Find Time to Reach Top of Beaker (10 cm)

[Orlo]

Suppose I have a ball which is submerged in a fluid at height h0

The mass (m) is given

The density of the fluid (ρw) is given

The density of the object (ρo) and it's volume (4/3pi*r^2) are also given

We can define all forces pushing the object downwards :

Gravity
m*g
Ignore altitude dependancy, gravity is negative.

Pressure from the water above the ball
pi *r^2*m*ρw*(h0-d)
where d is the displacement of the sphere from the origin

The drag force, combined with the relationship between reynold's number and the drag force for a sphere is:
42μLu/r

where μ is the dynamic viscosity
L is the characteristic length
u is the velocity
r is the radius

Upwards there is only the buoyant force, which is -ρw*g*4/3pi*r^2How would I find the time taken to reach the top of the beaker, considering the beaker is 10 centimetres tall, thanks in advance for any help:)

 

[Baluncore]

Welcome to PF.

volume (4/3pi*r^2)

Check the volume of a sphere. Should it not be r^3 ?

 

[berkeman]

Welcome @Orlo -- Is this question for schoolwork? If so, we can move it to the schoolwork forums.

 

[kuruman]

It seems to me that you can use Newton's second law to find the acceleration and use that in a kinematic equation to find the time. I would assume that "to reach the top of the beaker" means that the top of the ball just breaks the surface. The buoyant force becomes a bit tricky once the ball breaks the surface.

 

[A.T.]

The buoyant force becomes a bit tricky once the ball breaks the surface.

It becomes tricky even before the ball breaks the surface. The rising ball can create turbulence which will affect its path. So it won't just rise vertically, but oscillate horizontally as well. This can be easily observed with balls released under water in a pool or bathtub.

Here a relevant paper:
https://journals.aps.org/prfluids/abstract/10.1103/PhysRevFluids.1.074501

 

[Orlo]

Welcome @Orlo -- Is this question for schoolwork? If so, we can move it to the schoolwork forums.

Yes, it is, I was unaware there was one my bad

It becomes tricky even before the ball breaks the surface. The rising ball can create turbulence which will affect its path. So it won't just rise vertically, but oscillate horizontally as well. This can be easily observed with balls released under water in a pool or bathtub.

Here a relevant paper:
https://journals.aps.org/prfluids/abstract/10.1103/PhysRevFluids.1.074501

For my purposes I ignore that turbulence as I have a work around

It seems to me that you can use Newton's second law to find the acceleration and use that in a kinematic equation to find the time. I would assume that "to reach the top of the beaker" means that the top of the ball just breaks the surface. The buoyant force becomes a bit tricky once the ball breaks the surface.

Yup, to just break the surface

 

[kuruman]

Yes, it is, I was unaware there was one my bad

For my purposes I ignore that turbulence as I have a work aroundYup, to just break the surface

Then you would have to write down a differential equation of the form $\dfrac{du}{dt}=\text{BF}-mg-ku$ where BF is the buoyant force and $ku$ is the drag force to get $v(t)$. From that, you can get $y(t)$ and then the time. It would be prudent to bundle the constant forces BF - mg under a single symbol, say F, and substitute the constants they depend on at the very end.

 

[Orlo]

Then you would have to write down a differential equation of the form $\dfrac{du}{dt}=\text{BF}-mg-ku$ where BF is the buoyant force and $ku$ is the drag force to get $v(t)$. From that, you can get $y(t)$ and then the time. It would be prudent to bundle the constant forces BF - mg under a single symbol, say F, and substitute the constants they depend on at the very end.

Yup, I have done that, I related the drag force to reynold's number for a sphere, as that is relevant for my paper. The resulting equation is:

F=-36πrμv-gρ_wπr²h+k

Where f is the total forces o the object, r is the radius, μ is the dynamic viscosity, ρ_w is the density of water, and h is the displacement from the origin. If i want to know the time the ball takes to break the surface of the water, point at which h=0.1, how would I go about that? k is a bunch of constant forces, including gravity, a part of the pressure force and the buoyant force. v is the velocity of the ball.

g is negative for my purposes, in case something didn't add up

 

[Steve4Physics]

F=-36πrμv-gρ_wπr²h+k

36πrμv doesn't look right. You may want to read up Stokes’ law to check you have the correct expression for the drag force.

gρ_wπr²h is the downwards force on the upper surface of a horizontal disc. So it is irrelevant here. You may want to read up on buoyancy to get the correct expression for the upthrust force

Some other random thoughts...

