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

[Chestermiller]

There is significantly more work that I have done on this problem since my last post.

From the frame of reference of an observer traveling down the ramp with the velocity of the center of mass of the fluid, the equations in the reference that I gave in a previous post can be expressed in terms of the angular velocity $\omega$ of the fluid by $$\frac{\partial \omega}{\partial t}=\nu\left(\frac{\partial ^2 \omega}{\partial r^2}+\frac{3}{r}\frac{\partial \omega}{\partial r}\right)\tag{1}$$where $\nu$ is the kinematic viscosity of the fluid $\mu/\rho$. In addition, the shear stress exerted by the fluid on the wall of the can $\tau_W$ is given by $$\tau_W=\mu R\left(\frac{\partial \omega}{\partial r}\right)_{r=R}\tag{2}$$where $\mu$ is the dynamic viscosity of the fluid.

The overall force balance on the fluid-filled can (neglecting the mass and rotational inertia of the thin cylindrical shell) is given by $$\rho \pi R^2 L\frac{dv}{dt}=\rho\pi R^2L g\sin{\alpha}-F\tag{3}$$where F is the frictional force exerted by the ramp on the can surface. The frictional force F is related to the wall shear stress of the fluid on the inner radius of the can is given by $$F=2\pi RL\tau_W\tag{4}$$ (This equation assumes that the viscous drag exerted by the fluid on the lid and base of the can is negligible, or, equivalently, that the length to diameter ratio of the can is very large; note that, this is not really the case for the actual can, but, for now we will neglect the variation in angular velocity of the fluid with axial position within the can).
If we combine Eqns. 2-4, we obtain $$\rho \pi R^2 L\frac{dv}{dt}=\rho\pi R^2L g\sin{\alpha}-2\pi R^2 L\ \mu \left(\frac{\partial \omega}{\partial r}\right)_{r=R}\tag{5}$$or, equivalently,$$\frac{dv}{dt}=g\sin{\alpha}-2\nu\left(\frac{\partial \omega}{\partial r}\right)_{r=R}\tag{6}$$Kinematically, the velocity of the center of mass v is related to the rotation rate of the can surface $\Omega=\omega(t,R)$ by $v=\Omega R$. Combining this with Eqn. 6 gives: $$\frac{d\Omega}{dt}=\frac{g\sin{\alpha}}{R}-2\frac{\nu}{R}\left(\frac{\partial \omega}{\partial r}\right)_{r=R}\tag{7}$$

Questions so far? After these are addressed, I will proceed further by reducing these equations to dimensionless form.

 

[Chestermiller]

Our two key equations for this system are Eqns. 1 and 7 of the previous post. Eqn. 1 is the transient fluid flow equation for the fluid angular velocity as a function of time and radial position inside the can. Eqn. 7 can be regarded as a boundary condition for Eqn.1, analogous to the thermal boundary conditions in transient heat conduction problems.

We can reduce Eqns. 1 and 7 to dimensionless form by introducing the following dimensionless variables: $$\bar{r}=\frac{r}{R}$$ $$\bar{t}=\frac{\nu t}{R^2}$$ and $$\bar{\omega}=\frac{\omega \nu}{Rg\sin{\alpha}}$$In terms of these dimensionless parameters, our two key equations become: $$\frac{\partial \bar{\omega}}{\partial \bar{t}}=\left[\frac{\partial^2 \bar{\omega}}{\partial \bar{r}^2}+\frac{3}{\bar{r}}\frac{\partial \bar{\omega}}{\partial \bar{r}}\right]$$and $$\frac{d\bar{\Omega}}{\bar{dt}}=1-2\left(\frac{\partial \bar{\omega}}{\partial \bar{r}}\right)_{\bar{r}=1}$$Note that there are no dimensionless groups acting as coefficients in these equations. These are what I like to call "once-and-for-all" equations since, once we solve them just once, that solution (in terms of the dimensionless parameters) will apply for all time to all possible geometric parameters and physical properties of the fluid that we could encounter. The key relationship that we will be looking for will be the dimensionless can rotation rate $\bar{\Omega}$ as a function of the dimensionless time $\bar{t}$ since the can began rolling down the ramp.

