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

[mostafaelsan2005]

Homework Statement

I am currently [in the process of] working on a physics report that investigates the effects of changing the fluid (and thereby the density) within a cylindrical shell on rotational motion, specifically focusing on the time it takes for an object to reach the bottom of a ramp when placed in this cylindrical shell. I would greatly appreciate your assistance and feedback to ensure that my research and report are comprehensive, coherent, and well-structured.

Relevant Equations

Equations for acceleration and velocity were derived for the particular situation with the fully solid cylinder

I am not sure if my report is complex enough as it should be at the undergraduate level preferably based on the requirements for it and it feels like it's all over the place as of now.

 

[Chestermiller]

Show us what you did so far. Is the cylinder full of fluid? what is the fluid viscosity?

 

[haruspex]

Isn't this a repost of https://www.physicsforums.com/threa...-inertia-for-hollow-cylinder-w-water.1053807/

 

[mostafaelsan2005]

Isn't this a repost of https://www.physicsforums.com/threa...-inertia-for-hollow-cylinder-w-water.1053807/

I reworked the experiment and it's the same general topic but a completely different subset because I realized that evaluating/discussing moment of inertia as a rate was not going to fit within the word count I was allotted

 

[mostafaelsan2005]

Show us what you did so far. Is the cylinder full of fluid? what is the fluid viscosity?

This is a link to the document: https://docs.google.com/document/d/...SNinjcOlh2Yh_N9Cr0nfHQ/edit#heading=h.30j0zll
All cylinders were completely filled with the liquid (e.g. sunflower oil, automated transmission fluid, water, honey) and treated as a solid though I do understand that assumption is very generalized and since I am not particularly discussing the MOI I could perhaps add more of a theoretical section concerning how the fluids would behave depending on their viscosity, density, and fluid characteristics

[Mentor Note: Document downloaded as PDF and attached to this post]

 

[Steve4Physics]

This is a link to the document: https://docs.google.com/document/d/...SNinjcOlh2Yh_N9Cr0nfHQ/edit#heading=h.30j0zll

Looked at your document briefly and spotted various issues.

The topic and approach (as they currently stand) are high-school level, not university undergraduate level. You should try to look at one or more exemplar reports to better understand the level/standard you need to aim for. Ask your supervisor.

The topic claims to be about ‘inviscid’ liquids; that means ones with negligible viscosity. But substances such as honey have large viscosities. So it doesn’t make sense.

You initially state the research question is about how the type of liquid affects the results. Later on, you change this to how the amount of liquid affects the results. You need to be absolutely clear and consistent about what you are investigating. (Note, if by ‘amount’ you mean volume, then it appears that the ‘amount’ is in fact constant for all your measurements.)

There appear to be many mistakes ranging from the minor to the serious. For example:
- you state that you used 549kg of honey (Table 4) when you probably means 549g;
- you state that the length of your ramp was 145m (table 6) when you probably mean 145cm;
- sometimes units have been completely forgotten (e.g. Table 4 column 3);
- you state “...the gravitational force component remains the same across all experiments”. But that’s wrong because the containers of different liquids have different weights.

These are just some of the items I picked up with a quick read-through. It appears that you have not carefully/critically checked your own work. You must not rely on others to do this for you.

A minor point: you swap between using the terms ‘liquid’ and ‘fluid’. ‘Fluid’ means liquid or gas; make sure you are using the appropriat term.

All cylinders were completely filled with the liquid (e.g. sunflower oil, automated transmission fluid, water, honey) and treated as a solid though I do understand that assumption is very generalized and since I am not particularly discussing the MOI. I could perhaps add more of a theoretical section concerning how the fluids would behave depending on their viscosity, density, and fluid characteristics

There are clear differences in the rolling-times for the various liquids. You need to provide an explanation for this.

 

[mostafaelsan2005]

Apologies for that. Admittedly I was reworking it from another document so there are some mistakes that carried on that I didn't catch. In the last investigation I was only working with water with the assumption that it was inviscid but I am not taking all the liquids as inviscid anymore but rather that when they are within a fully-filled container where the they are in the shape of the container they have the same center of mass that a solid cylinder would have of the same dimensions with the variable being the density of the liquid which would result in different effects to the rotational motion. I figured that this was typical high school work and not something to be done as a an undergraduate but I have been finding difficulties in how I should approach the relationship between density and moment of inertia as I do not want to factor in viscosity as it complicates the topic in such a way that it surpasses the scope of the investigation.

