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A ball floating in the ocean : reinstate?
A ball is floating in the ocean at latitudes south
of 60 degree south latitude. What is the density of the
ball? calculus or geometry derivation ok.
Thank you.
no formula
Density ocean water = 1027 kg/m^3
solid ball uniform density
@ Koshka
The original question teaches an important lesson.
This question will go to vote or BA.
Thanks to bill for answering.
@Anyone
There is one numerical answer. Pick a coordinate
system (or geometry) and derive away!
Koshka is absolutely on the right
track with the formula version.
Koshka my hero. This is what Scythian did:
The bottom 30 degrees of the ball are below water:
pf = 1027 kg/m^3
a = pi / 3
r = 1
h = r - r * Sin(a)
vc = (1 / 3) * pi * h ^ 2 * (3 * r - h)
vs = (4 / 3) * pi * r ^ 3
p_ball = pf*(vc/vs)
Thanks to all. Lesson: Confucius say:
Sometimes it's better to keep your head
out of the ocean, maybe ;)
I was trying to discourage the use of the
cap formula, without derivation, with my
"no formula" comment. I should have clarified this.
Here's the way I originally did this: I like
nuts and bolts, but to each their own.
The volume below water: "volume cap"
d (rcosθ)/dθ = -rsinθ
Vc = ∫ dvol = ∫ π(rsinθ)^2 * -rsinθdθ
Vc = -πr^3 ∫ sinθ^3dθ [π/6,0]
Vc = -πr^3 [cos3θ - 9cosθ]/12 |π/6,0
Vc = 0.0538708182 cubic units
Volume of the ball:
Vs = (4/3)πr^3
In equilivrium we have
ρ => density of ball
ρf => density of ocean water
Fw = Fb
m g = ρf Vc g
ρ Vs g = ρf Vc g
ρ = ρf (Vc/Vs) = 13.2079497 kg/m^3
5 Answers
- KoshkaLv 510 years agoFavorite Answer
No formula?
The part that is "immersed" is (pi*h^2)/3* (3a-h) where h represents the height of the cap and "a" the radius of the cap. The weight (mg) of the spherical cap is the same as the water displaced.
Volume of the sphere minus volume of the spherical cap
(4*pi*r^3)/3 - (pi*h^2)/3* (3a-h)
Is this going somewhere?
It is 11:43 PM here
Perhaps someone might take it from there?
-------------------------
Édited: Scythian is my hero!
At least someone understands the question!!
Athough, I still can not figure out how he got the 1/2 * sqrt of 3 part.
Goodby!
- Scythian1950Lv 710 years ago
Taking a cue from Koshka's interpretation of the problem, the ratio of the volumes of an unit spherical cap subtending 60° and an unit sphere is:
(π/3)(3 - h²)h / (4/3)π
which is equal to, for h = 1 - (1/2)√3
(1/2) - (9/32)√3 = 0.0128607
so that the density of the ball is 0.0128607 x 1027 = 13.2079 kg/m³
Practically a beach ball.
Edit: You should give Koshka the BA, since I merely provided the "mechanics", while she cracked the problem.
- FredLv 710 years ago
Is there Earth ocean that far south? Google Earth says, yes, and plenty of it.
No calculus or geometry required. The answer is an inequality.
0 < ρ[ball] < 1027 kg/m^3
Important lesson? Am I that dense? [pun intended!]
PS. Thanks, Er. H.
I was excluding the neutral buoyancy, "submerged" case, because I don't consider that to be "floating."
- ?Lv 410 years ago
Yeah, Fred is right.
If it is partially floating then density of ball would be <1027 kg/m^3
and if it floating fully submerged then density would be =1027 kg/m^3
Thanks
- billrussell42Lv 710 years ago
Impossible to answer. The density could be anywhere from slightly lower than salt water to zero.
And the location doesn't matter.
.
