Depth to which the sphere is submerged

[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?

 

[haruspex]

It would be reasonable that the amount submerged should also be 2/5 th the radius

No, 2/5 of the volume, so nearly half the diameter.

As with most problems where you are given numerical values, it is better to set those aside and work purely algebraically. There are many advantages, including for other readers.
In the present case, you would get
$\frac 43\pi r^3\rho_m=\frac{\pi h^2}3(3r-h)\rho_w$
$4r^3\frac{\rho_m}{\rho_w}=3rh^2-h^3$
The cancellation will simplify the arithmetic and tend to improve the numerical accuracy of your answer.
At this point you can do the sanity check for the easy cases $\rho_m=\rho_w$ and $2\rho_m=\rho_w$.

 

[haruspex]

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.

A tougher set of problems concerns stable 'postures'. For a uniform solid cylinder radius r, length h, in which orientations can it float?

Edit: responses to that challenge moved to https://www.physicsforums.com/threads/stability-of-a-floating-cylinder.992553/#post-6380998

 

[Mike_A]

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.

View attachment 267778I 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.
Here is the solution,

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.

View attachment 267778I 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.
Here is the solution.

 

[Charles Link]

One thing that might be helpful is if you can follow the calculation of post 28. Then you need to subtract out the volume of a cone whose altitude is $h=R-d$, and whose $r^2$ of the base is $r^2=R^2-(R-d)^2$. You should know the volume of a cone as $V=\pi r^2 h/3$. This subtraction gives the volume of the submerged part as a function of $d$. See also post 46. This may be an easier approach than the solution you posted in post 53.

 

[haruspex]

@Mike_A :
Since the problem is already solved in the thread, I assume you have posted this as a model solution.
I am forever having to drum into students to resist the urge to plug in numbers straight away. Much better to keep everything symbolic as long as possible. There are many advantages.
In that spirit, I would write
$W_{sphere}=\rho_s\frac{4\pi r^3}3g$
$V_{cap}=\frac{\pi r^3}3(2+x)(1-x)^2$ where $x=\cos(\theta)$.
$\rho_s\frac{4\pi r^3}3g=\rho_w\frac{\pi r^3}3(2+x)(1-x)^2g$
Then cancel, etc.
Having obtained $3x=0.4+x^3$, N-R is overkill. It should be clear that x is small, so its cube very small. We can do one simpler iteration to be sure:
$x=0.4/3=0.13333...$
$x=(0.4+0.1333...^3)/3=0.1341...$. Time to stop already.
Finally, we don't need to find the angle, something else students routinely waste time, and maybe lose precision, doing. We only want its cosine, and we have that.

 

[Charles Link]

Just one comment is that they don't derive the formula they use for the submerged volume (in the solution posted in post 53). By substituting $d=R(1-x)$ where $x=\cos{\theta}$, the formula agrees with what was previously derived in this thread. Meanwhile, a very good analysis by @haruspex in post 55 of the solution posted in post 53. :)