 

[erobz]

but I have been finding difficulties in how I should approach the relationship between density and moment of inertia as I do not want to factor in viscosity as it complicates the topic in such a way that it surpasses the scope of the investigation.

You are not going to be able to ignore viscosity across your results. You are using honey as one of the internal fluids.

 

[kuruman]

Maybe instead of ignoring viscosity, you should embrace it. This link shows the viscosities of substances most of which you can find in a grocery store. Measure the rolling time for a fixed distance and angle. See if you can deduce a power law relating the acceleration and the viscosity.

I, too, looked at your paper. One important piece of advice is to leave the introduction for last. First put down what you have done and what you make of it and then put all that into perspective in the introduction. You cannot predict where the research will take you or how relevant it is until you complete it. There is truth in the saying "If we knew what we're doing, it wouldn't be called research."

 

[Steve4Physics]

they have the same center of mass that a solid cylinder would have of the same dimensions with the variable being the density of the liquid which would result in different effects to the rotational motion.

Hmm. This might help…

Suppose you have a solid cylinder of aluminium and a solid cylinder of lead. Both cylinders are the exactly the same size; the lead (high density) cylinder is around 4 times heavier than the aluminium cylinder.

You place the cylinders side by side at the top of a ramp and release them at the same time so they roll down. Which cylinder reaches the bottom first?

Once you’ve correctly answered that, you might want to consider your own results. Why is it significant that some of your containers of liquid roll down quicker than others?
_________

On a different note, I’m not sure that rolling-time is the best metric. I’d intuitively use average acceleration. I’d also want sets of data for different values of slope (lab' time permitting).

If you can find the value of viscosity for each liquid, you could look for an empirical relationship between the accelerations and the viscosities.

 

[Chestermiller]

Apologies for that. Admittedly I was reworking it from another document so there are some mistakes that carried on that I didn't catch. In the last investigation I was only working with water with the assumption that it was inviscid but I am not taking all the liquids as inviscid anymore but rather that when they are within a fully-filled container where the they are in the shape of the container they have the same center of mass that a solid cylinder would have of the same dimensions with the variable being the density of the liquid which would result in different effects to the rotational motion. I figured that this was typical high school work and not something to be done as a an undergraduate but I have been finding difficulties in how I should approach the relationship between density and moment of inertia as I do not want to factor in viscosity as it complicates the topic in such a way that it surpasses the scope of the investigation.

The viscosity of the liquid is critical to the results of these experiments, even probably for water.

If the fluid had infinite viscosity, the can would behave as a rigid solid, while if the viscosity were low, there could still be a significant shear stress (torque) from the liquid.

I have not seen any posts in this thread or in the reference provided by the OP in any way analyzing the key feature of this system, the fluid mechanics of a viscous fluid filling a cylinder, initially at rest, and subjected to boundary rotation that is a function of time. The first step is to solve this for a suddenly applied rotation at constant rate. Solve for the fluid shear stress at the wall as a function of time. There is a short time solution that involves growth of a boundary layer adjacent to the wall. Then, there is a general analytic solution that applies at all times. For non-constant boundary velocity, one can use convolution. Are there no fluid mechanics guys out there? https://oceanofmath.blog/?p=212

Everyone looking at this problem so far has tried to do the problem all in one shot. A basic principle of good modeling practice is to subdivide a complicated problem into smaller bite-sized chunks. This is what I'm talking about.

 

[mostafaelsan2005]

Hmm. This might help…

Suppose you have a solid cylinder of aluminium and a solid cylinder of lead. Both cylinders are the exactly the same size; the lead (high density) cylinder is around 4 times heavier than the aluminium cylinder.

You place the cylinders side by side at the top of a ramp and release them at the same time so they roll down. Which cylinder reaches the bottom first?

Once you’ve correctly answered that, you might want to consider your own results. Why is it significant that some of your containers of liquid roll down quicker than others?
_________

On a different note, I’m not sure that rolling-time is the best metric. I’d intuitively use average acceleration. I’d also want sets of data for different values of slope (lab' time permitting).

If you can find the value of viscosity for each liquid, you could look for an empirical relationship between the accelerations and the viscosities.

