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The moment of inertia of circular sector

  1. Nov 7, 2012 #1
    Hello

    how can I find the moment of inertia of a circular sector about the X axis , which the sector is symmetrical about , with -θ down and θ above ?!

    I[itex]_{x}[/itex] = ∫y[itex]^{2}[/itex] dA = ∫y[itex]^{2}[/itex] *y *dx

    Or =∫∫ y[itex]^{2}[/itex] dy dx

    I don't know how to put the limits of integration , I turned it to polar double integral and put the limits of r from zero to R , and limits of θ from -θ to θ and it gave the correct answer but I don't know why :)
     
  2. jcsd
  3. Nov 7, 2012 #2
    That's correct if you assume a uniform density distribution of 1. In x and y, you could use two types of limits. If we define the bounding curves by x as a function of y, you can go from the V shape to the circle. The V shape would be from x = |cot(θ)y| and it would go to x = sqrt(R^2 - y^2) . y would then vary from 0 to R. You can also go the other way around and define the curves as functions of x. However, you can see that since your area has radial symmetry, it is easier to integrate in polar coordinates.
     
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