# Depth to which the sphere is submerged

## Homework Statement:

A solid sphere of volume 20 cm and uniform density ##\rho_B## = 400 ##\text{kg m}^{-3}## floats in a vessel containing pure water. Calculate the depth to which the sphere is submerged.

## Relevant Equations:

1. ##\text{Law of floatation}## : When a body floats in a liquid, the weight of the body is equal to that of the weight of the liquid displaced : ##w_B = \Delta w_L##.
2. The volume of a spherical body of radius ##r_B## is given by : ##V_B = \frac{4}{3}\pi r_B^3##.
3. The weight of a body is given by : ##w_B = m_B g = \rho_B V_B g## where ##\rho_B## is the density of the body and ##V_B,m_B## are the volume and mass of the body respectively.
I draw a sketch of the diagram to the right.
The body has a radius of ##r_B = 0.2## m.
The volume of the body is given by ##V_B = \frac{4}{3}\times \pi \times 0.2^3 = 0.034\;\text{m}^3##.
The mass of the body : ##w_B = \rho_B V_B = 400\times 0.034 = 13.37\,\text{kg}##.
Hence the mass of the liquid displaced : ##\Delta m_L = 13.37\, \text{kg}## . (Law of floatation can also be "read off" as ##m_B = \Delta m_L##)
The volume of liquid displaced : ##\Delta V_L = \frac{\Delta m_L}{\rho_L} = \frac{13.37}{1000} = 1.34\times 10^{-3}\,\text{m}^3##.
This is the volume of the sphere below the blue line inside the water, as shown to the right.

Question is, how to connect the volume of liquid displaced ##\mathbf{\Delta V_L}## to the volume of the body ##\mathbf{V_B}## ?

I could proceed no further. Any help would be appreciated.

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etotheipi
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You need to find an expression for the volume of a spherical cap, which can be done with a volume of revolution.

Any clue as to finding the volume of a spherical cap?

etotheipi
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Position the sphere such that its centre is at the origin of a Cartesian coordinate system. With ##z=0##, the projection onto the ##x##-##y## plane is ##x^2 + y^2 = r_B^2##. You can re-arrange that for ##y^2##, and then evaluate the integral$$V_{cap} = \int_{r_b - d_l}^{r_b} \pi y^2 dx$$

Lnewqban
haruspex
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On a pedantic note, there is only one 'a' in "flotation".

etotheipi
kuruman
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On a pedantic note, there is only one 'a' in "flotation".
Just like a collection of floating objects known as a "flotilla".

etotheipi
Just like a collection of floating objects known as a "flotilla".
I suppose I'd struggle with the English language for good. It is an object that "floats" and yet its mere floating becomes "flotation", with the "a" missing. *Grins*

Thank you all for your help. I will get back to etothepi's method of finding the volume of a spherical cap presently.

kuruman
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I suppose I'd struggle with the English language for good. It is an object that "floats" and yet its mere floating becomes "flotation", with the "a" missing. *Grins*

Thank you all for your help. I will get back to etothepi's method of finding the volume of a spherical cap presently.
My personal preference is to use spherical coordinate angle ##\theta## because it automatically introduces the trig substitution into the integral. A disk of radius ##r## and thickness ##dz## has volume ##dV=\pi r^2~dz##. For a sphere of radius ##R##, ##r=R\sin\theta## and ##dz=R~d(\cos\theta)##. Then $$dV=\pi R^2(1-\cos^2\theta)R ~d(\cos\theta)=\pi R^3 (1-u^2)du~~~~(u\equiv \cos\theta).$$

etotheipi
Orodruin
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On a pedantic note, there is only one 'a' in "flotation".
... and 20 cm is not a volume ...

... and 20 cm is not a volume ...
Here on physicsforums, there seems an edit option to your posts which lasts only for a few minutes after you have posted. Had that option existed for a longer time, I'd make the corrections.

Orodruin
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Here on physicsforums, there seems an edit option to your posts which lasts only for a few minutes after you have posted. Had that option existed for a longer time, I'd make the corrections.
Had that option existed the mentors would be swamped with restoring threads where people deleted the content of their homework posts ...

Chestermiller
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Can't the volume below the surface be calculated as the fraction of the sphere volume included in the lower solid angle minus the volume of the included overlying cone?

Orodruin
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Can't the volume below the surface be calculated as the fraction of the sphere volume included in the lower solid angle minus the volume of the included overlying cone?
Sure, but that still requires you to know the relationship between the depth and the solid angle.

Chestermiller
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Sure, but that still requires you to know the relationship between the depth and the solid angle.
Sure, but isn't that just simple geometry?

kuruman
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Sure, but isn't that just simple geometry?
Sure, the simple geometry solution is sketched out in post #8. With ##u=\cos\theta=\frac{z}{R}##, the relation between volume and depth is obtained directly from a simple integral.

