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

[mostafaelsan2005]

I don't understand what you are asking?

Is the formula that was derived in post #37 related to the asymptotic solution?

 

[Chestermiller]

Is the formula that was derived in post #37 related to the asymptotic solution?

Well, when I derived those equations, I was still assuming that the can had mass, and that its inner and outer radii of the can were significantly different. The equation at the end of #37 is for the frictional force in this case, when the time t is short.

 

[mostafaelsan2005]

Well, when I derived those equations, I was still assuming that the can had mass, and that its inner and outer radii of the can were significantly different. The equation at the end of #37 is for the frictional force in this case, when the time t is short.

Oh I see now, and for the asymptotic solution, you assumed that the can has negligible mass (since the can used is constant across the different liquids it would make the most sense to treat the mass as negligible)? If that is so, I can just explain that derivation (asymptotic solution) only, right? Would you recommend any fundamental processes I should explain before skipping to the derivation because it would seem a bit abrupt

 

[Chestermiller]

Oh I see now, and for the asymptotic solution, you assumed that the can has negligible mass (since the can used is constant across the different liquids it would make the most sense to treat the mass as negligible)?

It is my understanding that the mass and moment of inertia of the can in your experiments is negligible.

If that is so, I can just explain that derivation (asymptotic solution) only, right? Would you recommend any fundamental processes I should explain before skipping to the derivation because it would seem a bit abrupt

I would explain that, for inviscid fluids, the behavior would be the same as if the can were sliding down the ramp with no friction (i.e., no tangential frictional force), and, for highly viscous fluids, the behavior approaches that of a solid cylinder; at low finite viscosities and/or short times, we have the boundary layer solution; for high finite viscosities and long times, we have the asymptotic long-time solution.

 

[Chestermiller]

The development is posts #51-54 is entirely new, and has never appeared in the literature before. It describes the angular velocity, frictional force, and angular acceleration at long times, when the long-time asymptotic behavior of the system is attained. This gives the same down-ramp acceleration is if there were a rigid solid inside the can.

 

[mostafaelsan2005]

The development is posts #51-54 is entirely new, and has never appeared in the literature before. It describes the angular velocity, frictional force, and angular acceleration at long times, when the long-time asymptotic behavior of the system is attained. This gives the same down-ramp acceleration is if there were a rigid solid inside the can.

I figured, I didn't see anything mathematically or conceptually similar even though I've been researching for quite some time. By the way, was it explained by the mass and MOI were omitted? The MOI is constantly changing (as per my idea in the beginning) but then we said that it cannot be used in this context, is that why it is omitted?

 

[Chestermiller]

I figured, I didn't see anything mathematically or conceptually similar even though I've been researching for quite some time. By the way, was it explained by the mass and MOI were omitted? The MOI is constantly changing (as per my idea in the beginning) but then we said that it cannot be used in this context, is that why it is omitted?

No. I'm omitting the mass and MOI of the metal can, not the fluid. The MOI concept doesn't apply to a fluid because a fluid is capable of shearing so that the fluid does not undergo a rigid body rotation.

The closest we can come to the MOI concept for this viscous fluid is with the asymptotic solution for long times. Here the angular velocity of the shearing fluid is a linear superposition of (1) a linear function of t, independent of r (which is basically a rigid body rotation of constant angular acceleration), plus (2) a quadratic function of r, independent of t (which is basically a shear flow with angular velocity that varies with r). The component (1) is what we get with a rigid solid, while the component (2) provides the shear stress at the boundary that is responsible for the frictional force.

 

[mostafaelsan2005]

No. I'm omitting the mass and MOI of the metal can, not the fluid. The MOI concept doesn't apply to a fluid because a fluid is capable of shearing so that the fluid does not undergo a rigid body rotation.

The closest we can come to the MOI concept for this viscous fluid is with the asymptotic solution for long times. Here the angular velocity of the shearing fluid is a linear superposition of (1) a linear function of t, independent of r (which is basically a rigid body rotation of constant angular acceleration), plus (2) a quadratic function of r, independent of t (which is basically a shear flow with angular velocity that varies with r). The component (1) is what we get with a rigid solid, while the component (2) provides the shear stress at the boundary that is responsible for the frictional force.