 

[Chestermiller]

ASYMPTOTIC SOLUTION AT LONG TIMES

As the can accelerates, the fluid within the can is lagging the angular velocity at the can surface, and is struggling to keep up. To do this, it must develop a radial angular velocity gradient and a corresponding shear stress profile (which varies with radius).

This situation is very much analogous to a transient conductive heating problem for a solid cylinder under the action of a constant heat flux at its surface. In the heat transfer situation, the temperatures within the solid cylinder are struggling to keep up with the rising surface temperature and, to do this, a corresponding radial temperature gradient and corresponding radial heat flux profile must develop. Eventually, the temperature at each radial location within the solid must be increasing linearly with time (because of the constant heat flux at the surface), but, superimposed on this, there must also be a radial temperature profile (independent of time) such that the rate of temperature rise at each radial location is the same.

If we employ this analogy to determine the asymptotic long-time solution to our two key dimensionless rolling cylinder equations of the previous post (Eqns. 1 and 2 of post #52), we obtain the simple result that, at long times, the angular velocity profile in the fluid approaches $$\bar{\omega}=\frac{2}{3}\bar{t}+\frac{1}{12}\bar{r}^2$$One can readily verify that this relationship satisfies Eqns. 1 and 2 of post #52 exactly. Furthermore, it satisfies all required boundary conditions at $bar{r}=1$ and $\bar{r}=0$. In fact, the only condition it does not satisfy is the initial condition $\bar{\omega}=0$ at $\bar{t}=0$; of course, satisfying the initial condition is not a requirement of the long-time asymptotic solution. $$\omega=\frac{2}{3}\frac{g\sin{\alpha}}{R}t+\frac{g\sin{\alpha}}{ R}\frac{r^2}{12\nu}$$ This equation shows that, at long times, the entire mass of fluid in the can experiences the same angular acceleration: $$\frac{\partial \omega}{\partial t}=\frac{2}{3}\frac{g\sin{\alpha}}{R}$$And the acceleration of the center of mass of the can at long times approaches: $$a=\frac{2}{3}g\sin{\alpha}$$. These are exactly the same angular- and linear accelerations that a solid cylinder would experience (at all times). Thus, we have shown that, in terms acceleration down the ramp, the viscous-fluid filled can behaves at long times as if the fluid were solid.

 

[Chestermiller]

SUMMARY OF RESULTS

SHORT TIMES ($\frac{\sqrt{ \nu t}}{ R}<<1$):

Acceleration: $$a(t)=g\sin{\alpha}\left(1-\frac{4}{\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}\right)$$
Frictional Force on can: $$F=M_Lg\sin{\alpha}\left(\frac{4}{\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}\right)$$According to this, at short times, the acceleration down the ramp of a can filled with a viscous liquid starts out with a value of $g\sin{\alpha}$ and decreases linearly with $\sqrt{t}$ as time progresses.

The frictional force on the can starts out at zero, and increases in proportion to $\sqrt{t}$.

LONG TIMES ($\frac{\sqrt{ \nu t}}{ R}>>1$):
Acceleration: $$a=\frac{2}{3}g\sin{\alpha}$$Frictional Force on can: $$F=\frac{1}{3}M_Lg\sin{\alpha}$$According to this, at long times, the can acceleration and frictional force for a can filled with a viscous liquid approach constant values that are exactly the same as those for a rigid cylinder.

The key dimensionless group for the effects of a viscous fluid on a can rolling down a ramp is $\frac{\sqrt{ \nu t}}{ R}$. The transition from short time behavior to long time behavior is monotonic, and varies only with this dimensionless group.

 

[erobz]

@Chestermiller, since you've gone through all this trouble maybe you could compile this in an insight complete with diagrams?

To be frank, the OP has apparently shifted gears to pendulum damping ( given the complexity of the analysis - its a reasonable choice IMO ).

https://www.physicsforums.com/threa...needed-information-in-physics-report.1055698/

 

[mostafaelsan2005]

@Chestermiller, since you've gone through all this trouble maybe you could compile this in an insight complete with diagrams?

To be frank, the OP has apparently shifted gears to pendulum damping ( given the complexity of the analysis - its a reasonable choice IMO ).

https://www.physicsforums.com/threa...needed-information-in-physics-report.1055698/

Haha don't worry, that's the internal assessment that I'm also doing, I am still fully on-board with this analysis and truly appreciate all the help that @Chestermiller has provided to me. I'm just in the middle of university application season and have been completely swamped with deadlines because I'm doing early action. I'll read through all the information now and see if I have any questions.

