Dropping Balls in tube with liquid

[bbq2014]

Homework Statement

So, I am having trouble trying to understand my results. The experiment was marble (1.5cm & 5gram) was dropped on the surface of water, in a 40 cm cylindrical tube roughly 4cm diameter. The graph is negative, since the I used a program to analyse the velocity.

Homework Equations

I don't understand why the velocity decreased after it hit max velocity.

The Attempt at a Solution

http://tinypic.com/view.php?pic=2n9cjue&s=8#.U_2AmLySyed

The graph is a velocity of marble vs. time graph

It makes sense that the marble accelerates therefore had increase in velocity after it was released. But why would it the velocity be parabolic?
It should be a square root looking graph since the marble would continue to accelerate until the weight is equal to the force of buoyancy and drag, therefore terminal veloicity. But if the graph is correct, the stationary point is the maximum velocity, is that the be the terminal velocity? If it is, it doesn't sound realistic, it reaches terminal velocity in 0.3sec.

I calculated the Weight of the marble = 0.05317 N
The buoyant force = 0.0173 N
The drag force at the maximum velocity of the marble from the graph (0.564m/s) = 0.0797 N
Therefore 0.05317 = 0.09703, the mass should be floating? which sort of makes sense since the marble's velocity is decreasing? But why did the marble reach max velocity so fast?

But it might be due to experimental error.

Thank you in advance!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[BvU]

You can start with drawing a straight line for the contribution that you know about: gravitational acceleration.
The drag force you calculate is based on a few assumptions (you don't mention them ?) that might not apply (yet).
It's a bit unfortunate that the speed doesn't stabilize before the end of the graph, though. I mean: the thing keeps falling, doesn't it ?

 

[bbq2014]

Yeah it does. It was only a 40cm tube. The drag force was from stoke's law i think. Fd=6πμvr. I don't think it should be a straight line either. Since its gaining drag the ball should decelerate from 9.81m/s/s, therefore as i mentioned above, it should be a square root looking graph.

 

[bbq2014]

Also what assumptions do you mean for drag force? Is it terminal velocity or velocity relative to sphere? thanks !

 

[haruspex]

Since its gaining drag the ball should decelerate from 9.81m/s/s,

You can decelerate from a speed, but 9.81m/s/s is an acceleration. Do you mean accelerate more slowly than 9.81m/s/s?

marble (1.5cm & 5gram) was dropped on the surface of water

Where was it in relation to the surface of the water when released? Some distance above? On the surface? Partly submerged?
The words and numbers on the graph are very hard to read. Can you estimate where the marble was in relation to the surface of the water when at max velocity?

 

[bbq2014]

Yeah, sorry i meant accelerate more slowly than 9.81m/s/s.

We tried to keep the marble on the surface, but sometimes the bottom of the marble is partly submerged. Yes I think I can find where the marble is at a point in time.

http://tinypic.com/view.php?pic=30vk8x1&s=8#.U_3Hr7ySyec here is a better quality i hope.

The x-axis is 0, 0.2, 0.4 etc , y-axis top is -0.3, -0.4 -0.5, just in the negative direction

 

[bbq2014]

You can decelerate from a speed, but 9.81m/s/s is an acceleration. Do you mean accelerate more slowly than 9.81m/s/s?


Where was it in relation to the surface of the water when released? Some distance above? On the surface? Partly submerged?
The words and numbers on the graph are very hard to read. Can you estimate where the marble was in relation to the surface of the water when at max velocity?

Also here is the corresponding ordered pairs.

http://tinypic.com/view.php?pic=2h6d92r&s=8#.U_3IqrySyec

 

[BvU]

OK, looked into this a little more. Found a picture in Wiki that has .5 m/s for 15 mm diameter quartz spheres as terminal velocity. So who knows...
But the deceleration after 0.4 s has me puzzled too. Do we need to know more details of the setup to find out what happens ?

---

Can't say I follow all the calculations. A 5g marble weight 0.005 * 9.8 would be .049 N, wouldn't it ?

Buoyant force I can reverse engineer as $\rho_{water}\, \pi {D^2\over 6}\, g$ with $\rho_{water} = 1000, \; g=9.8$

Stokes' law applies when the Reynolds number $\rho \, v \, D\over \mu$ is << 1. Now I'm puzzled what you use for the viscosity of water ? I thought it would be something like 0.001 Pa.s ?