This is what I have so far for the theoretical section. I was wondering if you had any input on whether or not I should include the findings from posts #51-54 as what I have in terms of solving for the time function seems sufficient from an experimental point of view as the report focuses on the experimental phase? From my understanding of the developments you have made in those posts, it was to prove more clearly that the liquids in the cans that took a longer time to reach the bottom of the ramp behave like rigid bodies. Would you say that is important to the understanding of a reader who would most likely not have experience in fluid dynamics? I'm inclined to say that it is necessary as it is better to include it than rely on talking about the viscosity boundary conditions as a purely theoretical concept perhaps.

 

[Chestermiller]

This is what I have so far for the theoretical section. I was wondering if you had any input on whether or not I should include the findings from posts #51-54 as what I have in terms of solving for the time function seems sufficient from an experimental point of view as the report focuses on the experimental phase? From my understanding of the developments you have made in those posts, it was to prove more clearly that the liquids in the cans that took a longer time to reach the bottom of the ramp behave like rigid bodies. Would you say that is important to the understanding of a reader who would most likely not have experience in fluid dynamics? I'm inclined to say that it is necessary as it is better to include it than rely on talking about the viscosity boundary conditions as a purely theoretical concept perhaps.

This is what I have so far for the theoretical section. I was wondering if you had any input on whether or not I should include the findings from posts #51-54 as what I have in terms of solving for the time function seems sufficient from an experimental point of view as the report focuses on the experimental phase? From my understanding of the developments you have made in those posts, it was to prove more clearly that the liquids in the cans that took a longer time to reach the bottom of the ramp behave like rigid bodies.

They approach the solution at large values of the key dimensionless group, which involves viscosity, time, and can radius. It isn't that they behave as rigid bodies; they don't; the fluids in the cans are still exhibiting large shearing. They just give the same shear stress at the can wall at long times as for a rigid body. At short times and lower viscosities, the results are closer to those for an inviscid fluid. So there is a transition between these two extreme limits.

Would you say that is important to the understanding of a reader who would most likely not have experience in fluid dynamics? I'm inclined to say that it is necessary as it is better to include it than rely on talking about the viscosity boundary conditions as a purely theoretical concept perhaps.

How you present it is as big a decision and job as how to solve the problem. This is a judgment call. if I were you, I would put the derivation in an appendix, and only allude to the results in the appendix within the main body of the report. Too much arithmetic in the main body puts everyone to sleep.

 

[mostafaelsan2005]

They approach the solution at large values of the key dimensionless group, which involves viscosity, time, and can radius. It isn't that they behave as rigid bodies; they don't; the fluids in the cans are still exhibiting large shearing. They just give the same shear stress at the can wall at long times as for a rigid body. At short times and lower viscosities, the results are closer to those for an inviscid fluid. So there is a transition between these two extreme limits.How you present it is as big a decision and job as how to solve the problem. This is a judgment call. if I were you, I would put the derivation in an appendix, and only allude to the results in the appendix within the main body of the report. Too much arithmetic in the main body puts everyone to sleep.

Yes that's what I decided to do (that is put the derivation for the dimensionless parameter and asymptotic solution in the appendix as I saw an exemplar do the same for a hefty derivation). I was thinking of referring to it in either the graphical comparison between the theoretical and experimental results in order to show that the viscosity boundary conditions can be proven quantitatively or in the conclusion as I explain the validity/accuracy of the results. Also, so the liquids (i.e. honey) that exhibit properties closer to that of fluids with infinite viscosities do not behave as rigid bodies but rather have similar characteristics in their behavior ( in terms of the shear stress on the walls) ? Is that the distinction? Hopefully I gave your derivations justice in the report in terms of how thorough the explanations are of the equations. Once again, I appreciate your help .

 

[Chestermiller]

Yes that's what I decided to do (that is put the derivation for the dimensionless parameter and asymptotic solution in the appendix as I saw an exemplar do the same for a hefty derivation). I was thinking of referring to it in either the graphical comparison between the theoretical and experimental results in order to show that the viscosity boundary conditions can be proven quantitatively or in the conclusion as I explain the validity/accuracy of the results. Also, so the liquids (i.e. honey) that exhibit properties closer to that of fluids with infinite viscosities do not behave as rigid bodies but rather have similar characteristics in their behavior ( in terms of the shear stress on the walls) ? Is that the distinction? Hopefully I gave your derivations justice in the report in terms of how thorough the explanations are of the equations. Once again, I appreciate your help .