Note that velocity is $v= \frac {dh}{dt}$ and acceleration is $a = \frac {d^2h}{dt^2}$. With a correct equation for the net force you can use ‘F=ma’. Replace F with $m\frac {d^2h}{dt^2}$. And replace v by $\frac {dh}{dt}$. Then you have a second order differential equation. The solution gives the relationship between h and t.

A diagram helps, e.g. is h measured from the top, centre or bottom of the ball?

 

[Orlo]

36πrμv doesn't look right. You may want to read up Stokes’ law to check you have the correct expression for the drag force.

gρ_wπr²h is the downwards force on the upper surface of a horizontal disc. So it is irrelevant here. You may want to read up on buoyancy to get the correct expression for the upthrust force

Some other random thoughts...

Note that velocity is $v= \frac {dh}{dt}$ and acceleration is $a = \frac {d^2h}{dt^2}$. With a correct equation for the net force you can use ‘F=ma’. Replace F with $m\frac {d^2h}{dt^2}$. And replace v by $\frac {dh}{dt}$. Then you have a second order differential equation. The solution gives the relationship between h and t.

A diagram helps, e.g. is h measured from the top, centre or bottom of the ball?

Thanks for the reply, well 36πrμv is what I get combining the equation for drag force and the relationship between reynold's number and the drag coefficient of a sphere.
gρ_wπr²h do I remove this? The buoyancy force is included in k...

Yup, ik that, but idk how to solve that system...

 

[kuruman]

I see @Steve4Physics got there before me. It seems you are confused about the buoyant force which is basically the net sum of all the "pressure times area" forces exerted on the object by the fluid. See here for a simplified description.

As I indicated to you in post #7, you need to solve an equation of the form $$\frac{dv}{dt}=F-kv$$To solve, you need to separate variables and integrate.

 

[Orlo]

I see @Steve4Physics got there before me. It seems you are confused about the buoyant force which is basically the net sum of all the "pressure times area" forces exerted on the object by the fluid. See here for a simplified description.

As I indicated to you in post #7, you need to solve an equation of the form $$\frac{dv}{dt}=F-ku$$To solve, you need to separate variables and integrate.

My bad, I didn't understand, that simplifies it a lot, now I can solve it myself, thanks for your help :)

 

[kuruman]

My bad, I didn't understand, that simplifies it a lot, now I can solve it myself, thanks for your help :)

Note that I edited my post #11 to correct the typo that had $v$ on the left side and $u$ on the right side for the velocity. It should be the same on both sides.

 

[Orlo]

Note that I edited my post #11 to correct the typot hat had $v$ on the left side and $u$ on the right side for the velocity. It should be the same on both sides.

Thanks :)

 

[Steve4Physics]

... 36πrμv is what I get combining the equation for drag force and the relationship between reynold's number and the drag coefficient of a sphere.

Note that Stoke’s law applies when the Reynolds number is very small. Stoke’ law basically gives the drag force on a sphere moving slowly with laminar flow through a viscous medium. This drag force is proportional to v.

But the drag coefficient (which you mention) is usually applied when the Reynolds number is larger. Crudely, this corresponds to some turbulence behind the moving object. In these situations, drag force is often taken to be proportional to v².

I recommend you decide what regime you are in, and carefully examine how you got 36πrμv!

 

[Orlo]

Thanks for the reply, I did some more research, I will be using stokes' law, I had not heard of it up till now, thank you very much, this simplifies a lot of things.

 

[Delta2]

It becomes tricky even before the ball breaks the surface. The rising ball can create turbulence which will affect its path. So it won't just rise vertically, but oscillate horizontally as well. This can be easily observed with balls released under water in a pool or bathtub.

Here a relevant paper:
https://journals.aps.org/prfluids/abstract/10.1103/PhysRevFluids.1.074501

This sounds probably correct, but just for the record I am afraid you are considering too much realism for this college/high school level problem. Many such problems are simplified from details of the complex reality like turbulence.