Depth to which the sphere is submerged

[brotherbobby]

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.

 

[etotheipi]

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

 

[brotherbobby]

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

 

[etotheipi]

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$$V_{cap} = \int_{r_b - d_l}^{r_b} \pi y^2 dx $$

 

[haruspex]

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

 

[kuruman]

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

Just like a collection of floating objects known as a "flotilla".

 

[brotherbobby]

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]

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 $$dV=\pi R^2(1-\cos^2\theta)R ~d(\cos\theta)=\pi R^3 (1-u^2)du~~~~(u\equiv \cos\theta).$$

 

[Orodruin]

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

... and 20 cm is not a volume ...

 

[brotherbobby]

... and 20 cm is not a volume ...

Sorry about that. It should read "a solid sphere of radius 20 cm".
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]

Sorry about that. It should read "a solid sphere of radius 20 cm".
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]

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]

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]

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]

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.

 

[Lnewqban]

This problem of connecting the volume of liquid to the height of the surface is common in industrial horizontal cylindrical and spherical tanks.

Please, see:
https://en.m.wikipedia.org/wiki/Spherical_cap

 

[Chestermiller]

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]

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.

 

[hutchphd]

I'll look it up again:
https://en.wikipedia.org/wiki/Solid_angle

seems OK to me ?

 

[kuruman]

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 $\Omega=2\pi(1-\cos\theta)$.

 

[brotherbobby]

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 https://www.physicsforums.com/members/etotheipi.664237/ . 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]

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.

 

[Lnewqban]

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?

 

[kuruman]

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.

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

 

[Chestermiller]

... ant the relationship between cone angle and solid angle, which is most easily found through integration.

If $\theta$ is the half-angle of the cone and d is the submerged depth, I get $$\cos{\theta}=\frac{R-d}{R}$$and a submerged volume of $$V=\frac{4}{3}\pi R^3\frac{\theta}{\pi}-\pi[R^2-(R-d)^2]\frac{(R-d)}{3}$$
Is this not correct?

 

[kuruman]

Is this not correct?

In order to have only one variable, it can be changed to $$V=\frac{4}{3}\pi R^3\frac{\cos^{-1}(\frac{R-d}{R})} {\pi}-\pi[R^2-(R-d)^2]\frac{(R-d)}{3}$$but in this form it less useful than the other equation for solving the force balance equation for $d$.

As to whether it is correct, I have to say that it is trying to be correct but is not quite there. Shown below is a plot of the submerged volume fraction as a function of depth d. The blue line is the expression from OP's post #22 and the brown line is the expression above normalized to the sphere's volume. The two curves match only the end points and the midpoint. I coudn't find an obvious fix; perhaps you can.

 

[hutchphd]

I believe the volume of the spherical sector (your first term) should be, in these variables,
$$dV_{SECTOR}=\Omega R^2 dR$$ and using the expression for solid angle previously quoted $\Omega=2\pi\frac d R$ so $$V_{SECTOR}=\frac {\Omega R^3} 3$$$$V_{SECTOR}=\frac {2\pi R^2d} 3$$

 

[etotheipi]

using the expression for solid angle previously quoted $\Omega=2\pi\frac d R$

Wouldn't it be $2\pi(1-\frac{d}{R})$?

The volume of the spherical cap (under water) can be found by "revolving" this circle about the z axis

Really you are revolving the arc defined by $y = \sqrt{R^2 - x^2}$ by one full rotation around the $x$ axis (i.e. rotating the circle defined by $x^2 + y^2 = R^2$ around the $z$ axis does nothing to the curve!). And that is, in practice, a summation of cylindrical volume elements.

 

[hutchphd]

If d is the sagittal depth shouldn't $\Omega=0$ for d=0 and $\Omega=2\pi$ for d=R ?

 

[etotheipi]

If d is the sagittal depth shouldn't $\Omega=0$ for d=0 and $\Omega=2\pi$ for d=R ?

Sorry, I misunderstood. I thought $d$ was from the centre to the water. With $d$ instead as the sagittal depth, you're right. Sorry!

 

[hutchphd]

There are several definitions running about...hope I got the right one.

 

[Chestermiller]

@hutchphd, @etotheipi, @Nugatory, @kuruman: I owe you guys an apology. I don't know what I was thinking. What I said in my post was obviously wrong. Sorry for generating so much froth.

Chet

 

[haruspex]

The volume of liquid displaced : $\frac{13.37}{1000} = 1.34\times 10^{-3}\,\text{m}^3$.

