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Homework Help: CM of an arc

  1. Jul 18, 2009 #1
    1. The problem statement, all variables and given/known data

    There is an arc with an angle of [tex]2\alpha[/tex] and radius [tex]R[/tex]
    What is its center of mass. (It is implied that it is to be measured from the center of the circle the arc is a part of.

    2. Relevant equations
    Integration, trigonometry.

    3. The attempt at a solution
    Well, the formula for a center of mass of [tex]i[/tex] objects is:
    [tex]\vec r_{cm}=\frac{\Sigma m_i \vec r_i}{M_{tot}}[/tex]

    Since the arc has symmetry with respect to the y axis, all we need to find is the y coordinate of its center of mass.

    We'll simply replace the discrete summation with an integral (I'm not too hot with these).

    The mass of every tiny element of the arc length, [tex]dS[/tex] is [tex]\sigma dS=\sigma R d\theta[/tex]
    Where [tex]\sigma \equiv \frac{M}{S}=\frac{M}{2R\alpha}[/tex]

    The distance of 'each' of these masses from the x axis, their y value, is provided by [tex]R\sin{\theta}[/tex]

    What I want my integral to do is 'sweep' across the arc. Would the correct boundaries for such an integral be:
    [tex]\frac{\pi}{2}-\alpha \leftrightarrow \frac{\pi}{2}+\alpha[/tex]

    The integral would need to sum up the product of each of the mass elements corresponding to a particular angle theta with its corresponding height (The sine of that theta).

    Would the correct integral be (I can't quite get the hang of boundaries in LaTeX, so refer to the boundaries I wrote above):
    [tex]\int \sigma R*R\sin{\theta}\ d\theta[/tex]

    I'm not in the least bit skilled with integrals so I'd like to know if that's the correct expression, and if so, where the [tex]d\theta[/tex] goes from the expression for the tiny mass element. Or if I got it wrong, I'd love a hint as to how to construct the proper integral. (A link with the basics of such integration would be greatly appreciated as well. :))

    The textbook's answer is [tex]\frac{R\sin{\alpha}}{\alpha}[/tex]
    The way I solved the integral gave me that answer as well.
    Last edited: Jul 18, 2009
  2. jcsd
  3. Jul 18, 2009 #2
    There is all correct in your solution. The [tex]d\theta[/tex] is not goes from the expression for the tiny mass element. It goes out after integrating:
    [tex]\int\limits_{\frac \pi 2 - \alpha}^{\frac \pi 2 + \alpha} \sigma R^2 \sin \theta d\theta = \left . -\sigma R^2 \cos \theta \right |_{\frac \pi 2 - \alpha}^{\frac \pi 2 + \alpha} = [/tex].

    Read this book about integrating and methods of applied mathematics: Zeldovich Ya.B., Yaglom I.M.
    Higher math for beginners (mostly physicists and engineers)
  4. Jul 18, 2009 #3
    Alright, thank you very much! :)
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