Yes, that is right. The shear stress at the wall in a fluid is equal to the viscosity times the fluid tangential velocity gradient at the wall. For high viscosity fluids, the long time behavior sets in at shorter times than for low viscosity fluids.

 

[mostafaelsan2005]

For the frictional force F in post # 37, if we assume that the can has negligible mass and moment of inertia, this force becomes, $$F=M_L\kappa^2\left(\frac{2\sqrt{\pi \nu}}{\pi R}\right)\int_0^t{\frac{a(\xi)}{\sqrt{t-\xi}}d\xi}$$Furthermore, if the inner and outer radii of the can shell are nearly equal, this equation reduces further to $$F=M_L\left(\frac{2\sqrt{\pi \nu}}{\pi R}\right)\int_0^t{\frac{a(\xi)}{\sqrt{t-\xi}}d\xi}$$And, for the case of thin outer boundary layers in the rotating fluid, the acceleration of the fluid as the can rolls down the ramp is going to be approximately constant. Under this approximation, our equation for the frictional force becomes $$F=\frac{4}{\sqrt{\pi}}M_L\frac{\sqrt{ \nu t}}{ R}a$$With these approximations, our force balance equation becomes $$M_Lg\sin{\alpha}-F=M_L a$$or $$a(t)=\frac{dv}{dt}=\frac{g\sin{\alpha}}{1+\frac{4}{\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}}\approx g\sin{\alpha}\left(1-\frac{4}{\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}\right)$$Integrating once to get the velocity, we have: $$v=g\sin{\alpha}\left(1-\frac{8}{3\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}\right)t$$Integrating again to get the distance then give us $$L\approx g\sin{\alpha}\left(1-\frac{32}{15\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}\right)t^2/2$$or, to the same level of approximation, $$L\left(1+\frac{32}{15\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}\right)\approx (g\sin{\alpha})t^2/2$$
A first approximation to the solution the this equation is the value obtained for a totally inviscid fluid: $$t\approx t_0=\sqrt{\frac{2L}{g\sin{\alpha}}}$$A second (better) approximation can then be obtained by substituting $t_0$ into the term in parenthesis and then re-solving for t:
$$t\approx t_0\sqrt{1+\frac{32}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}}\approx t_0\left(1+\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}\right)$$This equation is expected to describe the behavior in our system only if the mass- and moment of inertia of the metal can are negligible, and only in the limit of $\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}<<1$. The OP should make a plot of t as a function of $\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}$ to see how the predictions from this equation compare with the experimental data. In our system, $t_0=1.31\ sec$, R = 3.5 cm, and $\nu$ is the kinematic viscosity (cm^2/sec) of each individual fluid.

In regards to what is said here, I calculated it for water with the kinematic viscosity being 0.01 and t_0 being 1.31 (I have a question about this, how was 1.31 determined to be the initial time in order to find the maximum time for infinitely viscous fluids; the values for the boundaries basically) and I got 0.0196798321832. This answer is confusing to me as it is obviously not within the boundaries, any idea why this is happening or if I'm doing something wrong? For context, I am trying to plot

$\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}$

Also, if the equation:

$t_0=\sqrt{\frac{2L}{g\sin{\alpha}}}$ is used to find the time for a completely inviscid fluid, then how do we determine the angular acceleration and how is it used to plot the times? My issue in understanding is that I'm unsure if the minimum time is the one that pertains to water or not?

 

[Chestermiller]

In regards to what is said here, I calculated it for water with the kinematic viscosity being 0.01 and t_0 being 1.31 (I have a question about this, how was 1.31 determined to be the initial time in order to find the maximum time for infinitely viscous fluids; the values for the boundaries basically) and I got 0.0196798321832. This answer is confusing to me as it is obviously not within the boundaries, any idea why this is happening or if I'm doing something wrong? For context, I am trying to plot

The 1.31 is the approximate time for going down the ramp for the inviscid case. What do you mean that it is obviously not within the boundaries? You should be plotting the data points of $t/t_0$ vs $\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}$ for all the 4 fluids on the same graph. There should be one data point for each fluid. At low values of the square root term, t/to should approach $1+\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}$ (show this analytic curve on the graph). At large values of the square root term, t/to should approach $\sqrt{3/2}$ (show this asymptote on your graph).