 

[bbq2014]

The experiment looks basically, a ~60cm cylindrical tube diameter of 3.8cm. Water is filled to 40cm and the marble is dropped above the surface of the water, sometimes the bottom of the marble is partially submerged in the water when release. The time for it to drop 40cm was recorded. And the drop was recorded with slow motion.Yeah the mass was 5.42g, therefore W = 5.42 * 9.81
The diameter was 1.5cm

For the drag force, the viscosity of water i used was 1000kg/m/s, my bad! You are right, it was suppose to be 0.001.

 

[bbq2014]

OK, looked into this a little more. Found a picture in Wiki that has .5 m/s for 15 mm diameter quartz spheres as terminal velocity. So who knows...
But the deceleration after 0.4 s has me puzzled too. Do we need to know more details of the setup to find out what happens ?

---

Can't say I follow all the calculations. A 5g marble weight 0.005 * 9.8 would be .049 N, wouldn't it ?

Buoyant force I can reverse engineer as $\rho_{water}\, \pi {D^2\over 6}\, g$ with $\rho_{water} = 1000, \; g=9.8$

Stokes' law applies when the Reynolds number $\rho \, v \, D\over \mu$ is << 1. Now I'm puzzled what you use for the viscosity of water ? I thought it would be something like 0.001 Pa.s ?

I think Reynolds' number would be greater than 1 for water which would mean that stokes law would not be very reasonable.

I calculated Re to be 8475. Re=1000*0.565*0.015/0.001 =8475. Therefore Re>>1

 

[Chestermiller]

It looks like there is something wrong with the results in the figure. For one thing, it looks like there is an initial downward velocity. Also, irrespective of the Reynolds number, the downward velocity should never be decreasing. What does the data on distance vs time look like, and how did you use this data to estimate instantaneous velocity.

Chet

 

[haruspex]

It looks like there is something wrong with the results in the figure. For one thing, it looks like there is an initial downward velocity.

Since the ball was released when just touching the water, it would have reached 0.35 m/s in roughly 0.05 seconds, so it could just be that there's a small offset in the time base.

Also, irrespective of the Reynolds number, the downward velocity should never be decreasing.

By the time it is fully submerged it could be traveling faster than terminal velocity, so it would only decrease. What it certainly should not do is increase (after full submersion) and then decrease.

Two possibilities come to mind:
- there is something inhomogeneous about the water column (a hot layer on top?)
- the marble is initially spinning
but neither is very persuasive.

 

[Chestermiller]

By the time it is fully submerged it could be traveling faster than terminal velocity, so it would only decrease. What it certainly should not do is increase (after full submersion) and then decrease.

Right. We'll know more when we see the distance vs time data, and when we find out how he calculated (or directly measured?) the instantaneous velocities.

Chet

 

[bbq2014]

It looks like there is something wrong with the results in the figure. For one thing, it looks like there is an initial downward velocity. Also, irrespective of the Reynolds number, the downward velocity should never be decreasing. What does the data on distance vs time look like, and how did you use this data to estimate instantaneous velocity.

Chet

Yeah i think so to. Chart of distance vs time looks fine though, it is linear so there probably wasn't any mistake when plotting the data. The link to distance vs time graph: http://tinypic.com/view.php?pic=344qo1t&s=8#.U_5RuLySyec ignore the red, it is the x-axis displacement.

i did notmeasured the instantaneous velocity myself. I used a program called logger pro which you set the time equal zero and a scale and you plot the points every frame or so.

 

[bbq2014]

Since the ball was released when just touching the water, it would have reached 0.35 m/s in roughly 0.05 seconds, so it could just be that there's a small offset in the time base.


By the time it is fully submerged it could be traveling faster than terminal velocity, so it would only decrease. What it certainly should not do is increase (after full submersion) and then decrease.

Two possibilities come to mind:
- there is something inhomogeneous about the water column (a hot layer on top?)
- the marble is initially spinning
but neither is very persuasive.