 

[mostafaelsan2005]

I'm having trouble understanding some of the times you had in your experiment. If the inviscid limit for the time to roll down the ramp is 1.31 seconds and the infinite viscosity limit is $\sqrt{3/2}$ times this, or 1.6 seconds, how could you have gotten times greater than 1.6 seconds in your experiments for 3 out of the 4 samples? Please run a test where you have something totally rigid in the can in place of the fluid, like concrete or jello. I would like to determine if these give 1.6 seconds or not.

I have finally come back to the experiment and wanted to answer this question. I have reason to believe that the times are rather distorted because the ramp was slightly lubricated with the oil when I conducted the experiment. I wiped down the ramp but based on the theoretical results it does seem like it affected the results quite drastically. As requested, I redid the experiment with a fully-rigid can that I filled with sand and the time was 1.58 seconds. I redid the experiment with the different fluids again and they were within that range with 1.6 acting as a maximum time. Furthermore, I was wondering if I should include discussion of the velocity as well as I do have the measured time(s) and dimensions of the ramp or if constricting it to time would be wiser?

 

[Tom.G]

Mentioning the problem with the oily ramp in the report could be a plus.
It is, after all, a learning experience (even if it is embarrissing).

 

[Chestermiller]

I have finally come back to the experiment and wanted to answer this question. I have reason to believe that the times are rather distorted because the ramp was slightly lubricated with the oil when I conducted the experiment. I wiped down the ramp but based on the theoretical results it does seem like it affected the results quite drastically. As requested, I redid the experiment with a fully-rigid can that I filled with sand and the time was 1.58 seconds. I redid the experiment with the different fluids again and they were within that range with 1.6 acting as a maximum time. Furthermore, I was wondering if I should include discussion of the velocity as well as I do have the measured time(s) and dimensions of the ramp or if constricting it to time would be wiser?

I'm having trouble understanding. Are you saying that, when you redid the experiments, the maximum time was 1.6 seconds for, presumably the transmission fluid and the honey? Before, they were both about 2.0 seconds. I hope you realize that the change from the case of an inviscid fluid to the case of a rigid fluid is only a factor of 1.22. So, in these time experiments, accuracy is very important.

Regarding your last question, I would say that the times should be the focus, not velocity.

 

[mostafaelsan2005]

I'm having trouble understanding. Are you saying that, when you redid the experiments, the maximum time was 1.6 seconds for, presumably the transmission fluid and the honey? Before, they were both about 2.0 seconds. I hope you realize that the change from the case of an inviscid fluid to the case of a rigid fluid is only a factor of 1.22. So, in these time experiments, accuracy is very important.

Regarding your last question, I would say that the times should be the focus, not velocity.

As per requested, I did the experiment with a sand-filled can in order for it to act like a rigid body as a placeholder. As a result, the time it took to reach the bottom of the ramp was approximately 1.6 seconds as was theoretically determined. The problem lied in the fact that I had experimented with the oil first which leaked down the ramp and affected the subsequent results (and its own results as well). The reason for this is because the lubrication of the ramp caused the cylinder to sometimes slip off of its translational path partially and that lead to times that were greater when the can was almost vertical. I meant that the time-value for the other liquids was within the boundaries of x<1.6<x. Apologies if I am not clear enough I don't actually mean that it's a 'maximum'.

 

[mostafaelsan2005]

I'm having trouble understanding some of the times you had in your experiment. If the inviscid limit for the time to roll down the ramp is 1.31 seconds and the infinite viscosity limit is $\sqrt{3/2}$ times this, or 1.6 seconds, how could you have gotten times greater than 1.6 seconds in your experiments for 3 out of the 4 samples? Please run a test where you have something totally rigid in the can in place of the fluid, like concrete or jello. I would like to determine if these give 1.6 seconds or not.

