Varying fluid (density) in a cylinder rolling along an inclined plane

[Chestermiller]

I redid the calculations then and the range of values is 0.0183-0.2087. Also, I was graphing the parameter for ##t=\sqrt{\frac{\nu t_0}{\pi R^2}}## not ##t=t_0\left(1+\frac{16}{15}\frac{\nu t_0}{\sqrt{\pi R^2}}\right)##.

The correct equation for short time behavior is $$t=t_0\left(1+\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}\right)$$You should be plotting the data points as ##y=t/t_0## vs ##x=\sqrt{\frac{\nu t_0}{\pi R^2}}##

In regards to the dimensionless parameters, I read it quite a few times to determine where the sqrt(3/2) came from but it was never explicitly stated in the post, I may have misunderstood it though or maybe it was inferred piece of information?

Post #53 shows that the asymptotic viscous case acceleration of the center of mass at long times is $$a=\frac{2}{3}g\sin{\alpha}$$This compares with the inviscid case center of mass acceleration of $$a=g\sin{\alpha}$$Note that the asymptotic long time viscous case acceleration is 3/2 the inviscid case acceleration. The time to roll down the ramp is inversely proportional to the square root of the acceleration. So at very long times, the time to roll down the ramp for the viscous case is ##\sqrt{3/2}## the time for the inviscid case.

 

[mostafaelsan2005]

The correct equation for short time behavior is $$t=t_0\left(1+\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}\right)$$You should be plotting the data points as ##y=t/t_0## vs ##x=\sqrt{\frac{\nu t_0}{\pi R^2}}##

Post #53 shows that the asymptotic viscous case acceleration of the center of mass at long times is $$a=\frac{2}{3}g\sin{\alpha}$$This compares with the inviscid case center of mass acceleration of $$a=g\sin{\alpha}$$Note that the asymptotic long time viscous case acceleration is 3/2 the inviscid case acceleration. The time to roll down the ramp is inversely proportional to the square root of the acceleration. So at very long times, the time to roll down the ramp for the viscous case is ##\sqrt{3/2}## the time for the inviscid case.

I see, that makes much more sense when speaking about the factor 3/2 in terms of the acceleration. In terms of the equation for the short time behavior, we know that ##t_0## is 1.29; however, what would I be inputting as ##t## in the equation to divide by ##t_0##? I just want to make sure if that is the actual time it takes for the cylinder to reach the bottom of the ramp (based on the fluid within)

 

[Chestermiller]

I see, that makes much more sense when speaking about the factor 3/2 in terms of the acceleration. In terms of the equation for the short time behavior, we know that ##t_0## is 1.29; however, what would I be inputting as ##t## in the equation to divide by ##t_0##? I just want to make sure if that is the actual time it takes for the cylinder to reach the bottom of the ramp (based on the fluid within)

If you are plotting the data points, then in t/to, you put the actual measured time down the ramp for t, and the inviscid time 1.29 sec, for to.

The plot should crudely look something like this:

 

[mostafaelsan2005]

If you are plotting the data points, then in t/to, you put the actual measured time down the ramp for t, and the inviscid time 1.29 sec, for to.

The plot should crudely look something like this:
View attachment 335775

Using the values of kinematic viscosities of

0.01 (water)

0.53 (sunflower oil)

82.75 (molasses honey)

1.37 (transmission fluid)

Based on these values, I got these data points for the liquids in the same order:

1.155	0.018
1.108	0.133
1.023	1.66
1.077	0.214

I cannot make any sense of these numbers as they are nowhere near the boundaries (1.29 to 1.57). The graph looks like this (only the points attached below). Why do you think that the graph looks like this? I am using the value of 3.5 cm as the radius and the others are constant values.

 

[Chestermiller]

Using the values of kinematic viscosities of

0.01 (water)

0.53 (sunflower oil)

82.75 (molasses honey)

1.37 (transmission fluid)

Based on these values, I got these data points for the liquids in the same order:

1.155	0.018
1.108	0.133
1.023	1.66
1.077	0.214

I cannot make any sense of these numbers as they are nowhere near the boundaries (1.29 to 1.57). The graph looks like this (only the points attached below). Why do you think that the graph looks like this? I am using the value of 3.5 cm as the radius and the others are constant values.

I have no idea what the numbers in the columns represent or how they were obtained. Please provide a sample calculation for the sunflower oil data point.