The water is homogeneous. After replying the video 15 times or so. I've noticed that the marble has some slight spin on its way down, but I sort of think the spinning become a little faster near the end. But I don't think it is would be that significant to cause my data to be like this. Also At two point in time during the drop the marble hit the wall of the tube slightly but it bounced right off at ~0.3sec and ~0.55sec

 

[BvU]

i did notmeasured the instantaneous velocity myself. I used a program called logger pro which you set the time equal zero and a scale and you plot the points every frame or so

I can understand you want some program to do the work, but I can't understand how you can believe the velocity plot is correct if you have seen the distance versus time plot. Didn't you check any of the points on the velocity plot with $\Delta\,X\over \Delta \, t$ at all ? Just looking at the X,t graph at an angle: only the first and the last two intervals might deviate a little from an otherwise constant speed !

 

[Chestermiller]

I can understand you want some program to do the work, but I can't understand how you can believe the velocity plot is correct if you have seen the distance versus time plot. Didn't you check any of the points on the velocity plot with $\Delta\,X\over \Delta \, t$ at all ? Just looking at the X,t graph at an angle: only the first and the last two intervals might deviate a little from an otherwise constant speed !

Yes. As BvU points out, this solves the mystery. The distance vs time graph looks exactly the way we expected it to look. One can see the initial curvature from the first 3 grid points, corresponding to the initial acceleration, followed by a long region of constant terminal velocity. At the very end, the velocity seems to decrease slightly, possibly suggesting the effect of approaching the bottom of the column (the final grid point is only 4cm (<1") from the bottom).

The velocities obtained from logger pro are totally inconsistent with the distance vs time data. Not even close. Apparently, this fact did not immediately jump out at you, bbq2014, the way it did for those of us with more experience. Either logger pro was not applied properly, or it is defective. Either way, you (bbq2014) don't need it. Just draw your best straight line on the d vs t graph (in the region where the slope is constant), and evaluate the slope.

Chet

 

[haruspex]

The velocities obtained from logger pro are totally inconsistent with the distance vs time data. Not even close.

Are you sure? Looks to me that the velocity is pretty steady at 0.55 m/s from 0.2 to 0.45 seconds, in fair agreement with loggerpro. The initial acceleration and final deceleration also look about right. (It's hard to tie them up exactly without knowing how loggerpro calculates the velocity at a given instant - is it based on the next distance interval, the previous distance interval, or some sliding average?)

bbq2014, you say it's a 40cm cylinder, but what is the water depth? It does look as though it's encountering a bit extra resistance right at the end. Is the tube a uniform width all the way to the bottom? You say it hit the side at 0.55 seconds. Maybe being close to the side can slow it down - I don't know enough about hydrodynamics to answer that.

 

[Chestermiller]

Are you sure? Looks to me that the velocity is pretty steady at 0.55 m/s from 0.2 to 0.45 seconds, in fair agreement with loggerpro. The initial acceleration and final deceleration also look about right. (It's hard to tie them up exactly without knowing how loggerpro calculates the velocity at a given instant - is it based on the next distance interval, the previous distance interval, or some sliding average?)

bbq2014, you say it's a 40cm cylinder, but what is the water depth? It does look as though it's encountering a bit extra resistance right at the end. Is the tube a uniform width all the way to the bottom? You say it hit the side at 0.55 seconds. Maybe being close to the side can slow it down - I don't know enough about hydrodynamics to answer that.

Being close to the side can definitely slow it down.

I would like to see the raw distance vs time data, and do some hand calculations to check out loggerpro. Maybe I spoke too quickly. I still don't think that I see a factor of ~ 2 variation in the slope during the major part of the overall time interval. The hand calculations will answer all this.

Chet

 

[bbq2014]

I can understand you want some program to do the work, but I can't understand how you can believe the velocity plot is correct if you have seen the distance versus time plot. Didn't you check any of the points on the velocity plot with $\Delta\,X\over \Delta \, t$ at all ? Just looking at the X,t graph at an angle: only the first and the last two intervals might deviate a little from an otherwise constant speed !

I didn't say the velocity plot is correct. The whole purpose of this post is that the velocity graph is a bit strange! And if anyone could provide some pointers why the velocity is so.

