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Old Aug 20th 2015, 02:59 PM   #1
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kinetic energy of a falling body in water

Hi

I am considering the impact energy of an object hitting me on the head travelling vertically downward with impact velocity v in water. The mass of the object is m, and the hydrodynamic added mass from the displaced water is "ma". I believe the kinetic energy at impact is KE=1/2 (m+ma) v*v

However I am slightly troubled by the buoyancy force F from the displaced volume V of water F=pgV, where p is the density of water and g is acceleration due to gravity. The greater the buoyancy force, the slower the object will accelerate downwards, and the easier it will be to lift. So the object can be considered to dynamically have an effective mass of m'=m-pV.

So is the KE in fact KE=1/2 (m'+ma) v*v ? I don't think so, but if I was sure I wouldn't be asking.

Thanks
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Old Aug 20th 2015, 03:33 PM   #2
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Buoyancy doesn't affect the mass, only the weight of a body.

The KE of a body is simply 1/2 mv^2.
There are no hydrodynamic additions or subtractions to make.
Further it doesn't matter how the body achieves the velocity v.

A further thought.

You say the KE of the body hitting you.
Unless you physically stop this body falling with the water, it will only transfer some of its energy to you on impact.
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Old Aug 20th 2015, 10:18 PM   #3
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Thanks. Yes, on reflection, you are clearly correct with regard buoyancy. Clearly I can inject a velocity to a body with no effective weight (i.e. buoyant) and it will have kinetic energy.

With regard added mass, my question was maybe unclear. The object is not impacting the water. It is in water (as am I) and it is about to hit me with velocity v. You are right the object has kinetic energy based on m (no added mass), however the impacting kinetic energy that I will feel will have an added contribution due to the water displaced by the object. I am fairly sure I will feel this additional kinetic energy too. So to assess the collision energy, the kinetic energy of the system is 1/2(m+ma)v*v.
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Old Aug 21st 2015, 05:56 AM   #4
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Originally Posted by Tim1 View Post
So to assess the collision energy, the kinetic energy of the system is 1/2(m+ma)v*v.
I assume you are thinking of the water that makes up the pressure wave, or "bow wave," being pushed forward by the object. The problem with this is two-fold:

1. The mass of the bow wave is difficult to calculate - its not zero, but it's also not equal to the mass of the object.

2. The velocity of the bow wave of water does not go to zero when the water presses against you - it mostly flows around you, losing only a bit of its KE. So your body does not absorb the full amount of KE present in the wave. Hence the energy you absorb from the weave is not (1/2)m_a v^2, but somewhat less.

We can make some assumptions about the force of the bow wave against your body, and from that the work that is performed by your body to slow the bow wave, and from that the energy you absorb. But it would be only a very rough estimate. How big is the object, and what is its velocity?

Last edited by ChipB; Aug 21st 2015 at 12:28 PM.
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Old Aug 21st 2015, 12:00 PM   #5
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Originally Posted by ChipB View Post
I assume you are thinking of the water that makes up the pressure wave, or "bow wave," being pushed forward by the object.
Yes, the pressure wave from displacing the water. I think a "bow wave" is a good way to view it. You could also view it as a "following wave" filling the void behind the object. Either way I will feel the momentum from the displaced fluid. (https://en.wikipedia.org/wiki/Added_mass)

I agree it does feel intuitively correct that you will not feel all the added mass; I guess it depends on the size of the falling body relative to myself. Anyway thanks for the interest. I have some estimates for the added mass of this object and so for the design of my helmet I will conservatively assume that I feel 100% the added mass.
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