$\tfrac{\pi d_L^3}{3} (0.6-d_L)\Rightarrow d_L^3 -0.6 d_L + 1.28\times 10^{-3} = 0$.

Check the exponents in both.

 

[brotherbobby]

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

Thank you and apologies. Of course those two did not match for me and I have since corrected my results. I will do those corrections in a different post which will appear towards the end.

 

[brotherbobby]

Check the exponents in both.

Yes thank you. It is a careless mistake on my part : $\tfrac{13.37}{1000} = 0.0134\, \text{m}^3 \neq 1.37\times 10^{-3} \,\text{m}^3$. I will correct my solution above. Obviously it will show up in the last post.

 

[haruspex]

Yes thank you. It is a careless mistake on my part : $\tfrac{13.37}{1000} = 0.0134\, \text{m}^3 \neq 1.37\times 10^{-3} \,\text{m}^3$. I will correct my solution above. Obviously it will show up in the last post.

And the two exponent errors in the other extract I quoted?

 

[brotherbobby]

Wouldn't it be 2⁢π⁢(1−dR)?
Really you are revolving the arc defined by y=R2−x2 by one full rotation around the x-axis (i.e. rotating the circle defined by x2+y2=R2 around the z axis does nothing to the curve!). And that is, in practice, a summation of cylindrical volume elements.

Yes am sorry for that elementary mistake. The circle being in the x−y plane remains as it is if revolved about the z axis. Revolving it about the x axis, the y coordinate acts as the "radius" of the elemental cylinder while the incremental length dx acts as its "height". However I suppose it would make no difference if the circle was revolved about the y axis.

And the two exponent errors in the other extract I quoted?

The other error follows from the first one. So that must be wrong too. But they will form my corrected solution on which am working at the moment.

 

[haruspex]

The other error follows from the first one.

No, look at all the exponents (including "1") on d_L in

$\tfrac{\pi d_L^3}{3} (0.6-d_L)$.

and in

$d_L^3 -0.6 d_L + 1.28\times 10^{-3} = 0$.

 

[brotherbobby]

No, look at all the exponents (including "1") on d_L in

and in

Those exponents are correct - far as you confine yourself to that bit. The error lies ahead in my finding of the volume of the sphere submerged $\Delta V_L = V_{cap}$. This is what I had written earlier (post #22) :
$$\Delta V_L = V_C = 1.34\times 10^{-3} = \tfrac{\pi d_L^3}{3} (0.6-d_L)\Rightarrow d_L^3 -0.6 d_L + 1.28\times 10^{-3} = 0$$
I took the $\pi$ and the 3 to the other side and got the expression you mention. Clearly the coefficient of $d_L^3$ is 1 and that of $d_L$ is 0.6. It is the constant term $\cancel{1.28\times 10^{-3}}$ which is wrong and this comes with the mistaken volume of the solid submerged in the liquid $\Delta V_L \neq 1.37\times 10^{-3}$.

Thanks a lot.

 

[brotherbobby]

Wouldn't it be $2\pi(1-\frac{d}{R})$?
Really you are revolving the arc defined by $y = \sqrt{R^2 - x^2}$ by one full rotation around the $x$ axis (i.e. rotating the circle defined by $x^2 + y^2 = R^2$ around the $z$ axis does nothing to the curve!). And that is, in practice, a summation of cylindrical volume elements.

Yes am sorry for that elementary mistake. The circle being in the $x-y$ plane remains as it is if revolved about the $z$ axis. Revolving it about the $x$ axis, the $y$ coordinate acts as the "radius" of the elemental cylinder while the incremental length d$x$ acts as its "height". However I suppose it would make no difference if the circle was revolved about the $y$ axis. In that case we will have $V_{cap} = \pi \int_{r_B-d_L}^{r_B} x^2 dy$.

However, I'd like to correct the mistake I made in my solution (post #22 and post #1).

The volume of the body submerged in the liquid is $$\Delta V_L = \frac{\Delta m_L}{\rho_L} = \frac{13.37}{1000} = 0.01337 \,\text{m}^3 \approx 0.0133\,\text{m}^3 \color{red}{\neq 1.34\times 10^{-3}\,\text{m}^3}$$ (This was my mistake, marked in red).

Taking the new value of $\Delta V_L = V_C$ and continuing from my solutions in post#22, we have
$$\Delta V_L = V_C = 0.0133 = \tfrac{\pi d_L^2}{3} (0.6-d_L)\Rightarrow d_L^3 -0.6 d_L^2 + 0.013 = 0$$.