The lead cylinder will reach the bottom first as it has a lower moment of inertia. I've also observed through the experiments that the liquids with more mass reached the bottom of the ramp faster. The liquids that I have used have known viscosities. I have not been able to find a relationship between acceleration and viscosity (at least a direct one) but I had the idea of calculating the acceleration using diagonal length of the ramp and the varying times it takes to reach the bottom (a=2s/t^2). Although I couldn't find a direct relationship I could start by plotting acceleration against viscosity in order to find a general equation based on the equation of the graph.

 

[Steve4Physics]

The lead cylinder will reach the bottom first

No. The aluminium and lead cylinders will reach the bottom together.

The acceleration of the centre of mass of a uniform solid cylinder, rolling downhill without slipping, is $\frac 23 g \sin \theta$. It is independent of the cylinder's mass (and also radius). You should prove the equation for yourself.

as it has a lower moment of inertia.

No. The lead cylinder's moment of inertia is about 4 times bigger than the aluminium cylinder's.

I've also observed through the experiments that the liquids with more mass reached the bottom of the ramp faster.

If all the liquid rotates with the same angular speed as the container, it's behaving like a solid cylinder; differences in times would not be explained by differences in mass.

[EDIT. If the mass of the can is not negligible then we can not treat the can+liquid as a uniform cylinder (because the can and liquid have different densities). Changing the density of the liquid will then effect the rolling time - as noted by @haruspex in Post #15.]

The liquids that I have used have known viscosities. I have not been able to find a relationship between acceleration and viscosity (at least a direct one) but I had the idea of calculating the acceleration using diagonal length of the ramp and the varying times it takes to reach the bottom (a=2s/t^2). Although I couldn't find a direct relationship I could start by plotting acceleration against viscosity in order to find a general equation based on the equation of the graph.

Using $a =\frac {2s}{t^2}$ is fine but note that it will give you average acceleration. (The acceleration might not be constant. Can you think of any reasons for this?)

A graph of acceleration vs. viscosity might help you to identity a pattern. And you might be able to fit an empirical equation to the data.

You have only a small amount of data. Can you collect more using other angles and/or other liquids?

Minor edits.

 

[erobz]

The lead cylinder will reach the bottom first as it has a lower moment of inertia.

See: "Theoretical Background" in the report you are writing. You seem to be getting substantially mixed up on the underlying ideas.

Can you describe what you think are some of the theoretical differences you could expect to see when the cylinder is filled with honey vs. water in terms of acceleration down the incline?

Firstly, you need to compute the moment of inertia of the cylinder that contains the liquid by itself. Then you calculate the MOI of the fluid that fills the cylinder. A point to note is that the MOI of the fluid that fills the cylinder may or may not be relevant to theoretically predicting the acceleration of the system depending on what fluid is inside it, and that would be related mainly to the fluids viscosity. Density and viscosity are different properties.

 

[haruspex]

observed through the experiments that the liquids with more mass reached the bottom of the ramp faster.

That makes sense, even if the liquid were to rotate at the same rate as the cylinder. Increasing the liquid's mass reduces the radius of rotation of the cylinder+liquid combination. Less radius of rotation means great descent speed, independently of total mass.
If we allow also that the liquid's rotation rate will be less than that of the cylinder, that increases the descent rate, and it is still the case that, the viscosities being equal, the denser fluids will descend faster.
But if we fix the density and increase viscosity then the descent will be slower.

 

[Chestermiller]

See: "Theoretical Background" in the report you are writing. You seem to be getting substantially mixed up on the underlying ideas.

Can you describe what you think are some of the theoretical differences you could expect to see when the cylinder is filled with honey vs. water in terms of acceleration down the incline?

Firstly, you need to compute the moment of inertia of the cylinder that contains the liquid by itself. Then you calculate the MOI of the fluid that fills the cylinder.

The concept of MOI of the fluid that fills the cylinder makes no sense to me. MOI assumes that the body rotates as a rigid body (the r factor in calculating MOI), and this is not the case with a fluid that is shearing radially. Please see my previous post in this thread which explains in detail exactly how to include the fluid shear in the calculation.

 

[erobz]

The concept of MOI of the fluid that fills the cylinder makes no sense to me. MOI assumes that the body rotates as a rigid body (the r factor in calculating MOI), and this is not the case with a fluid that is shearing radially. Please see my previous post in this thread which explains in detail exactly how to include the fluid shear in the calculation.

It's not my intention to confuse the OP further, sorry if you think that is happening. I agree the problem in its true rigor is clearly complex, but it seems to me we have a high school student having trouble with the basic physics, not a graduate student studying fluid mechanics. My goal was to get them to try to find "simple problem" boundary models, i.e. what we expect if the liquid is approximated as inviscid (like water), and what we expect if the liquid has high viscosity like honey, in hopes that almost everything else tested would fall somewhere in between.