Also, if the equation:

The minimum curve (analytic) should lie below the 4 data points. The maximum asymptote should lie above the 4 data points.

 

[mostafaelsan2005]

The 1.31 is the approximate time for going down the ramp for the inviscid case. What do you mean that it is obviously not within the boundaries? You should be plotting the data points of $t/t_0$ vs $\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}$ for all the 4 fluids on the same graph. There should be one data point for each fluid. At low values of the square root term, t/to should approach $1+\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}$ (show this analytic curve on the graph). At large values of the square root term, t/to should approach $\sqrt{3/2}$ (show this asymptote on your graph).

The minimum curve (analytic) should lie below the 4 data points. The maximum asymptote should lie above the 4 data points.

I understand that 1.31 seconds is an approximation for inviscid fluids and that the graph should have two vertical asymptotes at 1.31 and 1.6 with them being the boundaries for the time taken to reach the bottom of the ramp. I do not quite understand though how the approximation for 1.31 was made, was it an analytical approximation? I also understand that we need to plot that function for each fluid; however, when I calculated it for water using the value that t_0 is 1.31 and the kinematic viscosity is 0.01 poise, I was getting a value that would not make sense as a time value in the given scenario. Am I misunderstanding something? Is the formula not used to find the approximated time to reach the bottom of the ramp based on the fluid in question? For example: \frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}=\frac{16}{15}\sqrt{\frac{0.01(1.31)}{\pi(0.035)^2}}=1.96798. What would this number represent, it is certainly too large to be that of a time value in the physical system being observed since the infinite viscosity limit is 1.6? As for the plot that I should make where \frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}} would be on the y-axis and t/t_0 is on the x-axis, what would this relationship show. I think I am also confused about the difference between t and t_0, I know t_0 is 1.31 but what is t? Is it the actual experimental value? Furthermore, why would the ratio t/t_0 approach 1+\frac{16}{15}\sqrt{\frac{\nu t_0}{\pi R^2}}? As this value again for example would be 2.96798 which I doubt that t/t_0 would give (in the case of water). Sorry for all the questions, I still have some uncertainty on this part of the investigation.

 

[erobz]

I understand that 1.31 seconds is an approximation for inviscid fluids and that the graph should have two vertical asymptotes at 1.31 and 1.6 with them being the boundaries for the time taken to reach the bottom of the ramp. I do not quite understand though how the approximation for 1.31 was made, was it an analytical approximation?

This seems to be an issue you can't seem to flush out... How are you expecting to pass all this analysis off as your own work if you are unable to understand the very simplest part of all this?

 

[mostafaelsan2005]

This seems to be an issue you can't seem to flush out... How are you expecting to pass all this analysis off as your own work if you are unable to understand the very simplest part of all this?

I understand every aspect of the investigation other than the approximation for the boundary conditions, what do you mean? I have genuine questions about results not adding up so I'm not sure where the condescension is coming from. It was only ever stated that 1.31 was the time for inviscid fluids and so I was confused about it and if the values I'm getting are not within the boundaries so I have the concern that the relationship that t/t_0 would not approach that expression

 

[erobz]

I understand every aspect of the investigation other than the approximation for the boundary conditions,

It is just not believable to me that you cannot figure out how to solve the inviscid fluid condition, given you "understand" everything else perfectly, which is orders of magnitude more difficult...

Try applying conservation of energy.

 

[mostafaelsan2005]

It is just not believable to me that you cannot figure out how to solve the inviscid fluid condition, given you "understand" everything else perfectly, which is orders of magnitude more difficult...

Try applying conservation of energy.

t=sqrt(2L/gsin(θ))=sqrt(2*1.45/9.81*sin(10.3)=1.28 seconds which can't be it and it would only equal 1.31 if the angle of inclination was 9.91 degrees which it isn't and why I have been confused.