What do you mean by inviscid limit? Is this the time limit assuming negligible viscosity? So from my understanding, you found a maximum and minimum in the form of the time limits for fluids with infinite viscosity and inviscid fluids? That would make sense to me; however, the only fluid that is still confusing me is honey which still results in it reaching the bottom of the ramp in more than less than 1.6 seconds while the others are near to or greater than 1.6. Furthermore, shouldn't an object with infinite viscosity roll down faster (which is what I observed) as fluids behave similarly to a rigid body has viscosity increases (with less viscosity resulting in more time to reach the bottom of the ramp)? For example, the automatic transmission fluid was the least viscous and subsequently took the most time to reach the bottom of the ramp whereas honey is extremely viscous and arrived at the bottom faster. Nevertheless, all the data points except honey (although it was only slightly lower) were greater than 1.6 when I redid the experiment and made sure that the ramp was not lubricated at all.

 

[Chestermiller]

What do you mean by inviscid limit? Is this the time limit assuming negligible viscosity?

Yes.

So from my understanding, you found a maximum and minimum in the form of the time limits for fluids with infinite viscosity and inviscid fluids?

Yes.

That would make sense to me; however, the only fluid that is still confusing me is honey which still results in it reaching the bottom of the ramp in more than less than 1.6 seconds while the others are near to or greater than 1.6. Furthermore, shouldn't an object with infinite viscosity roll down faster (which is what I observed) as fluids behave similarly to a rigid body has viscosity increases (with less viscosity resulting in more time to reach the bottom of the ramp)?

I don't think you found this, and it isn't correct. Cans with infinite viscosity have higher moments of inertia (longer times) than cans with zero viscosity (the latter have zero moment of inertia). For intermediate viscosities, the parameter $\frac{\nu t_0}{R^2}$ determines where you are relative to the two limits, where $t_0=\sqrt{\frac{2L}{g\sin{\alpha}}}$. The larger this dimensionless group, the more it approaches the higher viscosity-high moment of inertia limit.

For example, the automatic transmission fluid was the least viscous and subsequently took the most time to reach the bottom of the ramp whereas honey is extremely viscous and arrived at the bottom faster. Nevertheless, all the data points except honey (although it was only slightly lower) were greater than 1.6 when I redid the experiment and made sure that the ramp was not lubricated at all.

Water, the least viscous, should have come out close to 1.3 seconds; sunflower oil, the next higher viscosity should have come out less than 1.6 sec., but close to it. Transmission fluid, according to my research was still higher viscosity, and come out close to 1.6 sec. Honey, the most viscous, should also have come out close to 1.6 sec.

Lubricating the ramp would have allowed the can to slip on the ramp, thus lowering the friction force, and reducing the time (rather than increasing it). But, if a residue on the ramp made it sticky, it would have provided additional resistance to resist the can from rotating, and would have resulting in longer times.

 

[mostafaelsan2005]

Yes.

Yes.

I don't think you found this, and it isn't correct. Cans with infinite viscosity have higher moments of inertia (longer times) than cans with zero viscosity (the latter have zero moment of inertia). For intermediate viscosities, the parameter $\frac{\nu t_0}{R^2}$ determines where you are relative to the two limits, where $t_0=\sqrt{\frac{2L}{g\sin{\alpha}}}$. The larger this dimensionless group, the more it approaches the higher viscosity-high moment of inertia limit.

Water, the least viscous, should have come out close to 1.3 seconds; sunflower oil, the next higher viscosity should have come out less than 1.6 sec., but close to it. Transmission fluid, according to my research was still higher viscosity, and come out close to 1.6 sec. Honey, the most viscous, should also have come out close to 1.6 sec.

Lubricating the ramp would have allowed the can to slip on the ramp, thus lowering the friction force, and reducing the time (rather than increasing it). But, if a residue on the ramp made it sticky, it would have provided additional resistance to resist the can from rotating, and would have resulting in longer times.

I see what is happening; I am confusing the idea of comparing the moment of inertia of hollow/solid cylinders with this concept which is entirely different. This makes sense and I did get experimental results that were close to what you said other than the honey and automatic transmission fluid trials which I will redo tomorrow morning. I have to admit that I am still having a hard time understanding why cans filled with [closer to] infinite viscosities have higher moment of inertias than cans filled with inviscid fluids.

 

[Chestermiller]

I see what is happening; I am confusing the idea of comparing the moment of inertia of hollow/solid cylinders with this concept which is entirely different. This makes sense and I did get experimental results that were close to what you said other than the honey and automatic transmission fluid trials which I will redo tomorrow morning. I have to admit that I am still having a hard time understanding why cans filled with [closer to] infinite viscosities have higher moment of inertias than cans filled with inviscid fluids.

