Calculating Water Splash Depth of Wooden Ball

In summary, the ball will splash in and dip a little before buoyancy forces take over and bring it to the surface. There is no way to calculate how deep it will go after the initial splash before the buoyancy forces take over.
  • #1
shak
3
0
Hi!

I have a problem.

Say I have a wooden ball, 5 cm in diameter, 40 grams, I drop it from a height of 1 meter into some still water. Obviously, it will splash in and dip a little before buoyancy forces take over and bring it to the surface,

I have never heard of a way, but maybe the pros here know a formula or something,

Is their any way to calculate how deep it will go after the initial splash before the buoyancy forces take over??

Cheers!
Shak
 
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  • #2
I suppose you could consider a simplified verson using conservation of energy. The kinetic energy of the ball before it impacts the water plus the change in gravitational potential energy, must equal the work done against the buoyancy force.

~H
 
  • #3
There is also the complication of the huge drag factor while the ball moves through the water, and the flow will probably be turbulent.

Also the impact at the surface of the water will involve a combination of displacement and drag.
 
  • #4
Thanks guys!

I tried using the energy converting from potential into work done, but the figure comes out obviously wrong...I think you're right about the drag and other factors which are hard to account for...

is their a single complete formula for working this out? or could someone please post a quick and dirty formula which may work?
 
  • #5
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  • #6
wow! This is getting way over my head, I was expecting something simple like f=ma... :(

I also found this - http://www.owlnet.rice.edu/~phys111/Lab/expt02.pdf [Broken] (Its a PDF!)

Any chance someone who knows the high level maths of double differentials and cosh functions could solve this for the displacement?

As I have said, the ball, of mass m and diameter r, dropped from rest at a height h, top of water can be at h=0,...I guess at the deepest dip point (h=d), the vertical velocity will be 0 (?) sort of like a vertical pendulum experiment we did in school with a spring and weight at the end...

Anyone?!

Thanks in advance!
 
Last edited by a moderator:

1. How do you calculate the water splash depth of a wooden ball?

To calculate the water splash depth of a wooden ball, you will need to know the mass of the ball, the height from which it is dropped, and the surface tension of the water. You can then use the formula: d = (3mgh)/(4πrρg), where d is the splash depth, m is the mass of the ball, g is the acceleration due to gravity, h is the height of the drop, r is the radius of the ball, and ρ is the surface tension of the water.

2. Can the shape of the wooden ball affect the water splash depth?

Yes, the shape of the wooden ball can affect the water splash depth. A spherical ball will create a deeper splash compared to a flat or elongated ball with the same mass and height of drop. This is because a spherical shape has a smaller surface area and therefore displaces more water in a smaller area.

3. Is the water temperature a factor in calculating the splash depth?

Yes, the water temperature can affect the surface tension of the water, which in turn can affect the water splash depth. Warmer water has a lower surface tension, which means the wooden ball would sink deeper in the water compared to if it was dropped in colder water with a higher surface tension.

4. How does air resistance impact the water splash depth?

Air resistance does not have a significant impact on the water splash depth of a wooden ball. This is because the ball is falling through the water, which has a much higher density compared to air. Therefore, the effect of air resistance is negligible in this scenario.

5. Can this calculation be applied to other objects dropped in water?

Yes, this calculation can be applied to other objects dropped in water, as long as the necessary variables are known. However, the shape and density of the object may affect the accuracy of the calculation. Additionally, the formula may need to be adjusted based on the specific properties of the object being dropped.

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