 

[erobz]

t=sqrt(2L/gsin(θ))=sqrt(2*1.45/9.81*sin(10.3)=1.28 seconds which can't be it and it would only equal 1.31 if the angle of inclination was 9.91 degrees which it isn't and why I have been confused.

Where is the MOI, or the mass of the can factoring in on that? The can is rolling without slipping, the inviscid liquid inside is translating without rotating...

 

[mostafaelsan2005]

Where is the MOI, or the mass of the can factoring in on that? The can is rolling without slipping, the inviscid liquid inside is translating without rotating...

As established in post 64, we can use the equation I showed above but with the idea of the viscous vector, f, which is 0 in the case of a completely inviscid liquid. In terms of the MOI, the motion of the inviscid fluid down the ramp is not rotational so we do not take into account the MOI and in other cases we have neglected the mass and friction to reach the solution.

 

[erobz]

As established in post 64, we can use the equation I showed above but with the idea of the viscous vector, f, which is 0 in the case of a completely inviscid liquid. In terms of the MOI, the motion of the inviscid fluid down the ramp is not rotational so we do not take into account the MOI and in other cases we have neglected the mass and friction to reach the solution.

The can containing the inviscid liquid has mass, it also has rotational inertia. I fail to see why it should be neglected in this limiting case. It seems to me you are unable to solve the problem without Chester spoon feeding you the answers.

If I'm wrong, just solve the problem and see what it works out to be? It's a basic physics problem that someone who "understands" the advanced fluid mechanics in this thread should consider child's play... Why you are putting up resistance is beyond me.

 

[mostafaelsan2005]

The can containing the inviscid liquid has mass, it also has rotational inertia. I fail to see why it should be neglected in this limiting case. It seems to me you are unable to solve the problem without Chester spoon feeding you the answers.

If I'm wrong, just solve the problem and see what it works out to be? It's a basic physics problem that someone who "understands" the advanced fluid mechanics in this thread should consider child's play... Why you are putting up resistance is beyond me.

I don't object that the can itself has a MOI and mass; however, in the solution, we have treated it as negligible, and so why would we treat it as important in this case. It would just be counter-intuitive

 

[erobz]

I don't object that the can itself has a MOI and mass; however, in the solution, we have treated it as negligible, and so why would we treat it as important in this case. It would just be counter-intuitive

What's counter intuitive to me, is why it would be ignored when its quite relatively simply to account for? We are talking about a fraction of a second disagreement...

 

[mostafaelsan2005]

What's counter intuitive to me, is why it would be ignored when its quite relatively simply to account for? We are talking about a fraction of a second disagreement...

I solved it using conservation of energy and I get: t = √((3/4) * m * r^2 * (4π^2) / I) where I is the MOI (1/2MR^2)

 

[erobz]

I solved it using conservation of energy and I get: t = √((3/4) * m * r^2 * (4π^2) / I) where I is the MOI (1/2MR^2)

Please show you work. I don't think its correct. Also use LaTeX Guide, at this point I shouldn't even have to ask that you format your math with it.

 

[Chestermiller]

$t_0$ is the time that an inviscid fluid would roll down the length of the ramp: $$t_0=\sqrt{\frac{2L}{g\sin{\theta}}}$$Your graph for the data points should have t as the vertical axis and $\frac{\nu t_0}{\sqrt{\pi R^2}}$ as the value plotted on the horizontal axis. The intercept at $\frac{\nu t_0}{\sqrt{\pi R^2}}=0$ on the vertical axis should be $t_0$ and the asymptote at large values of $\frac{\nu t_0}{\sqrt{\pi R^2}}$ should be the horizontal line $t = t_0\sqrt{1.5}$. On this same graph, you can also show a plot (for comparison) of the curve at small values of $\frac{\nu t_0}{\sqrt{\pi R^2}}$ as $$t=t_0\left(1+\frac{16}{15}\frac{\nu t_0}{\sqrt{\pi R^2}}\right)$$

 

[mostafaelsan2005]