Solving as before using the internet for cubic equations, I find that the acceptable of the three answers for the length of the sphere submerged in the liquid is $$\boxed{\color{blue}{d_L = 0.175\,\text{m} = 17.5\,\text{cm}}}$$.

Problem is, I don't know if this is correct either. The sphere has a radius $r_B = 20 cm$ and almost 18 cms of it is submerged in water when its density is $400 \, \text{kg m}^3$, that is 2/5 ths that of water. It would be reasonable that the amount submerged should also be 2/5 th the radius $\approx 8$ cm.

One error am making could be the volume of sphere submerged. The volume is at the lower end. However, using the integral $V_{cap} = \pi \int_{r_B-d_L}^{r_B} y^2 dx$, is it that the submerged volume is calculated at both ends, leading to double the answer?

 

[kuruman]

I got 17.3 cm. Close enough if you substitute numbers at the very end to minimize round-off errors.

 

[brotherbobby]

I got 17.3 cm. Close enough if you substitute numbers at the very end to minimize round-off errors.

Yes, I forgot that it's only the radius that is 20 cm. The diameter which is the length of the whole sphere end-to-end is 40 cm. This makes the answer 17 cm reasonable.

 

[haruspex]

Those exponents are correct

They are not. In

$\tfrac{\pi d_L^3}{3} (0.6-d_L)$.

the exponent 3 should be a 2, but that is probably just a typo in the post since it was not propagated,
and in

$d_L^3 -0.6 d_L + 1.28\times 10^{-3} = 0$.

The second d_L should be squared.
That was not merely a typo since in post #22 you filled in the cubic equation parameters as b=0, c=-.6, but I see you now have them the other way around.

 

[brotherbobby]

They are not. In

the exponent 3 should be a 2, but that is probably just a typo in the post since it was not propagated,
and in
The second d_L should be squared.
That was not merely a typo since in post #22 you filled in the cubic equation parameters as b=0, c=-.6, but I see you now have them the other way around.

Yes. I saw my error and thought of responding to you separately. However, I have corrected them in my solution in post#41 above. Let me know if it's ok when you can.
Thanks a lot.

 

[Charles Link]

I've been following this somewhat , and there has been a fair amount of errors. I don't agree with the cubic expression that is being used. (I could be wrong). In any case, I get $V=\frac{\pi}{3}d(d^2-r_od+2r_o^2)=.0134$. [Edit: yes, my expression here is incorrect]. My expression does give the correct $(2/3)\pi r_o^3$ for $r_o=d$.
Edit: I agree with post 28. I next computed the volume of the cone that gets subtracted out. The rest is algebra. My mistake=I found my error=yes, I agree with $V=(\pi d^2/3)(.6-d)$. Very good.

 

[etotheipi]

If you are bored, now that you have the equilibrium depth $d_0$, you can consider displacing the sphere by $\varepsilon$ so that $d = d_0 + \varepsilon$. If $f_B(d)$ is the buoyant force, say$$f_B(d) - mg = -m\ddot{d} \implies f_B(d_0 + \varepsilon) - mg = -m\ddot{\varepsilon}$$and find the period of small oscillations about $d_0$. Neglect terms of $\varepsilon^2$ or higher order.

 

[hutchphd]

I think you are slightly evil...shall we put in viscosity next? This is a very nice problem actually. I always wanted to structure a course with one continually unfolding problem...for me it is a great way to learn.

 

[brotherbobby]

I think there are more things to do before ethiothepi's suggestion (post # 47).

(1) Calculate the depth of the sphere submerged $d_L$ using spherical polar coordinates and the method of solid angles. (These have been suggested already, but the student in me is yet to carry them out.)

(2) If the sphere was not of uniform density but say had some density profile $\rho(r) = Ar$ where A is a constant, to what depth would it sink? Would the volume of sphere submerged $\Delta V_L$ into water be the same?
I am confused.
On one hand, it has to be, for the sphere has the same mass and hence the same mass and volume of water would have to be displaced (flotation law). On the other hand, the sphere would require less volume to displace the same mass (and volume) of water, because its "outer" regions would have more mass as a result of increasing density. (Density increasing with distance from the center).

I'd think about it and answer (2).

 

[hutchphd]

For (2) it is sufficient to consider all the density in the outside shell (that is the extreme). And what if the density gradient is not radial?