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What is dm? (moment of inertia)

  1. Apr 3, 2004 #1


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    Mass moment of inertia 'I' is given by

    [tex]I = \int r^2 dm[/tex]

    r is the distance from the axis and dm is delta mass.

    What is dm? How do I figure that out?

    For a cylinder, the moment of inertia around the axis that goes from flat end to flat end is [itex]\frac{1}{2}mr^2[/itex].
    The r^2 part is self explanatory but where does (1/2)m come from?
  2. jcsd
  3. Apr 3, 2004 #2
    [tex] dm = \rho (\vec{r}) dV[/tex]
    where [itex]\rho(\vec{r})[/itex] is the mass density as a function of position,

    which yields a more general equation

    [tex]I = \int_V r^2 \rho(\vec{r}) \, dV[/tex]

  4. Apr 5, 2004 #3


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    Your 'differential mass' (dm) is the rate that the mass changes in response to a change in the radius - i.e. the density times the differential volume (rate volume changes).

    To find the density you need to how to find the volume of the object (for a cylinder, find the area of one of the ends and multiply by the height of the cylinder). And density is just mass divided by volume.

    The differential volume is just the rate that the volume changes in response to an increase in the radius. In other words, its the derivative of the formula used to find the volume.

    It's the same basic procedure for any shape, except, obviously, the formula for finding the volume changes depending upon the shape.

    For the more painful shapes, the key parameter is the density. If you get the density right, there's a lot of shortcuts you can take for finding the moment of inertia (finding I for a fictional solid object and then subtracting out I for the portion you don't need - much like the easiest way to find the area on a CD is to find the area of a solid disk and then subtract out the area of the hole).
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