Terminal Velocity of Water

Did the water fall with bucket ? as the water falls, it will break in drops and then droplets. The shape will keep on changing as the velocity changes (due to air drag)

First we have to find which drop reaches first ? the leading one or is overtaken by the other from behind (the advantages of the drag you know as in middle distance running)

Are we on a doctoral thesis ? why not a god ol' solid falling who does not change shape (till it hits the ground )

It had been poured out of the bucket. Doctorial not hardly I teache science and math in Native american high schools

Some of the water may have hit the side and was flowing down.

The bucket (or water) must be seen as plunger and would have to displace the volume of air below it, the pressure below the bucket will increase and slow it down. In a smaller diameter bore the descend may even be slower.

A couple of things to consider.

1. Sound: Is this how you timed the event? If so, an irregular shaped long tube, with temperature and humidity gradients will likely result in false signals from assorted near-resonance influences.

2. Electrostatic: As water falls and as it makes brief encounters with the sides, sections will become positively or negatively charged. The resulting interactions with the sides and with the other masses of falling water could result in additional drag. Check out Lord Kelvin's Water Dropper.

3. Electromagnetic: Whether there is naturally occurring magnetic material near the well, or even if it is just the earths magnetic field passign through iron in the crust, if the water is not DI water, then Lenz forces would be set up opposing the downward acceleration. In additioin to that you have the charged masses from above moving in a magnetic field.....

Drop a glass ball next time, and see if you get any closer to the predicted time.

The well may be only half as deep as you think it is. Falling water, being liquid, probably does not behave like a falling solid object with the same mass. As has been mentioned previously, the initial mass of the water in the bucket changes, as soon as it is poured out, breaking up into droplets and other amorphous shapes. Suggest turning water to ice and repeating the experiment.

It is a question
" how the terminal velocity was determoined to be 9-10 m/s."

By measurement, which agrees very closely with theory (fluid dynamics).

Btw, a falling water droplet bears little resemblance to the teardrop shape traditionally assumed. This droplet was photographed after falling eight stories down an elevator shaft:

Teardrops are much more streamlined. Wonder if anyone has photographed any falling down an elevator shaft. :-)

Would you assume that the flat bottom surface of the falling water drop is due to the effect of the air blanket resisting the fall which overcomes the surface tension of the droplet?

Yes, that and also because of the low-pressure region above the droplet:

What would the maximum velocity of a teardrop be if it dropped from an average height of 5.5 ft? It wouldn't have time to reach terminal velocity at that distance would it? Thanks, Kev

Hello Kev,

Welcome to CR4!

The terminal velocity of a water droplet is around 10 meters per second. In a vertical distance of 5.5 feet a water droplet - nor even a dropped stone - will reach this velocity. However, we can calculate the maximum velocity a dropped object (conveniently, a rock) can reach in that distance using a little algebra.

The velocity can be computed by v = gt, where

v = velocity, in feet per second;

g = acceleration due to gravity, or 32.2 ft/s²;

t = time, in seconds.

As this equation does not account for distance, ie, 5.5 ft, we can combine it with the equation for distance covered by a dropped object in a uniform gravitational field, and so:

d = ½ gt² , where

d = distance, in feet; ie, 5.5 ft.

g = 32.2 ft/sec², as before

t = time, in seconds, as before.

By combining these two equations, we can solve for velocity, since that is what we want:

from v = gt we have t = v/g,

and

d = ½ g (v/g)²

d = ½ v²/g

v² = 2dg

v = (2dg)^½

v = (2 × 5.5 × 32.2)^½

v = 18.8 ft/sec, or about 5.7 meters/sec, well under the terminal velocity of a 5 mm drop of water (about 10 m/s or 32.8 ft/sec).

BTW, by substituting the velocity back into t = v/g, we can figure the time it takes for the droplet to fall through 5.5 ft:

t = 18.8 / 32.2, or about 0.6 seconds.

Now, all of the above is what a dropped rock - dropped in a perfect vacuum, no less - would do. With a droplet of water things are a bit more complicated because the drop's acceleration is not uniform. Not only, but the coefficient of drag changes with velocity. Spheres can be unpleasant creatures, even at low velocities.

