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Whenever you drop an object in water, it takes some time to get fully immersed. I was wondering if this depends upon the buoyant force of the water on the object, slowing it down. However, I was not able to come up with a general formula for this.

I wanted to know, what would be the time taken for an object of density $d_1$, dropped from a height $h$ to be fully immersed in a liquid of density $d_2$. Take gravity as $g$. Part of the reason I was not able to do this was because I thought it might involve calculus (since the volume of water displaced changes over time and so does the buoyant force). I'm not competent enough with calculus to be able to attempt this sort of problem.

I know friction will affect the time, but what if there were $no$ friction? Can this be further generalized to gases as well?

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It's pretty tough to derive it without using calculus. And you should also specify the geometry of the object because the time will greatly depend on that. –  udiboy1209 Jul 6 '13 at 12:12

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up vote 1 down vote accepted

As you say yourself, there will be a dependence on time and you'll need a differential equation to solve the problem. Let's make some simplifying assumptions. We'll assume that our body is spherical and has a radius $R$. We'll assume that its density is homogeneous and that it is denser than the fluid in which it will be immersed. We'll assume it drops from a height $h_0$ from the surface of the liquid, i.e. its center of mass is at height $h_0$. We'll neglect friction and splashing and any phenomenon that will complicate our analysis. We'll only take gravity and the Archimedean force into account.

The variable component here is the Archimedean force which depends on the immersed volume which will change in time. The formula for this force is

$$F_A=\rho_{f} g V(h)$$

where $\rho_{f}$ is the fluid density, $g$ is the gravitational acceleration in the vicinity of the Earth's surface and $V(h)$ is the immersed volume. The formula for the immersed volume is

$$V(h)=\begin{cases}0 \; \text{ if } h\geq R \\ \frac{\pi}{3}(R-h)\left[2R^2+Rh+h^2\right] \; \text{ if } -R<h<R \\ \frac{4}{3}\pi R^3 \; \text{ if } h\leq -R \end{cases}$$

We therefore have the following equation of motion

$$-g\left[\rho_0 \frac{4}{3}\pi R^3-\rho_f V(h)\right] = \rho_0 \frac{4}{3}\pi R^3 \frac{d^2 h}{dt^2}$$

Where $\rho_0>\rho_f$ is the density of our sphere. This is a second order nonlinear ODE because $V(h)$ is of third order in $h$ for part of its range.

Introducing the new variables $x=h/R$ and $\tau=\sqrt{g/R}t$ and a new function

$$P(x) = \begin{cases}0 \; \text{ if } x\geq 1 \\ \frac{2-x-x^3}{4} \; \text{ if } -1<x<1 \\ 1 \; \text{ if } x\leq -1 \end{cases}$$

we can rewrite our equation as

$$\frac{d^2x}{d\tau^2}=\frac{\rho_f}{\rho_0}P(x)-1$$

We can integrate this equation w.r.t. $x$ once to reduce the order of our equation. This gives us

$$\frac{1}{2}\left(\frac{dx}{d\tau}\right)^2=(h_0/R-x)+\frac{\rho_f}{\rho_0}Q(x)$$

in which I introduced $Q(x)$ which is the integral of $P(x)$,

$$Q(x) = \begin{cases}0 \; \text{ if } x\geq 1 \\ \frac{-5+8x-2x^2-x^4}{16} \; \text{ if } -1<x<1 \\ x \; \text{ if } x\leq -1 \end{cases}$$

We can therefore solve for the time $\tau$ in function of $x$

$$\tau=\frac{1}{\sqrt{2}}\int_{-1}^{h_0/R}\frac{dx}{\sqrt{h_0/R-x+\frac{\rho_f}{\rho_0}Q(x)}}$$

I've computed a first order approximation in $\rho_f/\rho_0$ for the case of the sphere dropping from just above the surface until the moment its gets fully immersed:

$$\tau = 2+\frac{43}{210}\frac{\rho_f}{\rho_0}+O((\rho_f/\rho_0)^2)$$

or

$$t=2\sqrt{\frac{R}{g}} +\frac{43}{210}\frac{\rho_f}{\rho_0}\sqrt{\frac{R}{g}}+O((\rho_f/\rho_0)^2)$$

The zeroth order term is what you'd expect if there was no fluid, the next term gives a correction for a very dense sphere compared to the density of the fluid.

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doesn't the OP mention that s/he doesn't want calculus to be involved? –  udiboy1209 Jul 6 '13 at 12:56
    
And the general method would have been easier to understand with a simpler geometrical object, maybe a cylinder or cuboid. –  udiboy1209 Jul 6 '13 at 13:01
    
@udiboy: Yes, but as you can see, calculus is unavoidable. The final formula, whatever the method you use to arrive at it will contain an integral. Of course, you could maybe partly guess a simplified formula by some dimensional analysis I suppose. –  Raskolnikov Jul 6 '13 at 13:03
    
@udiboy: I think the sphere is somewhat harder but more realistic. With the other forms, I would have to make an additional assumption of how the piece drops into the fluid to be able to compute things. With the sphere, it obviously doesn't matter. –  Raskolnikov Jul 6 '13 at 13:05

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