 

[mostafaelsan2005]

I have no idea what the numbers in the columns represent or how they were obtained. Please provide a sample calculation for the sunflower oil data point.

Sure, based on the kinematic viscosity of sunflower oil being approximately 0.53, then using ##\sqrt{\frac{\nu t_0}{\pi R^2}}##, which is sqrt(0.53*1.29/pi*3.5^2) = 0.133288.

 

[mostafaelsan2005]

Sure, based on the kinematic viscosity of sunflower oil being approximately 0.53, then using ##\sqrt{\frac{\nu t_0}{\pi R^2}}##, which is sqrt(0.53*1.29/pi*3.5^2) = 0.133288.

Could this be an issue with the values for kinematic viscosity? The values calculated are correct but perhaps the poise values used for the viscosity and/or the density values calculated have some issue unless the equation could somehow not be correct for this situation? Is there another way to calculate it?

 

[mostafaelsan2005]

I have no idea what the numbers in the columns represent or how they were obtained. Please provide a sample calculation for the sunflower oil data point.

I've been tweaking the numbers for some time now and I've come to the conclusion that even if the values for kinematic viscosity were such that we get the expected value from the boundary condition parameter, its values would not make sense theoretically based on the actual liquid. Is there something we're missing?

 

[Chestermiller]

Let’s see your graph.

 

[mostafaelsan2005]

This graph shows the viscosity boundary condition parameter where the kinematic viscosity is x in order to show the general shape of the curve. I've highlighted the first data point where x (kinematic viscosity) is 0.01 and you can see that the value is 0.018 which does not make sense in the context of the experiment.

 

[Chestermiller]

View attachment 336672
This graph shows the viscosity boundary condition parameter where the kinematic viscosity is x in order to show the general shape of the curve. I've highlighted the first data point where x (kinematic viscosity) is 0.01 and you can see that the value is 0.018 which does not make sense in the context of the experiment.

I can't see a thing from your photo.. Print it out and then take the photo.

 

[mostafaelsan2005]

I can't see a thing from your photo.. Print it out and then take the photo.

This is a scatterplot of what we find when graphing t/t_0 against the viscosity boundary condition parameter with data points that I have outlined in post #109. As you can see, there is little meaning attached to these data points as they are outside of the boundary conditions set. I have made sure to keep the radius in centimeters although it doesn't change much if it is in meters as the results still do not add up. I can add the asymptotes for the time into this graph, but it is surpassing it from both the minimum and maximum so it would not add anything.

 

[Chestermiller]

This is a scatterplot of what we find when graphing t/t_0 against the viscosity boundary condition parameter with data points that I have outlined in post #109. As you can see, there is little meaning attached to these data points as they are outside of the boundary conditions set. I have made sure to keep the radius in centimeters although it doesn't change much if it is in meters as the results still do not add up. I can add the asymptotes for the time into this graph, but it is surpassing it from both the minimum and maximum so it would not add anything.

This doesn't make any sense to me. Please make me a table of the times vs the kinematic viscosities. Previously, in your experiments, the shortest time was water and the longer times were the more viscous fluids.

 

[mostafaelsan2005]

	Density () ± 0.6 gcm3	Viscosity () in poise (approximated)	Kinematic viscosity ()
Water	1.0	0.01	0.01
Sunflower oil	0.92	0.49	0.53
Molasses honey	1.45	120.00	82.75
Transmission fluid	0.87	1.20	1.37

	Average time taken to reach bottom of ramp (s) ± 0.03 seconds
Water	1.49
Sunflower oil	1.43
Molasses honey	1.32
Transmission fluid	1.39

 

[Chestermiller]

	Density () ± 0.6 gcm3	Viscosity () in poise (approximated)	Kinematic viscosity ()
Water	1.0	0.01	0.01
Sunflower oil	0.92	0.49	0.53
Molasses honey	1.45	120.00	82.75
Transmission fluid	0.87	1.20	1.37

	Average time taken to reach bottom of ramp (s) ± 0.03 seconds
Water	1.49
Sunflower oil	1.43
Molasses honey	1.32
Transmission fluid	1.39

The time to roll down the ramp was largest for water and decreased monotonically with increasing kinematic viscosity. This is the exact opposite of the results in your original experiments. Why?