For inviscid fluids, the friction force doesn't have to cause them to rotate. For highly viscous fluids, the friction force does have to cause them to rotate.

The way this plays out is as follows: $$t=\sqrt{\frac{2Lf}{g\sin{\alpha}}}$$where the viscous vector f is given by: $$f=1.0+0.5G\left(\frac{\nu t_0}{R^2}\right)$$where got function G varies monotonically with its argument from 0 at $\frac{\nu t_0}{R^2}\rightarrow 0$ to 1.0 at $\frac{\nu t_0}{R^2}\rightarrow \infty$

 

[mostafaelsan2005]

For inviscid fluids, the friction force doesn't have to cause them to rotate. For highly viscous fluids, the friction force does have to cause them to rotate.

The way this plays out is as follows: $$t=\sqrt{\frac{2Lf}{g\sin{\alpha}}}$$where the viscous vector f is given by: $$f=1.0+0.5G\left(\frac{\nu t_0}{R^2}\right)$$where got function G varies monotonically with its argument from 0 at $\frac{\nu t_0}{R^2}\rightarrow 0$ to 1.0 at $\frac{\nu t_0}{R^2}\rightarrow \infty$

Thank you for assistance with this problem, I'll keep you updated on my progress with the paper. Do you have any recommendations for what I should include in my theoretical background? My word count is limited so I wasn't sure which topics to narrow them down to (since there are two approaches to the solution, would you recommend I utilize the asymptotic solution as it was identified or start with the initial formula for the shear stress and work up to it?). I plan to split it up into the following sections where the first two sections (I will probably change the name or add a subsequent section specifically called deriving the approximation for the time it will take to reach the bottom of the ramp) will include everything until what you determined in post #34:
2.1 Viscosity boundary conditions
2.2 Boundary layer thickness
2.3 Asymptotic solution at long times

 

[Chestermiller]

Thank you for assistance with this problem, I'll keep you updated on my progress with the paper. Do you have any recommendations for what I should include in my theoretical background? My word count is limited so I wasn't sure which topics to narrow them down to (since there are two approaches to the solution, would you recommend I utilize the asymptotic solution as it was identified or start with the initial formula for the shear stress and work up to it?). I plan to split it up into the following sections where the first two sections (I will probably change the name or add a subsequent section specifically called deriving the approximation for the time it will take to reach the bottom of the ramp) will include everything until what you determined in post #34:
2.1 Viscosity boundary conditions
2.2 Boundary layer thickness
2.3 Asymptotic solution at long times

This looks good if you can really explain it adequately. Another approach would be to say that dimensional analysis tells us that the key parameter controlling the transition from short-time/low-viscosity behavior to long-time/high-viscosity behavior is $\frac{\nu t}{R^2}$, where $\nu$ is the kinematic viscosity. At small values of this parameter, inviscid behavior is approached with negligible friction force and viscosity factor f=1, while at high values of this parameter, the rolling time is the same as if the cylinder were a rigid solid with f = 3/2.

 

[mostafaelsan2005]

This looks good if you can really explain it adequately. Another approach would be to say that dimensional analysis tells us that the key parameter controlling the transition from short-time/low-viscosity behavior to long-time/high-viscosity behavior is $\frac{\nu t}{R^2}$, where $\nu$ is the kinematic viscosity. At small values of this parameter, inviscid behavior is approached with negligible friction force and viscosity factor f=1, while at high values of this parameter, the rolling time is the same as if the cylinder were a rigid solid with f = 3/2.

I see, I'll work on it then. My final question would be if the equation derived on post #34 was directly related to the asymptotic solution (If I should combine them into one section essentially)?

 

[Chestermiller]

I see, I'll work on it then. My final question would be if the equation derived on post #34 was directly related to the asymptotic solution (If I should combine them into one section essentially)?

No. The development in post #34 is strictly for small values of the key parameter.

 

[mostafaelsan2005]

No. The development in post #34 is strictly for small values of the key parameter.

So I should focus on the various parts of the asymptotic solution?

 

[Chestermiller]

So I should focus on the various parts of the asymptotic solution?

I don't understand what you are asking?

 

[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.