 

[haruspex]

It's not my intention to confuse the OP further, sorry if you think that is happening. I agree the problem in its true rigor is clearly complex, but it seems to me we have a high school student having trouble with the basic physics, not a graduate student studying fluid mechanics. My goal was to get them to try to find "simple problem" boundary models, i.e. what we expect if the liquid is approximated as inviscid (like water), and what we expect if the liquid has high viscosity like honey, in hopes that almost everything else tested would fall somewhere in between.

The concern, I believe, is that MoI of a fluid is undefined. Judging from the above, you intended only two extreme cases to be analysed: an inviscid (and therefore non rotating) fluid, and a fluid so viscous it is effectively solid for the duration of the experiment. This would put bounds on the behaviours of each actual fluid.

 

[erobz]

Judging from the above, you intended only two extreme cases to be analysed: an inviscid (and therefore non rotating) fluid, and a fluid so viscous it is effectively solid for the duration of the experiment. This would put bounds on the behaviours of each actual fluid.

That is why I was focusing on the end cases. What's happening in the middle seems to require quite advanced theoretical explanation. If I were writing the paper at the level I believe the OP is this is approaching this from (who am I kidding- this is probably where I would stop too with my current level of understanding) I would stop trying to quantitatively explain the middle and pivot toward the empirical results of the experiment. But over the last two threads about the problem @mostafaelsan2005 has not been very convincing in conveying their level of understanding. I'm trying to lead them to consider conservation of energy at the end cases to get them some theoretical foundation (leaving the middle alone for the time being) to frame the experimental results.

 

[Chestermiller]

Question for the OP: What are typical times for the cylinder to roll down the incline, and what is the inside and outside diameter of the cylinder?

 

[mostafaelsan2005]

The concept of MOI of the fluid that fills the cylinder makes no sense to me. MOI assumes that the body rotates as a rigid body (the r factor in calculating MOI), and this is not the case with a fluid that is shearing radially. Please see my previous post in this thread which explains in detail exactly how to include the fluid shear in the calculation.

After reading all the posts here, I now understand that I cannot assume the chosen liquids as inviscid (obviously) and must come back to the viscosity for an answer for the differences in speed at the bottom of the ramp; however, I do have difficulty with understanding the viscosity equations as I have not gotten that far into fluid mechanics.

As for the research on this topic I was able to find, I found this one to be of particular interest: https://www.researchgate.net/public..._filled_cylinder_rolling_on_an_inclined_plane

In terms of the moment of inertia, I do also understand I cannot use that idea in this case as the cylinder (filled with various liquids) is not a rigid body and thus cannot be treated as such. (My level of understanding is quite limited, my apologies, I was given this topic and my supervisor for it has left on a sick leave for some months now so I have been doing it entirely on my own.) In terms of what I understood, I can set the honey as an infinitely viscous fluid and the water as inviscid in order to set some boundary conditions for the paper; however, there still comes the issue of figuring out the MOI's in-between because I had trouble with this initially when I was varying the amount of water in the cylinder and was not able to figure out an effective method to get the MOI's in between of the fully-solid and hollow cylinders. I'll be honest to say that explanations using the Bessel equations are quite a stretch in terms of me attempting to understand them so I have been searching for an alternative method to a solution to the rq. The solution that makes the most sense to me is relating the average acceleration to the viscosity. Also, in order to explain how complex it must be so it can be understood, a high scoring exemplar made their essay on how the temperature of a steel cylindrical billet, subject to electromagnetic induction, change with time.

 

[mostafaelsan2005]

Question for the OP: What are typical times for the cylinder to roll down the incline, and what is the inside and outside diameter of the cylinder?

The outside diameter was 7 cm while the inside diameter was approximately 5.5 cm

 

[Steve4Physics]

The outside diameter was 7 cm while the inside diameter was approximately 5.5 cm

That would mean the wall-thickness of the can was over 7mm. The photo' in your report shows what seems to be a food can (though it's not clear); if so, the wall thickness should be much less than 7mm.

Edit: Maybe 7mm is the thickness of the can's corrugations, not the thickness of the sheet material.