 

[bbq2014]

Yes. As BvU points out, this solves the mystery. The distance vs time graph looks exactly the way we expected it to look. One can see the initial curvature from the first 3 grid points, corresponding to the initial acceleration, followed by a long region of constant terminal velocity. At the very end, the velocity seems to decrease slightly, possibly suggesting the effect of approaching the bottom of the column (the final grid point is only 4cm (<1") from the bottom).

The velocities obtained from logger pro are totally inconsistent with the distance vs time data. Not even close. Apparently, this fact did not immediately jump out at you, bbq2014, the way it did for those of us with more experience. Either logger pro was not applied properly, or it is defective. Either way, you (bbq2014) don't need it. Just draw your best straight line on the d vs t graph (in the region where the slope is constant), and evaluate the slope.

Chet

No it didn't but but it should have! The distance vs time graph looks linear to me. I will try it right now!

 

[bbq2014]

Are you sure? Looks to me that the velocity is pretty steady at 0.55 m/s from 0.2 to 0.45 seconds, in fair agreement with loggerpro. The initial acceleration and final deceleration also look about right. (It's hard to tie them up exactly without knowing how loggerpro calculates the velocity at a given instant - is it based on the next distance interval, the previous distance interval, or some sliding average?)

I myself am not too sure, but I think it does.

bbq2014, you say it's a 40cm cylinder, but what is the water depth? It does look as though it's encountering a bit extra resistance right at the end. Is the tube a uniform width all the way to the bottom? You say it hit the side at 0.55 seconds. Maybe being close to the side can slow it down - I don't know enough about hydrodynamics to answer that.

Sorry i wasn't too clear. The height of the water is 40cm, and the tube is much longer. Yeah the tube is uniform throughout, and indeed it does not fall straight in the center but falls at an angle hence hitting the side of the tube.

 

[Chestermiller]

Please do me a favor and provide the distance vs time table (like you did for velocity vs time). I would like to play with the numbers. Thanks.

Chet

 

[bbq2014]

Please do me a favor and provide the distance vs time table (like you did for velocity vs time). I would like to play with the numbers. Thanks.

Chet

Sure no problem.

http://tinypic.com/view.php?pic=21989vk&s=8#.U_8Xi7ySyec

http://tinypic.com/view.php?pic=b8ssww&s=8#.U_8XvLySyec

These are different trials with different densities of spheres

 

[Chestermiller]

Sure no problem.

http://tinypic.com/view.php?pic=21989vk&s=8#.U_8Xi7ySyec

http://tinypic.com/view.php?pic=b8ssww&s=8#.U_8XvLySyec

These are different trials with different densities of spheres

I hand calculated the instantaneous velocities for these cases using finite differences (central difference approximation), and, in neither case did I obtain a velocity variation with time anything like what loggerpro gave you. After the initial transient, the velocities were nearly constant (of course, with some expected scatter). For the first case, I got a terminal velocity of about 0.4 m/s, and, in the second case, about 0.25 m/s.

Chet

 

[haruspex]

I hand calculated the instantaneous velocities for these cases using finite differences (central difference approximation), and, in neither case did I obtain a velocity variation with time anything like what loggerpro gave you. After the initial transient, the velocities were nearly constant (of course, with some expected scatter). For the first case, I got a terminal velocity of about 0.4 m/s, and, in the second case, about 0.25 m/s.

Chet

But neither of those tables match the distance-time graph linked at post #14 (which, in my view, does match the velocity-time graph and data originally posted). So I don't think these are from the right trial.

 

[Chestermiller]

Good point. So maybe bbq2014 can make that x-t data available to us?

Chet

 

[bbq2014]

Good point. So maybe bbq2014 can make that x-t data available to us?

Chet

Sorry I don't have the x-t table for the first graph, and i can't recreate the graph again since, i don't know how many fps i did when analysing it, so it would be different. Alternatively you could approximate the points. Sorry!

 

[Chestermiller]

Try it with the data sets you sent me. Use a spreadsheet like excel. To get the velocity vs time plot, calculate v=(Xi+1 - Xi)/(Ti+1 - Ti) at t=(Ti+1 + Ti)/2 at each i , and plot the v's vs the x's. Compare what you get with what logger pro gave. Share with us the comparison.

Chet