Moment of inertia of a solid sphere

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Homework Statement


Find the moment of inertia of a solid sphere of uniform mass density (like a billiard ball) about an axis through its center

Homework Equations


I = ∫rρdV

The Attempt at a Solution


I =ρ ∫r4πr2dr = ρ4π∫r4
Then I integrate this from 0 (the center) to R, so I = (ρ4π)*(R5/5)
And ρ = mv so ρ = M/(4/3)πR3 = 3M/4πR3. Put ρ into the equation for moment of inertia to get I = 3MR2/5.

My book tells me the answer is (2/5)MR^2. Where did I go wrong?
 

Answers and Replies

  • #2
PeroK
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Homework Statement


Find the moment of inertia of a solid sphere of uniform mass density (like a billiard ball) about an axis through its center

Homework Equations


I = ∫rρdV

The Attempt at a Solution


I =ρ ∫r4πr2dr = ρ4π∫r4
Then I integrate this from 0 (the center) to R, so I = (ρ4π)*(R5/5)
And ρ = mv so ρ = M/(4/3)πR3 = 3M/4πR3. Put ρ into the equation for moment of inertia to get I = 3MR2/5.

My book tells me the answer is (2/5)MR^2. Where did I go wrong?
Can you explain how you are setting up the integral? What shape are you integrating from 0 to R?
 
  • #3
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Can you explain how you are setting up the integral? What shape are you integrating from 0 to R?
The surface area of a spherical shell from the center of the sphere to the outer shell of radius R.
 
  • #4
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I figured it out. If I use spherical shells, the points from the shell aren't all the same distance form the axis of rotation. I was thinking about the distance from the center.
 
  • #5
PeroK
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I figured it out. If I use spherical shells, the points from the shell aren't all the same distance form the axis of rotation. I was thinking about the distance from the center.
Exactly!
 

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