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Confusion in centre of gravity and centre of mass differentiation

  1. Oct 12, 2013 #1
    Hi friends,
    I have [COLOR="Hi friends,
    I have an issue in solving a [COLOR="Blue"]Kinetic energy problem during collision[/COLOR].
    Please Help me in solving this.
    Thank you all in advance.

    The problem is as:

    https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-frc1/q71/s720x720/1380301_1432382870322152_49372184_n.jpg


    Attempt:

    https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn2/q79/s720x720/1385129_1432380796989026_538254057_n.jpg

    Please try to help me in this.
    I will appreciate the help.
     
  2. jcsd
  3. Oct 13, 2013 #2

    Simon Bridge

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    What did they say the correct answer was?
    I'd have interpreted it the same as you.

    However, if the big circle on the diagram represents the Earth, then the mass of the earth must be included in the estimate ... leading to R/2 as the best answer.
     
  4. Oct 13, 2013 #3

    haruspex

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    I think you are supposed to take the rod as lying on a radius of the Earth. The net gravitational force on the two mass system does not necessarily pass through the c.o.m. However, c.o.g. is not a well-defined concept. In principle, any point on the line of action of the net gravitational force will do. In the present case, you can take it to be the point where that line intersects the rod.
    EDIT: Maybe you can define the c.o.g. as a point where the object would experience exactly the same force if all its mass were concentrated there. But... I'm not sure such a point necessarily exists. It must exist in the present case since it's just a matter of finding the point on the rod where the force would have the right magnitude.
     
    Last edited: Oct 13, 2013
  5. Oct 15, 2013 #4
    Is there any case where we can define COM and COG as different points.
    Some where I read that moon revolves round earth in 28 days and about its own axis in 28 days also. And from any point on the surface of earth we can see only same face of the moon ever. Its happens because the COM and COG of earth - moon system do not co - inside with each other.
    How I can explain this?
     
  6. Oct 15, 2013 #5

    D H

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    Yes. Your long rod with two point masses. Orient the rod vertically, with one end on the surface of the Earth and the other end one Earth radius above the surface. The total gravitational force on the system is GMm/R2+GMm/(2R)2 = (5/4)GMm/R2. Suppose you replace those two masses with a single point mass with mass 2m and attach that single point mass on the place on the rod such that the gravitational force on the rod+point mass system is (5/4)GMm/R2. That point is at √(8/5) R from the center of the Earth, or a bit over 1/4 the way up the rod. That's the center of gravity of the vertically oriented rod + two point masses system. Compare with the center of mass of this system, which is obviously halfway up the rod.


    The first two sentences are true. That it happens because the COM and COG of the Earth-Moon system do not coincide with one another is nonsense. The phenomenon that is responsible for this is tidal locking.
     
  7. Oct 15, 2013 #6

    Simon Bridge

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  8. Oct 15, 2013 #7

    haruspex

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    Umm.. are you sure they are not the same thing? Tidal locking would not occur in a uniform field, right?
     
  9. Oct 15, 2013 #8

    D H

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    What does the "COG of the Earth-Moon system" even mean? Center of gravity has, as far as I know, three different meanings:
    • A synonym for center of mass.
    • The point about which, regardless of orientation, a rigid body experiences no torque due to some external gravitational field.
    • The point about which a point mass experiences the same gravitational force due to some external gravitational field as does the object in question.
     
  10. Oct 15, 2013 #9

    haruspex

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    Sorry, I didn't read the post you were objecting to carefully enough. I read what I expected to see: that it's because the Moon's COM and COG do not coincide in Earth's field. That could be a sensible explanation, but I haven't checked it.
     
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