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Moment of Inertia of Solid and Hollow Spheres using Disks and Rings Respectively

  1. May 31, 2009 #1
    Whilst following my textbooks advice and "proving to myself that the inertia listed are true" I considered the Moment of Inertia of a hollow sphere by adding up infinitesimally thin rings:

    dI = y^2 dm
    = y^2 \sigma dS
    = y^2 \sigma 2\pi y dz
    = 2\pi y^3 \sigma dz

    This didn't work however as it appears you have to consider the projection of [itex]dS[/itex] as "seen" from the z-axis; that is, [itex]dS = \frac{dz}{sin\theta}[/itex].

    I kind of understand this but I was wondering if there is anything similar to consider when treating a solid sphere as infinitesimally thin disks; a "projection of volume"? http://hyperphysics.phy-astr.gsu.edu/HBASE/isph.html#sph2 doesn't imply that there is but in that case can someone explain why not?
  2. jcsd
  3. Jun 2, 2009 #2


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    From what you say, it's not clear you really understand the appearance of the [tex]\sin \theta[/tex], so just in case let me go over that first.

    The important point is that you need to know the mass of each strip, which amounts to knowing their surface area. Now imagine you have a paper sphere and you draw lines of latitude around it dividing it into many small strips. The surface area of one of these strips is just length times width. The length is just [tex]2 \pi y[/tex], but the width depends on where the strip is. Along the equator, if you draw the bottom line at a height [tex]z=0[/tex] and the next line at a height [tex]z=dz[/tex] then the width of the strip so formed is just [tex]dz[/tex] (for small enough [tex]dz[/tex]) because the surface formed by the strip is parallel to the vertical ([tex]z[/tex]) direction. But if you imagine a strip around the sphere at an angle [tex]45^{\circ}[/tex] above the equator, then drawing the lines [tex]dz[/tex] apart will actually result in a strip whose physical width (i.e. the amount of paper) is [tex]dz\sqrt{2}[/tex] since the surface made by the strip of paper is at [tex]45^{\circ}[/tex] to the vertical and so more paper is traversed in increasing the height by [tex]dz[/tex]. The [tex]\sin{\theta}[/tex] term accounts for this effect.

    Now, for the case of a solid sphere you are interested instead in the volume of each disk in order to account for their mass. Once again we still have the problem that the edges of the disk are not in general parallel to the [tex]z[/tex] axis. An easy way to deal with this problem is to pretend first that the infinitesimal disks are vertical sided, and then to cut a small amount off their corner in order to make them the correct size, e.g.

    [tex]dV =[/tex] area [tex]\times[/tex] height - corner volume

    [tex] = \pi y^2\; dz - [/tex] corner volume

    The 'corner' that we cut off is basically a ring with a triangular cross-section. The volume of this ring is given by length times cross-sectional area. The length is just [tex]2\pi y[/tex], and the cross-sectional area is given by the usual area of a triangle:

    [tex] dA = 1/2 \times [/tex] base [tex] \times [/tex] height

    which if you work it out gives:

    [tex] dA = 1/2 \tan{\theta} (dz)^2 [/tex]

    Now, because we are already making [tex]dz[/tex] infinitesimally small, [tex](dz)^2[/tex] will be much much smaller still, and so we can completely ignore the volume of the corner. So we can just write the volume of each disk ([tex]dV[/tex]) as if we hadn't corrected it for the shape of the edges. This is why you can (luckily) get the correct result without considering the shape of the disks at the edges of the sphere.

    Depending on how familiar you are with calculus, this may seem like the answer must as a result just be an approximation. This is not so, the answer is still exact. The basic idea is that we are free to choose the increment [tex]dz[/tex] as small as we like as we go through evaluating our integral. Thus we can just keep making this increment smaller and smaller until we get to the point where the contribution from the 'corner correction' ([tex]\propto (dz)^2[/tex]) is infinitely smaller than the contribution from the rest of the integrand. In the limit as this increment tends to zero, so the contribution from the 'corner correction' becomes zero.
  4. Jun 2, 2009 #3
    Ah, thank you; I understood where the [itex]sin(\theta)[/itex] came from but didn't really phrase my enquiry very well. What I really wanted to ask was do you not have to consider the fact that the disks aren't vertically sided (as you do for the rings) which you have already answered; thanks! ^^
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