 

[mostafaelsan2005]

The time to roll down the ramp was largest for water and decreased monotonically with increasing kinematic viscosity. This is the exact opposite of the results in your original experiments. Why?

I did the experiments again, remember? I found that the results at first weren't accurate due to the lubrication of the ramp I was using. Once I did them again, and made sure that the ramp was not lubricated, I found this to be the set of values for the average time to go down the ramp. If it is needed, I can redo the experiment as I have all the equipment currently. Is the problem with the water data point?

small edit: Ill do the experiment again for all the data values, it should take me about 30 minutes or so and Ill relay the information. I'm not sure why the nearly inviscid liquid is coming out to be that with the highest time while the most viscous was the fastest. From what I understand theoretically, it should be the opposite due to less fluid resistance, right?

 

[Chestermiller]

I did the experiments again, remember? I found that the results at first weren't accurate due to the lubrication of the ramp I was using. Once I did them again, and made sure that the ramp was not lubricated, I found this to be the set of values for the average time to go down the ramp. If it is needed, I can redo the experiment as I have all the equipment currently. Is the problem with the water data point?

It is with the order of the points with respect to kinematic viscosity. Water should be the shortest time, and the time should increase monotonically with the kinematic viscosity of the fluid. In the original set of data, the time for water was only about 1.3 sec.

 

[mostafaelsan2005]

It is with the order of the points with respect to kinematic viscosity. Water should be the shortest time, and the time should increase monotonically with the kinematic viscosity of the fluid. In the original set of data, the time for water was only about 1.3 sec.

Apologies for the late response, I conducted the experiment again and I measured the times through 5 trials. Here are the average values for time:

Average time taken to reach bottom of ramp (s) ± 0.03 seconds
Water	1.33
Sunflower oil	1.35
Molasses honey	1.48
Transmission fluid	1.39

I'm not sure why there was a discrepancy in values but I have the suspicion that the timer I used in the previous set of data was damaged in some way. Here the value for water is coming out similar to the first conducted experiment so this set of data is the most accurate of the three with a functioning timer and minimal lubrication of the ramp.

 

[Chestermiller]

Apologies for the late response, I conducted the experiment again and I measured the times through 5 trials. Here are the average values for time:

Average time taken to reach bottom of ramp (s) ± 0.03 seconds
Water	1.33
Sunflower oil	1.35
Molasses honey	1.48
Transmission fluid	1.39

I'm not sure why there was a discrepancy in values but I have the suspicion that the timer I used in the previous set of data was damaged in some way. Here the value for water is coming out similar to the first conducted experiment so this set of data is the most accurate of the three with a functioning timer and minimal lubrication of the ramp.

Let’s see the graph now

 

[mostafaelsan2005]

Let’s see the graph now

The graph won't change though? The changing variable in the viscosity boundary condition parameter is the kinematic viscosity which is the same as I've said in Post #119.

 

[Chestermiller]

The graph won't change though? The changing variable in the viscosity boundary condition parameter is the kinematic viscosity which is the same as I've said in Post #119.

Of course it will. The times are different.

 

[mostafaelsan2005]

Of course it will. The times are different.

It has the same general shape but the values for t/t_0 are just different:

 

[Chestermiller]

What the hell are you talking about? This is not the same general shape at all. Now draw the two asymptotic approximations on this same graph for comparison.

 

[mostafaelsan2005]

What the hell are you talking about? This is not the same general shape at all. Now draw the two asymptotic approximations on this same graph for comparison.

The asymptotic approximations for t/t_0 will range from 1 to 1.217, correct? Otherwise it will be outside the boundary. Minimum = 1.29/1.29 and maximum = 1.57/1.29.

 

[Chestermiller]

The asymptotic approximations for t/t_0 will range from 1 to 1.217, correct? Otherwise it will be outside the boundary. Minimum = 1.29/1.29 and maximum = 1.57/1.29.

The asymptotes are y = 1.224 and ##y=1+\frac{16}{15}x##, where $$y=\frac{t}{t_0}$$and $$x=\sqrt{\frac{\nu t_0}{\pi R^2}}$$

 

[Chestermiller]

The asymptotes are y = 1.224 and ##y=1+\frac{16}{15}x##, where $$y=\frac{t}{t_0}$$and $$x=\sqrt{\frac{\nu t_0}{\pi R^2}}$$