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could you explain the process of finding Iz for sphere? why do we use the moment of inertia of a flat disk for which dm was p(2pir)dr? Whereas for a sphere dm = p(pir^2)dz , so the element disk here is with volume with thickness dz. 

Sphere
Let us find the moment of inertia of a uniform solid sphere of radius a and mass m about
an axis (the z-axis) passing through the center. We divide the sphere into thin circular discs,
as shown in Figure 8.3.3. The moment of inertia of a representative disc of radius y, from
Equation 8.3.7, isy'dm. But dm = pлy² dz; hence,
2
8
1₂ = Sª ½npy^ dz = [ª ½-¹р(a² − z²)² dz = ³пpa³
-0
(8.3.8)
The last step in Equation 8.3.8 should be filled in by the student. Because the mass m is
given by
we have
m =
nа³p
2
1₂ = ²/ma²
for a solid uniform sphere. Clearly also, I₁ = Iy ³
=
= 1₂
(8.3.9)
(8.3.10)
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Transcribed Image Text:Sphere Let us find the moment of inertia of a uniform solid sphere of radius a and mass m about an axis (the z-axis) passing through the center. We divide the sphere into thin circular discs, as shown in Figure 8.3.3. The moment of inertia of a representative disc of radius y, from Equation 8.3.7, isy'dm. But dm = pлy² dz; hence, 2 8 1₂ = Sª ½npy^ dz = [ª ½-¹р(a² − z²)² dz = ³пpa³ -0 (8.3.8) The last step in Equation 8.3.8 should be filled in by the student. Because the mass m is given by we have m = nа³p 2 1₂ = ²/ma² for a solid uniform sphere. Clearly also, I₁ = Iy ³ = = 1₂ (8.3.9) (8.3.10)
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