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Inertia problem help

  • Thread starter lilholtzie
  • Start date
  • #1
1. If the Earth's axis of rotation were located at a position tangent to the equator, how much more rotational intertia (moment of inerti) would the Earth have?



2. I (of sphere) = mr^2



3. I have absolutely no idea where to start.
 

Answers and Replies

  • #2
gneill
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20,793
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Look up "Parallel Axis Theorem" in your text.
 
  • #3


We don't use our books in class as far as the lessons, so I looked online and is this equation: I=Md^2 the same thing you are talking about?
 
  • #4
gneill
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20,793
2,773


We don't use our books in class as far as the lessons, so I looked online and is this equation: I=Md^2 the same thing you are talking about?
That looks incomplete. It should include the original centroidal moment of inertia (about a principle axis).

If Ic is the original moment of inertia about some axis, then the moment of inertia about an axis parallel to the original axis but offset by distance d would be I = Ic + Md2, where d is the offset distance and M the mass of the object.
 
  • #5


Okay so I think I'm starting to understand. In part 1 of the problem we found the moment of inertia using a mass of 6x10^24 and a radius of 6.378x10^3 and the formula I=mr^2 which gave me 9.76x10^31. For this part I would just take 9.76x10^31 + (6x10^24)(12756)^2 to get a final answer of 1.073893216E33?
 
  • #6
gneill
Mentor
20,793
2,773


You'll want to be careful about your units. What units was the radius given in? Also, check your formula for the moment of inertia of a sphere -- differently shaped bodies have different expressions (at least different constants of proportionality).

Other than those things, I don't see a problem with your method.
 

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