Chestermiller
Mentor
Sure, the simple geometry solution is sketched out in post #8. With ##u=\cos\theta=\frac{z}{R}##, the relation between volume and depth is obtained directly from a simple integral.
What I'm saying is I can do it without having to integrate. All I need to know are the formulas for the volume of a sphere and the volume of a cone.

hutchphd
What I'm saying is I can do it without having to integrate. All I need to know are the formulas for the volume of a sphere and the volume of a cone.
And the relationship of conical angle to solid angle$$\Omega=2\pi(1-cos(\theta))$$ which I always have to look up.

.

Chestermiller
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And the relationship of conical angle to solid angle$$\Omega=2\pi(1-cos(\theta))$$ which I always have to look up.

.
I don't. think that this is correct.

kuruman
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I don't. think that this is correct.
I think it is. An element of solid angle is
##d\Omega=\sin\theta d\theta~d\phi =-d(\cos\theta)~d\phi ##
If you go around at constant ##\theta##, the solid angle becomes
##d\Omega=-2\pi~d(\cos\theta)##
Without integration:
The solid angle element is linear in ##\cos\theta## which means that we can write
##\Omega=a ~\cos\theta +b##.
When ##\cos\theta=1##, ##\Omega=0## (we are the north pole)
When ##\cos\theta=-1##, ##\Omega=4\pi## (we are the south pole)
##0= a +b##
##4\pi=-a+b##
The solution of this system is ##b= 2\pi##, ##a=-2\pi##.
Hence ##\Omega=2\pi(1-\cos\theta)##.

hutchphd
I am through with the first of the solutions to, what seems to me still, a fairly tricky problem. The following method is the simplest for which I thank ethotheipi . I will present the other two solutions, one using spherical (polar) coordinates and the other using the idea of solid angles as soon as I can help it. Thank you for your discussions and help.

Solution 1
(using the method of cartesian coordinates to find the volume of the spherical cap) :

I begin by drawing a bird's eye-view of the situation. The sphere is seen from above. It's projection on to the x−y plane is a circle with the equation ##x^2+y^2=r_B^2##.The blue part of the sphere (-z) is where it is submerged in water to a depth of ##d_L## and is at the lowest. It is the volume of this part that I am interested in. Of course, there is an equally large "cap" on the top side of the sphere (+z) which is red and now shown.

The volume of the spherical cap (under water) can be found by "revolving" this circle about the z axis. The volume of the cap : $$V_{cap} = \pi \int_{r_B - d_L}^{r_B} y^2 dx = \pi \int_{r_B - d_L}^{r_B} (r_B^2 - x^2) dx = \pi \left[r_B^2 x - \tfrac{x^3}{3} \right]_{r_B-d_L}^{r_B} = \pi \left[\cancel{r_B^3} - \tfrac{r_B^3}{3} - r_B^2(\cancel{r_B} - d_L) + \tfrac{{\left(r_B-d_L \right)}^3}{3}\right]$$ $$= \pi \left[ \cancel{-\tfrac{r_B^3}{3}} + \bcancel{r_B^2 d_L} + \cancel{\tfrac{r_B^3}{3}} - \bcancel{r_B^2 d_L} + r_B d_L^2 - \tfrac{d_L^3}{3} \right] = \pi d_L^2 \left[r_B - \tfrac{d_L}{3} \right] = \pi \tfrac{d_L^2}{3}(3r_B - d_L)$$.

It is reassuring at this stage that my answer thus far matches that of the Wikipedia article posted above. I only put the picture alongside and the answer. This answer is the same as mine for ##r_B\rightarrow r## and ##d_L\rightarrow h##.

From my calculations earier (post#1), we had the volume of the sphere under the liquid given by ##\Delta V_L = V_C = 1.34\times 10^{-3} = \tfrac{\pi d_L^3}{3} (0.6-d_L)\Rightarrow d_L^3 -0.6 d_L + 1.28\times 10^{-3} = 0##.

An online solution to the above cubic equation yielded three answers of which only one is feasible to the problem above given the radius of the sphere is ##r_B = 0.2##m.

The answer is ##\boxed{\color{blue}{d_L = 0.00213\, \text{m} \approx 0.2\, \text{cm}}}##.

Do you think this answer is right?

For one, given the radius of the sphere (20 cm) and density (400 kg/m##^3##), less than a cm is submerged inside the liquid, which looks unreasonable.

Orodruin
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What I'm saying is I can do it without having to integrate. All I need to know are the formulas for the volume of a sphere and the volume of a cone.
... ant the relationship between cone angle and solid angle, which is most easily found through integration.

Do you think this answer is right?

For one, given the radius of the sphere (20 cm) and density (400 kg/m##^3##), less than a cm is submerged inside the liquid, which looks unreasonable.
Those 2 mm can't be the correct answer.
If the density of the sphere is about 40% of the density of water, about 40% of its volume should be submerged, in order to achieve buoyancy.
Something like 18 cm should be close to the correct answer.

Your calculated volume of liquid displaced in post #1 is incorrect.
It should be 0.01337 m^3?

Last edited:
kuruman
I get close to 18 cm. Check that your equation of force balance obeys Archimedes's principle $$\rho_{sphere}V_{sphere}=\rho_{water}V_{cup}.$$