac{zR}{h}

    Thus..

     \int_{0}^{h} z\cdot \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz = \frac{\pi R^{2}h^{2}}{4}
    z co-ordinate:
    \frac{ \int_{0}^{h} z\cdot \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz}{\int_{0}^{h} \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz}
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  5. #5
    Senior Member Twig's Avatar
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    Quote Originally Posted by malaygoel View Post
    z co-ordinate:
    \frac{ \int_{0}^{h} z\cdot \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz}{\int_{0}^{h} \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz}

    Yes, alright thanks. Here you divide by the entire volume of the cone, right, since it evaluates to \frac{\pi R^{2}h}{3} .

    The 'formula' is  Vz_{T}=\int z_{c}\; dV ?

    Where I let the subscripit T denote the center of mass and subscript C denote moment arm.
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  6. #6
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Twig View Post
    Yes, alright thanks. Here you divide by the entire volume of the cone, right, since it evaluates to \frac{\pi R^{2}h}{3} .

    The 'formula' is  Vz_{T}=\int z_{c}\; dV ?

    Where I let the subscripit T denote the center of mass and subscript C denote moment arm.
    Your formula is correct.

    have you got your answer??? and most importantly, understood it?

    As in this question, use symmetry wherever possible
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  7. #7
    Senior Member Twig's Avatar
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    Hi!

    Yes, I got \frac{3h}{4}

    I think I understand decently.
    Yes, the book also says USE SYMMETRY! =)
    makes life easier

    The hardest part is definetely setting up the integral, or the geometry of the figure.

    thanks a lot.
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