The drag coefficient for a sphere is given with a range of values because the drag on a sphere is highly dependent on Reynolds number. Flow past a sphere or a cylinder goes through a number of transitions with velocity. At very low velocity, a stable pair of vortices are formed on the downwind side. As velocity increases, the vortices become unstable and are alternately shed downstream. As velocity is increased even more, the boundary layer transitions to chaotic turbulent flow with vortices of many different scales being shed in a turbulent wake from the body. Each of these flow regimes produce a different amount of drag on the sphere.

The moment the droplet is released, it is accelerating much as a dropped rock would; it is not going fast enough to feel much air resistance (it doesn't feel any at first!). But by the time it reaches terminal velocity, it is not accelerating at all! This is because the air resistance just balances out the force of gravity and, after that, the droplet just falls at a constant velocity until it hits bottom.

One thing about a droplet of water: It is almost never teardrop-shaped! That's just someone's "artist's conception" that caught on somehow. You can see its actual shape clearly in the following sequence:

Notice that the droplet is teardrop-shaped at first not because it's falling, but because the surface tension between the droplet and the water above it elongates the droplet until it is free. Once the droplet "lets go" it becomes spherical.

So does it keep this shape? Yes, more or less, but it becomes flattened on the bottom as you can see in the pic below. Here the droplet has fallen eight storeys already and so it has been at terminal velocity for some time. It is not teardrop-shaped:

I hope this answers your question.

In a previous thread you can find some explanations which will allow you to estimate the velocity. I explained in this thread how a droplet falls and later I was able to determine why the rain drops have a limited volume. The final velocity depends on the drop dimensiosn since the drag is proportional to d^2 and the weight to d^3. If the velocity leeds to a pressure on the down surface so that the surface tension is > superficial tension the drop changes its form (flattens) and even breaks in smaller drops. According to measurements done by german scientist the maximal d is around 5 mm. If you consider a sphere of this diameter falling in air you can determine the final velocity setting that the weight is equal to the drag. Since you are teaching physics and mathematics you can do the rest on your own.

just wondering..... does anyone know if there is there a change in shape of the water drop as it accelerates to terminal velocity?

Yes, the droplet shape changes from spherical to the flattened shape seen in the photo. Falling droplets also oscillate as they fall.

At what temperature?

What is the diameter of the well?

Hi PW,

Your question brings up an interesting point: what if the droplets are frozen? Frozen as spheres before they flatten out? Would they fall faster or slower?

We could figure this out if we knew the Coefficient of Drag (C_d) on a flattened sphere as shown in the pix above. We know the C_d for a sphere (there is more than one, depending on the sphere's speed as I explained to Kev, above), but we also know that the sphere becomes flattened as its speed increases. Coefficients of Drag are radically influenced by the shape of the object. For example, the C_d for a sphere is at least 30 times greater than for a streamlined airfoil at the same flow rate. For a flat plate it is at least three times greater than for a sphere, or more than 90 times higher than for an airfoil.

This would be an interesting experiment, nonetheless.

As for the water behaving like a falling "piston" in a narrow well (someone mentioned this sort of thing earlier in the thread), this condition would not persist for long because the pressure beneath the water would build to the point where it pushed through the blob. At that point the water would break up into different-sized 'pieces' which would then fall at different rates. The biggest piece would reach the bottom first and would probably not be spherical, but would have, rather, a complex shape making calculations difficult. These latter sorts of problems are best handled by fluid dynamics supercomputers or empirically through experiment.

-e

Dear Mr. D.A.JESSEN,

What I understand from your expression is, the bucket itself is travelling in to the well, with water, as a whole. If so,

To find the terminal velocity, it is calculated from the formula ( we all studied in Physics Class) that, v^2 = u^2 + 2x(a)xS where v= terminal velocity, u= initial velocity ( in this case it is Zero), a= acceleration due to gravity ( it is 32.2 ft/sec^2) and S = distance travelled, in this case it is 350 feet as stated by you. By this calculation - the terminal velocity will be 150.133 Ft/Sec.

This equation/formula does not recognise friction caused by air, while falling.

Thanks,

RAJESWARI.