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Calculate the moment of inertia of this sphere

  1. Mar 2, 2007 #1
    1. The problem statement, all variables and given/known data
    A sphere consists of a solid wooden ball of uniform density 800kg/m^3 and radius 0.20 m and is covered with a thin coating of lead foil with area density 20kg/m^2

    Calculate the moment of inertia of this sphere about an axis passing through its center.

    2. Relevant equations
    I_cm = m(r^2)


    3. The attempt at a solution

    1. I took the volume formula of the sphere which is 4/3*pi*r^3 to get the volume and then multiply by its density to get the mass of the solid ball inside the sphere.

    2. Then I did the same with the lead coating, only using the surface area formula A = 4*pi*r^2

    (This I'm not sure about because I don't understand the picture. Is the solid ball centered in the sphere? Because I took the radius to be .20m assuming that it is and that it takes on the shape of the sphere...)

    3. I combined the mass to get total mass and then multiply with the radius (.20m) with respect to the x and y axis.

    But yeah, it's wrong. Someone pls guide me...somehow. :uhh:
     
  2. jcsd
  3. Mar 2, 2007 #2

    cepheid

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    1. Yeah, it's just a wooden ball with a coating of lead on it, that's all. A "ball" is just the solid volume enclosed by a sphere anyway.

    2. The formula [itex] I = mr^2 [/itex] only applies to a point mass of perp. distance r from the axis about which it is rotating! Here you have a continuous distribution of masses, not just one point mass. To me, that suggests that you have to integrate in order to calculate I (unless if you can use the spherical symmetry and uniform density to make a simplification, I don't know).
     
  4. Mar 2, 2007 #3

    cepheid

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    Edit:

    But for constant density, the integral would be trivial, wouldn't it? Hrmm...

    [tex] dI = r^2 dm = r^2 \rho dV [/tex]

    [tex] I = \int \! \! \! \int \! \! \! \int_V r^2 \rho dV = \rho \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{0.20} (r^2) r^2 \sin \theta dr d\theta d\phi [/tex]

    I hope I'm doing this right. Not sure what to do about the lead foil...it must be pretty simple but I'm tired right now...
     
    Last edited: Mar 2, 2007
  5. Mar 2, 2007 #4

    cepheid

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    Edit 2: I just realized that this is the introductory physics forum! Hrmmm...maybe the question is supposed to be simpler than this. Do you have any canned formulas for the moments of inertia of various shapes? That integral would lead me to believe that the result is

    [tex] I_{\textrm{sphere}} = \frac{3}{5}MR^2 [/tex]

    where M is the total mass of the sphere, and R is the radius. But my first year physics textbook says that:

    [tex] I_{\textrm{sphere}} = \frac{2}{5}MR^2 [/tex]

    Huh?! Close but no cigar...
     
  6. Mar 2, 2007 #5

    cepheid

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    Anyway, the point is that you need to find the formulae for the moments of inertia of a solid and a hollow sphere, and used them to solve this problem. As you can see, these can be calculated using integrals in 3 dimensions (as I have just attempted), but the results for several common shapes are usually given in any first year physics textbook.
     
  7. Mar 2, 2007 #6
  8. Mar 2, 2007 #7
    I got it! :D Thanks for all the help, really appreciate it. The formulas are in the book. I tried the formula for the sphere before but I just assumed that I could add the mass of the ball with the mass of the lead foil so I ignored the formula for the hollow sphere. :x
     
  9. Mar 4, 2007 #8

    cepheid

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