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Centre of Mass of an empty and filled goblet

  1. Jul 20, 2016 #1
    1. A goblet consists of a uniform thin hemispherical cup of radius r, a circular base of the same material thickness and radius as the cup, and an intervening stem of length r and whose weight one quarter of that of the cup.
    (a) Show that the height of the centre of gravity above the base is (13/14)r.
    (b) If the mass of the goblet is M and that of the amount of liquid that fills it is M' . Show that filling it raises the centre of gravity through a distance (39/65) r (M'/M+M')



    2. X = x1M1 + x2M2+....xnMn / M1+M2+....Mn
    where x is the centre of mass of each body that makes up the goblet and M is their respective mass



    3. I've managed to do part a where I've considered the surface densities of the hemispherical shell, the rectangular stem and since it's from the base, I considered the mass off the circle but the distance to the base would be 0.

    So C.O.M of the hemispherical shell = r/2 + r (from the start till the base)
    C.O.M of the stem is r/2 to the base
    C.O.M circle = 0 since it's to the base itself

    Then Mass = surface density x area ; surface density is represented by sigma
    Area of hemisphere = 2 pi r^2 so m= 2 pi r^2 (sigma) Area of circle = pi r^2 so m = pi r^2 (sigma)
    And m of stem = 1/4 (mass of hemisphere) = pi r ^2 /2

    Plug it into the equation and you get 13/14 r

    What I don't get is part b since they haven't mentioned whether how full the cup is? How do we find its centre of mass?
     
  2. jcsd
  3. Jul 20, 2016 #2

    Doc Al

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    Staff: Mentor

    They say the liquid fills the cup.
     
  4. Jul 20, 2016 #3
    So wouldn't the centre of mass of the liquid be the same as that of the cup? I tried that, fit it into my equation but it still doesn't give me the required answer.
    M( 13/14)r +M' (x) / (M + M') where x is the centre of mass of the liquid which I don't know.
    Am I right in using surface densities? I used this because it said a thin cup and it's hollow as well. Or should volume densities be used?
     
  5. Jul 20, 2016 #4

    Doc Al

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    Are the centers of mass of the solid and hollow hemispheres the same?
     
  6. Jul 21, 2016 #5
    The centre of mass of a hemispherical shell is r/2 from the flat side and for a solid hemisphere it is 3r/8 from the flat side.
     
  7. Jul 21, 2016 #6

    Doc Al

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    Good. Now make use of that fact.
     
  8. Jul 21, 2016 #7
    I've tried but it's not worked. I have just been using surface densities and not volume densities to find the mass.

    So for part a this is what I did: My diagram included a horizontal goblet with the x-axis as it's line of symmetry.
    C.O.M of the hemispherical shell = r/2 + r (from the start till the base)
    C.O.M of the stem is r/2 to the base
    C.O.M circle = 0 since it's to the base itself

    Then Mass = surface density x area ; surface density is represented by sigma
    Area of hemisphere = 2 pi r^2 so m= 2 pi r^2 (sigma) Area of circle = pi r^2 so m = pi r^2 (sigma)
    And m of stem = 1/4 (mass of hemisphere) = pi r ^2 /2 (sigma)

    Then my equation for an empty goblet became:

    (r/2 + r) ( 2 pi r^2 sigma) + (r/2)( pi r^2 /2 sigma) + 0( pi r^2 sigma) / ( 2 pi r^2 sigma + pi r^2 /2 sigma +( pi r^2 sigma)

    the pi r^2 sigma terms cancel out leaving me with:

    (r/2 + r) + (r/2)(1/2) / 2 + (1/2) + (1) = (13/4) r / (7/2) = (13/14)r

    For part b, when they put in the liquid, what am I supposed to take, the surface density or the volume density? I'm not sure how to go about it. Even if I take it as the completely filled and I take the surface density, all I get is:

    M(13/14)r + M' (r/2 +r) / (M+M') and simplifying this does not give (39/65) r (M'/M+M')


     
  9. Jul 21, 2016 #8

    Doc Al

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    You have the center of mass of the empty goblet and of the filled liquid. Now find the center of mass of the system. (No need for any densities.)
     
  10. Jul 21, 2016 #9
    For the centre of mass of the liquid, should I take it till the base if the goblet or just it's centre of mass alone?
    So for the latter case my equation would become:

    M (13/14)r +M'(r/2) / (M+M')
    but how do I simplify this equation? I can't add the variables in the numerator?
     
  11. Jul 21, 2016 #10
    Got it!
    Basically we assume the liquid to be a solid hemispherical shell and not hollow so the c.o.m of the shell from the curved surface is 5/8 r but since it's to the base it will be 5/8 r + r where r is the length of the stem. Then because it's asking for the difference we basically have to subtract it from that of the empty goblet.
    So the equation is
    M'(5/8 r + r) + M(13/14)r / M+M'
    - M (13/14)r/M
    = 39/56 r (M'/M+M')
     
  12. Jul 21, 2016 #11

    Doc Al

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    Good! :thumbup::approve:
     
  13. Jul 21, 2016 #12
    Thank you.

    Doc Al would you mind taking a look at the new question I posted please?
     
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