$t_0$ is the time that an inviscid fluid would roll down the length of the ramp: $$t_0=\sqrt{\frac{2L}{g\sin{\theta}}}$$Your graph for the data points should have t as the vertical axis and $\frac{\nu t_0}{\sqrt{\pi R^2}}$ as the value plotted on the horizontal axis. The intercept at $\frac{\nu t_0}{\sqrt{\pi R^2}}=0$ on the vertical axis should be $t_0$ and the asymptote at large values of $\frac{\nu t_0}{\sqrt{\pi R^2}}$ should be the horizontal line $t = t_0\sqrt{1.5}$. On this same graph, you can also show a plot (for comparison) of the curve at small values of $\frac{\nu t_0}{\sqrt{\pi R^2}}$ as $$t=t_0\left(1+\frac{16}{15}\frac{\nu t_0}{\sqrt{\pi R^2}}\right)$$

Thanks for the reply. Did you use 10.3 degrees as the angle? I have calculated it and it is coming out to be 1.28 seconds (specifically, 1.28581369737) rather than 1.31 seconds. Also, in terms of the graph, $\frac{\nu t_0}{\sqrt{\pi R^2}}$ is the parameter that quantifies the time taken to reach the bottom of the ramp between the two limits? Furthermore, when calculating the time taken the results are confusing. For example, let's take water with a kinematic viscosity of 0.01 poise, $t_0$ is 1.31 seconds, and the radius R is 0.035 m. The value I am getting is 1.84498426718 seconds which cannot be the case in this physical system. If I'm understanding correctly, when graphing $t/t_0$ then I should be able to find that what is right hand side of the formula is equal to $sqrt(3/2)$ at high viscosities and what would it show at low viscosities? Also where is the formula for shear stress found in the Transport Phenomena book, the problem is on page 115 but I cannot find the formula for shear stress (using velocity, boundary layer thickness, and viscosity) that you had mentioned in the initial attempt on the solution.

 

[Chestermiller]

Thanks for the reply. Did you use 10.3 degrees as the angle? I have calculated it and it is coming out to be 1.28 seconds (specifically, 1.28581369737) rather than 1.31 seconds.

OK. 1.29 seconds

Also, in terms of the graph, $\frac{\nu t_0}{\sqrt{\pi R^2}}$ is the parameter that quantifies the time taken to reach the bottom of the ramp between the two limits?

No. The time is a function of this parameter (dimensionless group) for a viscous fluid.

Furthermore, when calculating the time taken the results are confusing. For example, let's take water with a kinematic viscosity of 0.01 poise, $t_0$ is 1.31 seconds, and the radius R is 0.035 m. The value I am getting is 1.84498426718 seconds which cannot be the case in this physical system.

You are getting 1.84 seconds from the analytical approximation?

If I'm understanding correctly, when graphing $t/t_0$ then I should be able to find that what is right hand side of the formula is equal to $sqrt(3/2)$ at high viscosities and what would it show at low viscosities?

It would show values predicted by the analytical expression.

Also where is the formula for shear stress found in the Transport Phenomena book, the problem is on page 115 but I cannot find the formula for shear stress (using velocity, boundary layer thickness, and viscosity) that you had mentioned in the initial attempt on the solution.

The shear stress is equal to the viscosity time the velocity gradient at the wall.

 

[mostafaelsan2005]

OK. 1.29 seconds

No. The time is a function of this parameter (dimensionless group) for a viscous fluid.

You are getting 1.84 seconds from the analytical approximation?It would show values predicted by the analytical expression.

The shear stress is equal to the viscosity time the velocity gradient at the wall.

Yes, I am getting 1.84 for the calculation (for water) and the number increases/decreases based on the fluid inputted (in terms of kinematic viscosity in poise where the radius 0.035 m). OK, so it shows the values that are predicted by the analytical expression by looking at the changes in time divided by 1.29 seconds and that's where we can see the values predicted? Is that used to figure out that the timing for long times is given by $sqrt3/2*t_0$ where the values will approach but never be able to pass the horizontal line that gives that value? As for the second formula (second approximation for the solution), using the same values for the radius and kinematic viscosity of water gives the result that $t/t_0=2.376$ for water where that number is a dimensionless parameter presumably. I guess I'm having trouble understanding what this number represents?

 

[Chestermiller]

Yes, I am getting 1.84 for the calculation (for water)

What does the 1.84 represent.

and the number increases/decreases based on the fluid inputted (in terms of kinematic viscosity in poise

The kinematic viscosity $\nu=\mu/\rho$ is not in poise. It is in cm^2/sec. Dynamic viscosity $\mu$ is in poise. Watch your units!!!

where the radius 0.035 m). OK, so it shows the values that are predicted by the analytical expression by looking at the changes in time divided by 1.29 seconds and that's where we can see the values predicted?

