How would you calculate how deep an object would go in water?

[flamephoenix15]

Homework Statement

So given a ball, radius r, mass m ,an a known height y from the surface the ball is dropped from, how would you calculate the depth the ball goes to in water (including the water it displaces), with the density of the ball less than the density of water. Ignore surface tension and the viscous properties of water.

Homework Equations

equation[/B]
F_b = ρgAh (h being depth and A cross sectional area)
mgy = Fg

The Attempt at a Solution

I believe that the "Bouyant energy" is equal to ∫F_bdh, h being depth since Gravitational Energy is equal to ∫F_gdy, y being height. However, I do not know if I am approaching this correctly and I do not know how to solve for the displaced fluid which is incorporated into the depth the ball goes to.

 

[Andrew Mason]

Homework Statement

So given a ball, radius r, mass m ,an a known height y from the surface the ball is dropped from, how would you calculate the depth the ball goes to in water (including the water it displaces), with the density of the ball less than the density of water. Ignore surface tension and the viscous properties of water.

Homework Equations

equation[/B]
F_b = ρgAh (h being depth and A cross sectional area)
mgy = Fg

The Attempt at a Solution

I believe that the "Bouyant energy" is equal to ∫F_bdh, h being depth since Gravitational Energy is equal to ∫F_gdy, y being height. However, I do not know if I am approaching this correctly and I do not know how to solve for the displaced fluid which is incorporated into the depth the ball goes to.

Hi Flamephoenix15. Welcome to PF!

I gather that we are to assume that regardless of the energy of the impact of the ball with the water, the water does not gain any kinetic energy from that impact. So we are dealing only with the conversion of the ball's gravitational potential energy at height y to buoyant potential energy at depth d.

The problem is that the buoyant force on the ball while not completely submerged will depend on the amount of water displaced by it and that could vary from 0 to V_ball. That is not a simple calculation.

Addendum: If the ball is not dropped at all, work out the amount of water it will displace and, from that, determine the depth that the bottom of the spherical ball will reach. Then it would be a matter of determining the increased displacement as a function of y and converting that to depth.

AM

 

[kuruman]

The problem is that the buoyant force on the ball while not completely submerged will depend on the amount of water displaced by it and that could vary from 0 to Vball. That is not a simple calculation.

I think it is a simple calculation. Assume that the ball is submerged so that its lowest point is $z$ below the surface. Then if it is submerged by an additional amount $dz$, the additional displaced water will have the volume of a disk of area $A=\pi[R^2-(R-z)^2]$. This can be used to find the incremental buoyant force $d(BF)$ and the integration to find the total buoyant force should not be too hard.

 

[flamephoenix15]

I think it is a simple calculation. Assume that the ball is submerged so that its lowest point is $z$ below the surface. Then if it is submerged by an additional amount $dz$, the additional displaced water will have the volume of a disk of area $dA=\pi[R^2-(R-z)^2]$. This can be used to find the incremental buoyant force $d(BF)$ and the integration to find the total buoyant force should not be too hard.

So if we have $dA$, we could integrate it and multiply it as a function of depth to find the Volume of the displaced fluid. Then you could intengrate the buoyant force with respect to volume $dV$ to find the depth, or would you integrate it with respect to depth $dh$?

 

[kuruman]

So if we have $dA$, we could integrate it and multiply it as a function of depth to find the Volume of the displaced fluid. Then you could intengrate the buoyant force with respect to volume $dV$ to find the depth, or would you integrate it with respect to depth $dh$?

No. First you need to find by how much the buoyant force increases when the sphere sinks by an additional amount $dz$. This is related to the area of the disk $A$ (I mistakenly wrote $dA$ in post #3). Then the volume of the displaced water is $dV=A~dz$. What is $d(BF)$? That's what you should integrate.

 

[flamephoenix15]

Hi Flamephoenix15. Welcome to PF!

I gather that we are to assume that regardless of the energy of the impact of the ball with the water, the water does not gain any kinetic energy from that impact. So we are dealing only with the conversion of the ball's gravitational potential energy at height y to buoyant potential energy at depth d.

The problem is that the buoyant force on the ball while not completely submerged will depend on the amount of water displaced by it and that could vary from 0 to V_ball. That is not a simple calculation.

Addendum: If the ball is not dropped at all, work out the amount of water it will displace and, from that, determine the depth that the bottom of the spherical ball will reach. Then it would be a matter of determining the increased displacement as a function of y and converting that to depth.

AM

It is going to varry, however if I could experimentally measure h

No. First you need to find by how much the buoyant force increases when the sphere sinks by an additional amount $dz$. This is related to the area of the disk $A$ (I mistakenly wrote $dA$ in post #3). Then the volume of the displaced water is $dV=A~dz$. What is $d(BF)$? That's what you should integrate.

Ohhhh, I got it, that actually helps quite a bit assuming you are correct, and your calculations appear to make sense. Thanks!

 

[kuruman]

It is going to varry, however if I could experimentally measure h

Ohhhh, I got it, that actually helps quite a bit assuming you are correct, and your calculations appear to make sense. Thanks!

Perhaps you can post your results when you get them. Also make sure that your expression gives the correct $BF$ when the depth is $z=0$ (must be zero) and when it is fully submerged ($z=2R$).