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

In summary, the study of varying fluid density in a cylinder rolling down an inclined plane explores how the distribution and behavior of fluid within the cylinder affect its motion. As the cylinder rolls, the fluid's density can change due to factors such as gravitational forces and the inclination of the plane. This variation influences the cylinder's moment of inertia, rolling resistance, and overall dynamics, resulting in complex interactions between the fluid and the cylinder's motion. Understanding these effects is crucial for applications in engineering and physics where fluid dynamics and rotational motion are involved.
  • #106
mostafaelsan2005 said:
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}}##
mostafaelsan2005 said:
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.
 
Physics news on Phys.org
  • #107
Chestermiller said:
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)
 
  • #108
mostafaelsan2005 said:
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:
1700397569085.png
 
Last edited:
  • Like
Likes mostafaelsan2005
  • #109
Chestermiller said:
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:
1.1550.018
1.1080.133
1.0231.66
1.0770.214
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.
 

Attachments

  • Screenshot 2023-11-22 122542.png
    Screenshot 2023-11-22 122542.png
    3.9 KB · Views: 42
  • #110
mostafaelsan2005 said:
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:
1.1550.018
1.1080.133
1.0231.66
1.0770.214
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.
 
  • #111
Chestermiller said:
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.
 
  • #112
mostafaelsan2005 said:
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?
 
  • #113
Chestermiller said:
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?
 
  • #114
Let’s see your graph.
 
  • #115
IMG_3143.jpg

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.
 
  • #116
mostafaelsan2005 said:
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.
 
  • #117
Chestermiller said:
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.
 

Attachments

  • chart.pdf
    14.1 KB · Views: 50
  • #118
mostafaelsan2005 said:
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.
 
  • #119
Density () ± 0.6 gcm3Viscosity () in poise (approximated)Kinematic viscosity ()
Water 1.0 0.010.01
Sunflower oil 0.920.490.53
Molasses honey 1.45120.0082.75
Transmission fluid 0.871.201.37

Average time taken to reach bottom of ramp (s) ± 0.03 seconds
Water1.49
Sunflower oil 1.43
Molasses honey 1.32
Transmission fluid 1.39
 
  • #120
mostafaelsan2005 said:
Density () ± 0.6 gcm3Viscosity () in poise (approximated)Kinematic viscosity ()
Water1.00.010.01
Sunflower oil0.920.490.53
Molasses honey 1.45120.0082.75
Transmission fluid0.871.201.37

Average time taken to reach bottom of ramp (s) ± 0.03 seconds
Water1.49
Sunflower oil1.43
Molasses honey1.32
Transmission fluid1.39
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?
 
  • #121
Chestermiller said:
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?
 

Attachments

  • Fluid Rotational Motion EE Final.pdf
    1.5 MB · Views: 49
Last edited:
  • #122
mostafaelsan2005 said:
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.
 
  • #123
Chestermiller said:
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
Water1.33
Sunflower oil1.35
Molasses honey1.48
Transmission fluid1.39
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.
 
  • #124
mostafaelsan2005 said:
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
Water1.33
Sunflower oil1.35
Molasses honey1.48
Transmission fluid1.39
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
 
  • #125
Chestermiller said:
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.
 
  • #126
mostafaelsan2005 said:
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.
 
  • #127
Chestermiller said:
Of course it will. The times are different.
It has the same general shape but the values for t/t_0 are just different:
 

Attachments

  • chart (2).pdf
    14.1 KB · Views: 58
  • #128
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.
 
  • Love
Likes Tom.G
  • #129
Chestermiller said:
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.
 
  • #130
mostafaelsan2005 said:
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 $$x=\sqrt{\frac{\nu t_0}{\pi R^2}}$$
 
  • #131
Chestermiller said:
The asymptotes are y = 1.224 and ##y=1+\frac{16}{15}x##, where $$y=\frac{t}{t_0}$$and $$x=\sqrt{\frac{\nu t_0}{\pi R^2}}$$
1702211743752.png
 

Similar threads

Back
Top