 

[mostafaelsan2005]

Oh, I see what you mean. It's a standard food can with a thickness of 0.15mm

 

[erobz]

Oh, I see what you mean. It's a standard food can with a thickness of 0.15mm

Did you conduct the experiment with it in the state shown...with paper folds and electrical tape all over the can?

 

[mostafaelsan2005]

I was in a bit of a rush that day, but no that was just an initial position photo. Afterwards I removed the paper folds and made the duct tape covering better and less haphazard (as you can see it was dripping so i also cleaned the ramp so that there wouldn't be any effects on the friction) before I conducted the experiment to measure the time it took to reach the bottom of the ramp

 

[Chestermiller]

The outside diameter was 7 cm while the inside diameter was approximately 5.5 cm

And typical range of times to roll down the ramp are.....?

 

[mostafaelsan2005]

And typical range of times to roll down the ramp are.....?

I'm not sure I understand but I repeated the experiment three times for each liquid

 

[Chestermiller]

I'm not sure I understand but I repeated the experiment three times for each liquid

And what values of time did you get for reaching the bottom of the ramp?

 

[mostafaelsan2005]

They ranged from 1.25 seconds to 2 seconds

 

[Chestermiller]

They ranged from 1.25 seconds to 2 seconds

This tells me that, in the experiments on most of the fluids (certainly water) , the fluid angular velocity inside the can varies from zero only in a thin momentum boundary layer near the can surface. I'll provide a detailed analysis of this when I have a chance.

 

[Chestermiller]

In the book Transport Phenomena by Bird, Stewart, and Lightfoot, they solve the problem of viscous Flow Near a Wall Suddenly Set in Motion: A semi-infinite body of liquid with constant density and viscosity is bounded below by a horizontal surface. Initially, the fluid and solid surface are at rest. Then, at time t = 0, the solid surface is set in motion in the horizontal direction with velocity V.

Their analysis shows that, at time t, the shear stress exerted by the fluid on the surface is given by $$\sigma=\mu\frac{V}{\delta(t)}$$where $\mu$ is the fluid viscosity and $\delta$ is the effective "boundary layer thickness:" $$\delta = \sqrt{\pi \nu t}$$ where the kinematic viscosity of the fluid $\nu$ is given by $$\nu=\frac{\mu}{\rho}$$with $\rho$ representing the density of the fluid. At distance from the wall beyond the boundary layer thickness $\delta(t)$, the fluid has essentially not started moving yet at time t.

Can anyone think of how these results can be used to calculate the torque exerted by the fluid on the inner surface of our can in our problem of a can filled with fluid rolling down a ramp? (At first glance, it would appear that these two problems could not be related to one another in any way).

 

[erobz]

I don't have a formal fluid mechanics approach, but my idea (assuming rotational symmetry) would be to skirt the issue of forces and integrate the kinetic energy of concentric cylindrical shells of thickness $dr$ over $r$ as:

$$ d KE = dKE_{trans}+ dKE_{rot} $$

$$ d KE = \frac{1}{2}v^2_{c.o.m} dm + \frac{1}{2} \omega(r)^2 dI $$

The trick would still be to find the velocity distribution $v(r)$within the fluid though. I would want to propose integrating the shear stress $\tau (r) v(r) dA = \mu \frac{ dv}{dr} v dA$ to get the rate of waste heat generation withing the fluid.

But I'm certainly grasping at straws; if it has very little merit, I wouldn't be surprised.

 

[Chestermiller]

Let's compare the magnitude of the velocity boundary layer thickness $\delta(t)=\sqrt{pi \nu t}$ with the radius of the can 2.75 cm. For water, $\nu=0.01\ cm^2/sec$; and the amount of time for the can to roll down the ramp is nominally 2 seconds. So for water, the final surface boundary layer thickness would be $$\delta=\sqrt{\pi (0.01)(2)}=0.25\ cm$$This is only 9% of the radius of the can. So in the case of water, neglecting the curvature of the can surface would certainly be a good approximation. Even in the case of higher viscosity fluids where the boundary layer is thicker, in my judgment, using the approximation in post #32 would still be a reasonable approximation since the largest velocity gradients are going to be near the surface. So for a can with suddenly imposed constant rotational velocity V imposed on the can inner surface, the torque imposed by the fluid on the inside surface of the can would be given by: $$\tau=2\pi R^2L\sigma=2\pi R^2L\frac{\mu V}{\sqrt{\pi\nu t }}=2\rho R^2L\sqrt{\pi \nu}\frac{V}{\sqrt{t}}\tag{1}$$where R is the inside radius of the can.

