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Torque Question

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data

    The earth can be approximated as a solid, uniform sphere with r = 6380 km.
    a- Using 1 day = 24 hrs, find the rotational kinetic energy.
    b- It revolves around the sun at a distance of 1.5 x 10 ^ 11 m once every 365.25 days. Find kinetic relative to tsun.
    c- If we wanted to increase the day to 25 hours, what torque would be required in 1 year to do this?

    2. Relevant equations



    3. The attempt at a solution
    I am using mass = 5.97 x 10^24 kg

    a- 2.58 x10^29 J
    b- 2.66 x 10 ^33 J
    c- ?
     
  2. jcsd
  3. Nov 9, 2008 #2

    tiny-tim

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    Hi veronicak5678! :smile:

    (have an omega: ω :wink:)
    What is the formula for rotational KE?

    What is the moment of inertia of a sphere? :smile:
     
  4. Nov 9, 2008 #3
    RK = 1/2 I w^2

    I = 2/5 MR^2

    (w for omega, angular velocity)
     
  5. Nov 9, 2008 #4

    tiny-tim

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    Hi veronicak5678! :smile:

    (what happened to that ω I gave you?)

    ok, now torque = rate of change of angular momentum …

    so what is the change in angular momentum if the day is lengthened to 25 hours? :smile:
     
  6. Nov 9, 2008 #5
    Sorry! I didn't even see that little alphabet.
    So the desired ω is .0000698 rad/s to make a 25 hour day. I guess by change in momentum you mean the same as change in velocity? .0000727 - .0000698 rad/s ?
     
  7. Nov 9, 2008 #6

    tiny-tim

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    No, angular momentum is Iω.

    I mean the change in Iω.
     
  8. Nov 9, 2008 #7
    OK. Actual Iω is 9.77 x 10^37 kg m ^2 * .0000727 rad /s = 7.11 x 10^33.
    Desired Iω is 9.77 x 10^37 kg m ^2 * .0000698 rad /s = 6.82 x 10^33


    Difference of 2.90 x 10^32

    Not sure if that is what you meant...
     
    Last edited: Nov 9, 2008
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