Let's see the results for t or t/to vs the other dimensionless parameter involving kinematic viscosity.

Is that used to figure out that the timing for long times is given by $sqrt3/2*t_0$ where the values will approach but never be able to pass the horizontal line that gives that value?

No. The $\sqrt{3/2}$ comes from the asymptotic solution at large values of the dimensionless parameter involving kinematic viscosity.

As for the second formula (second approximation for the solution), using the same values for the radius and kinematic viscosity of water gives the result that $t/t_0=2.376$ for water where that number is a dimensionless parameter presumably. I guess I'm having trouble understanding what this number represents?

The analytic solution at short times can't give 2.376. It has got to be close to 1. I need to see your calculated results for t/to vs the dimsionless parameter $\frac{\nu t_0}{\sqrt{\pi R^2}}$ for each of the fluids. One data point for each fluid.

 

[mostafaelsan2005]

What does the 1.84 represent.

The kinematic viscosity $\nu=\mu/\rho$ is not in poise. It is in cm^2/sec. Dynamic viscosity $\mu$ is in poise. Watch your units!!!

Let's see the results for t or t/to vs the other dimensionless parameter involving kinematic viscosity.

No. The $\sqrt{3/2}$ comes from the asymptotic solution at large values of the dimensionless parameter involving kinematic viscosity.

The analytic solution at short times can't give 2.376. It has got to be close to 1. I need to see your calculated results for t/to vs the dimsionless parameter $\frac{\nu t_0}{\sqrt{\pi R^2}}$ for each of the fluids. One data point for each fluid.

1.84 should represent the time it takes for water to reach the bottom of the ramp but that cannot be the case. I have done the calculation several times with a calculator and have gotten the same answer. It is weird that it comes out to 2.376, though, I thought that the value was supposed to be < < 1, is that not the case? Though I do note that in the solution you stated that it's much less than 1 for when we are calculating $sqrtvt/R$ and not $\frac{\nu t_0}{\sqrt{\pi R^2}}$ though I'm not quite sure of the difference and which one I should focus on. For the asymptotic solution at long times, how was sqrt(3/2) determined from the dimensionless parameters. On the section involving the dimensionless parameters I can't seem to find the derivation for it? I'll plot the rest of the values and see if there is any meaning to the data then using $\frac{\nu t_0}{\sqrt{\pi R^2}}$ for the other liquids I used.

 

[Chestermiller]

We lost something in transferring the results from post to post. The correct results are in post #45. The horizontal axis parameter should be $$\sqrt{\frac{\nu t_0}{\pi R^2}}$$ rather than $$\frac{\nu t_0}{\sqrt{\pi R^2}} $$

Try redoing the calculations with the correct parameter and let's see what you get. The incorrect parameter results should not have come out dimensionless.

Also, I’d like to see your calculations for the case of water.

 

[mostafaelsan2005]

We lost something in transferring the results from post to post. The correct results are in post #45. The horizontal axis parameter should be $$\sqrt{\frac{\nu t_0}{\pi R^2}}$$ rather than $$\frac{\nu t_0}{\sqrt{\pi R^2}} $$

Try redoing the calculations with the correct parameter and let's see what you get. The incorrect parameter results should not have come out dimensionless.

Also, I’d like to see your calculations for the case of water.

