# Dropping Balls in tube with liquid

1. Aug 27, 2014

### bbq2014

1. The problem statement, all variables and given/known data

So, im 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.

2. Relevant equations

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

3. 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.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Aug 27, 2014
2. Aug 27, 2014

### 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 ?

3. Aug 27, 2014

### bbq2014

Yeah it does. It was only a 40cm tube. The drag force was from stoke's law i think. Fd=6πμvr. I dont 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.

4. Aug 27, 2014

### bbq2014

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

5. Aug 27, 2014

### haruspex

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?

6. Aug 27, 2014

### 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

7. Aug 27, 2014

### bbq2014

Also here is the corresponding ordered pairs.

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

8. Aug 27, 2014

### 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 ?

9. Aug 27, 2014

### 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.

Last edited: Aug 27, 2014
10. Aug 27, 2014

### bbq2014

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

11. Aug 27, 2014

### Staff: Mentor

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

12. Aug 27, 2014

### haruspex

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 travelling 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.

13. Aug 27, 2014

### Staff: Mentor

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

14. Aug 27, 2014

### bbq2014

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.

15. Aug 27, 2014

### bbq2014

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

16. Aug 27, 2014

### BvU

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 !

17. Aug 27, 2014

### Staff: Mentor

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

18. Aug 27, 2014

### haruspex

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.

19. Aug 27, 2014

### Staff: Mentor

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

20. Aug 28, 2014

### bbq2014

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.