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Moment of inertia of a solid

  1. Jun 13, 2005 #1
    How do I find the moment of inertia of a solid formed by rotating the curve of y=sinx about the x-axis in the interval [0, pi]?

    I've tried to set up integrals by summing up cylinders parallel to the y-axis but to no avail.
  2. jcsd
  3. Jun 13, 2005 #2


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    I'm assuming it's the moment of inertia of the solid about the x-axis that the question is asking for.

    Let the volume density of the solid of revolution be [itex]\rho[/itex]. Then a cylindrical element of mass [itex]dm[/itex] is defined by [itex]\rho \pi y^2 dx[/itex]. The moment of inertia of that element about the x-axis is defined by [itex]\frac{1}{2}y^2dm = \frac{1}{2}\rho \pi y^4 dx[/itex]. Substitute [itex]y = \sin x[/itex] and integrate over the required bounds and you have the answer. To remove the [itex]\rho[/itex] term and leave your answer purely in terms of the total mass [itex]M[/itex], just calculate the volume of revolution [itex]V[/itex] the usual way and put [itex]\rho = \frac{M}{V}[/itex].
    Last edited: Jun 13, 2005
  4. Jun 13, 2005 #3
    Yeah, that worked!

    What I did was EXACTLY the same as your method, except I multipled dm by x^2 instead of y^2. Oops.. :shy:

    Thank you. :smile:
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