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Homework Help: Moment of Inertia Calculation

  1. Nov 14, 2006 #1
    Hey guys

    I was reading up on how to calculate I for various shapes on a website i often use and i understand their calc for a rod of length L but i dont get it for a sphere.

    This page here http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html#sph3
    shows [tex] dI = \frac{y^2dm}{2}[/tex] and i dont know where the half comes from, can anyone explain?

  2. jcsd
  3. Nov 14, 2006 #2


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    Because it is the moment of inertia of a think disk, so basicly you are calculating the moment of inertia of the sphere as summing the moments of thin disks.
  4. Nov 14, 2006 #3
    I understand that, you didnt read my question. I am asking about the half. As far as i know dI = r^2 dm not a half of that
  5. Nov 15, 2006 #4
    the differential disk is a solid disk. for a solid disk the moment of inertia is .5mr^2, if it was a ring, it would've been mr^2
  6. Nov 15, 2006 #5
    Err ok dont spose you can show me the working for that please?
  7. Nov 15, 2006 #6
    Okay -- you know dI = r^2 dm

    For the ring -- r is fixed. So each ring is I = r^2 m
    For a disk of radius R, you'll need to then integrate over rings. For this, you need the dm for each ring, which relates to the total mass of the disk.

    You need
    area density:
    [tex]\rho_A=\frac{M}{\pi R^2}[/tex]
    linear density within ring of radius r:
    [tex]\lambda=2\pi r \rho_A[/tex]
    Note [tex]\lambda=\frac{dm}{dr}[/tex]...

    So change variables from dm to dr, and integrate over the radius of the rings (from r=0 to R) to find the disk. You should get a factor of 1/2.

    Once you do this... try the same thing with disks to the sphere. It's a little harder.

    Alternatively (and I think this is easier):
    For the disk you can use the area density, and just do a double integral (using cylindrical coordinates) over the variables r and theta.
    For the sphere you can use volume density, and do a volume integral over the spherical coordinates r, theta, phi.

    Edit: Minor edits to the text for format
    A later edit to correct a latex coding
    Last edited: Nov 15, 2006
  8. Nov 15, 2006 #7
    ah ok i get ya. I think tho you can cut out the equating of different densities though if you write

    [tex] \frac{M}{R} = \frac{dM}{dR}[/tex] and then sub for dM in the intergral or is that cutting corners and it only works in this situation kinda thing?
  9. Nov 15, 2006 #8
    In my notation, M was the total mass of the disk, R was the radius of the disk... dm was a litte piece of mass somewhere, and r was a little radius variable we needed to integrate over. in that sense.. a dM/dR doesn't make any sense, and no -- M/R is probably not dm/dr.

    the dm's we wanted were the contributions of mass from each ring so we could integrate over the rings.

    I didn't really "equate" the densities so much as find the contribution of the area density to the ring's mass --- by multiplying by the circumference of the ring.

    I REALLY like the second suggestion better -- take the area density and do an area integral:
    [tex]\int r^2 \rho_A dA [/tex]
    [tex]\int_0^{2\pi} \int_0^R r^2 \rho_A r dr d\theta [/tex].

    You can do this in volume for the sphere -- much cleaner... although since you've been thinking about the other way, I'd do it too for practice.
  10. Nov 15, 2006 #9


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    I guess everyone must realize that the calculation performed on that website is valid not for the 2-sphere, but for the 2-ball.

    For the 2-sphere the whole story (including the result) is different.

  11. Nov 15, 2006 #10
    Ah ok the ring thing makes sense, so youre saying that given the overall density, you wanted to find the mass dm in each ring of width dr? and so treating those rings simply as lines with no depth you used a linear density times the circumferance...sound right?

    Another stupid question then, area density is mass on area right? then why have you used mass on 2A? ie 2pir^2
  12. Nov 15, 2006 #11
    right on both counts -- The first conceptually -- and the second in my latex typing (not in my own scribbled results)-- I'll go correct that in an "edit"... with the typo you would get a factor of 1/4!
    Last edited: Nov 15, 2006
  13. Nov 15, 2006 #12
    Ah ok that clears that up, thought i was going insane for a second. On the first bit though, your saying the linear density is dm/dr which is fine, from that angle. but coming from the fact you defined lamda as C times the area density, doesnt that simply give a dm value? not a dm/dr ratio?
  14. Nov 15, 2006 #13
    a length times a mass per unit area will give a mass per unit length.

    Edit: I'm certainly insane! :yuck:

    Note -- I'll probably be signing off soon seeing as it's something like 4 AM here! I'm insane! :yuck:
    Last edited: Nov 15, 2006
  15. Nov 15, 2006 #14
    Also -- your thought earlier:

    the "linear density" (although in retrospect it might be better to call it the "radial density"?) lambda is simply the mass dm per ring dr. -- yeah , I just defactoed to lambda for the units.

    Yeah -- radial density would have been better. sorry. :zzz:

    I'll check up on this thread again tomorrow. This is way too late US time.
  16. Nov 15, 2006 #15
    Err now im confused even more haha, yeh im kinda gettin brain fry from this aswell, cramming for 2 exams over last week, the waiting is the worst :( keep second guessing myself
  17. Nov 15, 2006 #16
    just unit analysis...

    circumference times area density is radial:
    [tex] m\left(\frac{kg}{m^2}\right)=\frac{kg}{m}[/tex]

    yeah -- brain fry is a common thing :yuck:
    Be sure to get plenty of good eats and sleep!

    Last night I dreamed about doing the disk problem with maybe an "angular density" too -- how much mass dm is in a little wedge [tex]d\theta[/tex]? Then you could integrate over the angle from 0 to 2 pi.
  18. Nov 15, 2006 #17
    Oh yeh, i wasnt thinking in metres squared :P

    Thats a pretty good idea actually, the radial density one.

    thanks for your help!
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