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 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.
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 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...
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...
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...
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...
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...
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...
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.
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...
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...
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)##. In regards to the dimensionless parameters, I read it quite a few times to...
##\sqrt{\frac{\nu t_0}{\pi R^2}} = sqrt((1.29*0.01)/(pi*0.035^2)) = 1.83##. Any idea why this is the case (this is the calculation for water)? Also, for how you expressed the change in angular velocity over time in post #51, is there a source for this formula or is this from your own knowledge...