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Angular momentum for a shrunken Earth

  1. Mar 4, 2017 #1
    1. The problem statement, all variables and given/known data
    If the Earth, with a radius of 6400 km, were collapsed into a sphere of the same mass, having a radius of 10 km, what would be its rotational period?

    2. Relevant equations
    L = Iw

    3. The attempt at a solution
    I can solve this if the moment of inertia is given but since it isn't I was wondering if there is a way to solve it without knowing the moment of inertia.
     
  2. jcsd
  3. Mar 4, 2017 #2

    gneill

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    If you at least know the form that the moment of inertia takes, then yes. Work out your result symbolically first using variables for the two moments of inertia.

    By the way, I've changed your thread title to be more specific. Titles that are too generic are not allowed.
     
  4. Mar 4, 2017 #3

    Orodruin

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    This is unsolvable unless you know the initial and final mass distributions. It is likely that the person constructing the problem intended for you to assume that the spheres have homogeneous mass distributions but it really is not clear and therefore an ill posed question.
     
  5. Mar 4, 2017 #4
    I assume that the earths are to be considered as uniform spheres

    if I take the moments of inertia as variables then:
    x= moment of inertia of large earth
    y=moment of inertia of small earth

    angular velocity of large earth = 1rev/day = 7.27 *10-5 rad/s
    x*7.27 *10-5 rad/s = y *w

    I don't know what to do from there
     
  6. Mar 4, 2017 #5

    Orodruin

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    So what is the moment of inertia of a uniform sphere in terms of its mass and its radius?
     
  7. Mar 4, 2017 #6

    gneill

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    Solve for w in terms of the other variables (and the constant angular velocity that you've introduced). What relationship do x and y have in the result?

    Then you can do as @Orodruin suggests and look up the moment of inertia formula for a uniform solid sphere and apply it twice, or you can consider the form that the moment of inertia takes for rotationally symmetric mass distributions: things like disks and hoops and spheres all have moments of inertia that share a common form. You may not need the precise formula if many parts of the formulas cancel out in your solution...
     
  8. Mar 4, 2017 #7

    Orodruin

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    To be honest, you do not need the rotationally symmetric part. As long as you have the same geometry and mass, there is only one form that the moment of inertia can take based on the size of the object. This follows directly from dimensional analysis. In fact, this entire problem can be done using dimensional analysis (assuming the same type of mass distribution, i.e., just scaling it along with the size of the Earth.
     
  9. Mar 5, 2017 #8
    I'm not sure, I would have to look it up.

    If I treat the spheres as point masses then their moments of inertia will be mr2.
    m(6400000)2 * 7.27 *10-5 = m (10000)2w
    w =29.78 rad/s

    so period is 2pi /29.78 = 0.21 s

    is this what you meant?

    i don't know the 'common form' you're referring to. How can i find this form?

    can you please show me how it can be done by dimensional analysis?
     
  10. Mar 5, 2017 #9

    Orodruin

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    Why don't you try it out yourself first? You have four dimensional variables and three independent physical dimensions. There will be only one dimensionless combination of those.
     
  11. Mar 5, 2017 #10
    I wonder whether we are talking about the same thing. The dimensional analysis that I know about is a method used to convert units.
    if I have a quantity in kg units then I can convert it to grams by using the conversion factor 1000g/kg so
    x kg * 1000g/kg = x000 g (the kg cancel out.)

    how can this solve the question?

    If I take the angular momentum equation i wrote earlier:
    x*7.27 *10-5 rad/s = y *w

    I insert the units for the variables. I think the units for moment of inertia are kg*m2. so:
    x (kg *m2) * 7.27 *10-5 rad/s = y (kg *m2) * w rad/s

    both sides have the same units, so I can divide both sides by the units (kg*m2 * rad/s) to get a dimensionless equation.
    (x * 7.27 *10-5 ) / y = w

    I don't see how this brings me any closer to solving the problem. Is this what you mean?
     
  12. Mar 6, 2017 #11

    Orodruin

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    That is not dimensional analysis, that is unit conversion. Dimensional analysis is using the physical dimension of the relevant quantities to deduce the possible functional form for their relationship.
     
  13. Mar 6, 2017 #12
    Okay, I haven't yet learnt dimensional analysis

    Is the method I used to solve the problem by treating the earths as point masses the other method that you were referring to?

     
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