$\sqrt{\frac{\nu t_0}{\pi R^2}} = sqrt((1.29*0.01)/(pi*0.035^2)) = 1.83$. 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 (the part before deriving the solution based on dimensionless parameters)? Would it be fair to say that I can omit the asymptotic solution and the solution of the dimensionless parameters from the investigation because an in-depth discussion into the viscosity boundary conditions and the parameters used to find the time taken to roll down the ramp should be sufficient based on my personal judgement, no? The only thing I would include perhaps is the derivation using the dimensionless parameters in the appendix so that I can refer to it in the theoretical background when I refer to how the long-time formula is:
$t=t_0*sqrt(3/2)$
Since the graphs I will include in the investigation will be based on the viscosity boundary conditions parameters that is what I based my personal judgement out of as to not let the theoretical background be too tedious for the examiner.
In that case, then is there a way that you derived the sqrt3/2 out of the dimensionless parameters because that it isn't explicitly mentioned how. I'm thinking of doing the experiment again today with more data points as it seems that the calculations aren't adding up, maybe they'll come out closer this time to the expression we have to calculate the time. I graphed the parameter as a function and you can see it here: https://www.desmos.com/calculator/fvlmni6hev
the calculations under the function and set hor. asymptotes and from top to bottom the fluids are: water, oil, honey, transmission fluid and as you can see the results are nonsensical. I'm not quite sure why this is happening (I am sure the calculations for the kinematic viscosity is correct as it is the viscosity in poise divided by the density of the liquid, maybe there is an issue in the densities? the numbers are ridiculously high

 

[Chestermiller]

$\sqrt{\frac{\nu t_0}{\pi R^2}} = sqrt((1.29*0.01)/(pi*0.035^2)) = 1.83$. Any idea why this is the case (this is the calculation for water)?

Yes. The units are screwed up. It should be 0.0183. The R should be 3.5, not 0.035. We are working with cgs.

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 (the part before deriving the solution based on dimensionless parameters)?

See Bird, Stewart, and Lightfoot, Transport Phenomena, Appendices

Would it be fair to say that I can omit the asymptotic solution and the solution of the dimensionless parameters from the investigation because an in-depth discussion into the viscosity boundary conditions and the parameters used to find the time taken to roll down the ramp should be sufficient based on my personal judgement, no? The only thing I would include perhaps is the derivation using the dimensionless parameters in the appendix so that I can refer to it in the theoretical background when I refer to how the long-time formula is:
$t=t_0*sqrt(3/2)$

This is a judgment call on your part. However, the derivation of the asymptotic solution is new, and has never appeared in the literature before. This would be a major contribution to the understanding and to the solution to this problem. You can take credit for it if you want. I don't care about being acknowledged.

Since the graphs I will include in the investigation will be based on the viscosity boundary conditions parameters that is what I based my personal judgement out of as to not let the theoretical background be too tedious for the examiner.

The theoretical development can be relegated to the Appendix. You can take credit for marshaling the resources to get the theoretical development implemented.

In that case, then is there a way that you derived the sqrt3/2 out of the dimensionless parameters because that it isn't explicitly mentioned how.

I don't follow. I thought I did this.

I'm thinking of doing the experiment again today with more data points as it seems that the calculations aren't adding up, maybe they'll come out closer this time to the expression we have to calculate the time. I graphed the parameter as a function and you can see it here: https://www.desmos.com/calculator/fvlmni6hev

I can't read the graph. t/to should be y, and the other dimensionless group should be x. Do you not have access to Microsoft Excel? In terms of these x and y parameters, the short term solution should be a straight line. $y=(1+\frac{16}{15}x)$

the calculations under the function and set hor. asymptotes and from top to bottom the fluids are: water, oil, honey, transmission fluid and as you can see the results are nonsensical. I'm not quite sure why this is happening (I am sure the calculations for the kinematic viscosity is correct as it is the viscosity in poise divided by the density of the liquid, maybe there is an issue in the densities? the numbers are ridiculously high

Like I said, the 0.035 should be 3.5.

 

[mostafaelsan2005]

Yes. The units are screwed up. It should be 0.0183. The R should be 3.5, not 0.035. We are working with cgs.

See Bird, Stewart, and Lightfoot, Transport Phenomena, Appendices

This is a judgment call on your part. However, the derivation of the asymptotic solution is new, and has never appeared in the literature before. This would be a major contribution to the understanding and to the solution to this problem. You can take credit for it if you want. I don't care about being acknowledged.

The theoretical development can be relegated to the Appendix. You can take credit for marshaling the resources to get the theoretical development implemented.

I don't follow. I thought I did this.

I can't read the graph. t/to should be y, and the other dimensionless group should be x. Do you not have access to Microsoft Excel? In terms of these x and y parameters, the short term solution should be a straight line. $y=(1+\frac{16}{15}x)$

Like I said, the 0.035 should be 3.5.

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 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?