Angular momentum of the earth

  • Thread starter vpea
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  • #1
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I'm stuck on the second part of a problem and can't seem to get the right answer:

Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles. Treat the earth as a uniform sphere of radius 6.38*10^6 that makes one revolution in 24.0 hours. (m=5.97*10^24)

I got the answer 1.77*10^34 but it's wrong :(
 

Answers and Replies

  • #2
Doc Al
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Show your work. Explain how you got your answer.
 
  • #3
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I found the moment of Inertia (mr^2) to be = 2.43*10^38. Then found the angular velocity to be (2pi)/(24*60*60). Using the equation I=m*r^2, I got 1.77*10^34
 
  • #4
Doc Al
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vpea said:
Using the equation I=m*r^2....
That's the moment of inertial of a particle, not a solid sphere about its center. The one you want is [itex]I = 2/5 m r^2[/itex].
 
  • #5
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Thanks for that! I tried looking on the internet for the moment of Inertia of the earth and was getting all sorts of weird numbers no wonder it wasn't working.
 
  • #6
jtbell
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vpea said:
I tried looking on the internet for the moment of Inertia of the earth

Every "General Physics" textbook I've ever seen has a table of formulas for moment of inertia of various shapes of objects. What kind of a course are you taking?
 

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