Questions so far?

 

[mostafaelsan2005]

Let's compare the magnitude of the velocity boundary layer thickness $\delta(t)=\sqrt{pi \nu t}$ with the radius of the can 2.75 cm. For water, $\nu=0.01\ cm^2/sec$; and the amount of time for the can to roll down the ramp is nominally 2 seconds. So for water, the final surface boundary layer thickness would be $$\delta=\sqrt{\pi (0.01)(2)}=0.25\ cm$$This is only 9% of the radius of the can. So in the case of water, neglecting the curvature of the can surface would certainly be a good approximation. Even in the case of higher viscosity fluids where the boundary layer is thicker, in my judgment, using the approximation in post #32 would still be a reasonable approximation since the largest velocity gradients are going to be near the surface. So for a can with suddenly imposed constant rotational velocity V imposed on the can inner surface, the torque imposed by the fluid on the inside surface of the can would be given by: $$\tau=2\pi R^2L\sigma=2\pi R^2L\frac{\mu V}{\sqrt{\pi\nu t }}=2\rho R^2L\sqrt{\pi \nu}\frac{V}{\sqrt{t}}$$where R is the inside radius of the can.

Questions so far?

So far I understood the question posed to solve the problem of viscous flow near a wall suddenly set in motion and I understand the working out generally but conceptually I do not understand the idea of 'boundary wall thickness'. After some research on it, I understand that it is 'the distance from this surface to the point where the velocity is 99% free-stream' and it depends on structural geometry which in this case is a cylinder and that increasing the boundary layer thickness increases the drag force experienced by the liquid in the can as an object moves. Is this a correct understanding up to now? Also, apologies for the late responses have been busy with interviews this week.

 

[Chestermiller]

So far I understood the question posed to solve the problem of viscous flow near a wall suddenly set in motion and I understand the working out generally but conceptually I do not understand the idea of 'boundary wall thickness'. After some research on it, I understand that it is 'the distance from this surface to the point where the velocity is 99% free-stream' and it depends on structural geometry which in this case is a cylinder and that increasing the boundary layer thickness increases the drag force experienced by the liquid in the can as an object moves. Is this a correct understanding up to now? Also, apologies for the late responses have been busy with interviews this week.

You're somewhat correct, but, in this case, the boundary layer thickness is the penetration depth of the velocity profile from a value of V at the rotating inner surface of the can to a value of zero in the bulk of the fluid away from the surface. Increasing the boundary layer thickness is the same as decreasing the velocity gradient in the fluid near the surface, which results in reduced drag on the surface as time progresses.

In Eqn. 1 of post #34, the velocity V is the tangential velocity of the inner surface of the can as reckoned by an observer who is traveling down the incline at the velocity of the center of mass of the can. As written, the equation assumes that the can rotation starts from rest and then stays constant for all subsequent time. But, in our system, the rotation rate of the can is not constant, but is changing as time progresses. The next issue to be addressed is how this equation can be modified to take this varying velocity history into consideration. Any ideas?

 

[Chestermiller]

Because the torque imposed by the shearing fluid on the inside surface of the can is a linear function of the tangential inside surface velocity of the can, the form of Eqn. 1 in post #34 for the case where the rotation rate of the can is variable in time can be obtained by linear superposition in terms of a so-called "convolution integral:" $$\tau(t)=2\rho R^2L\sqrt{\pi \nu}. \int_0^t{\frac{V'(\xi)}{\sqrt{t-\xi}}d\xi}$$where $V'=\frac{dV}{d\xi}$ and $\xi$ is a dummy (time) variable of integration. Note that we also have that $$V'(xi)=R\alpha(\xi)$$where $\alpha$ is the angular acceleration of the can. So, combining these equations, we get: $$\tau(t)=2\rho R^3L\sqrt{\pi \nu}. \int_0^t{\frac{\alpha(\xi)}{\sqrt{t-\xi}}d\xi}$$
MOMENT BALANCE ON CAN:$$FR_0-\tau=I_c\alpha$$where $R_0$ is the outside radius of the can, F is the frictional force, $I_c$ is the moment inertia of the can: $$I_c=M_c\frac{R_0^2+R^2}{2}$$with $M_c$ representing the mass of the can. Substituting and solving for the friction force, we obtain: $$F=2\rho \frac{R^3}{R_0}L\sqrt{\pi \nu}. \int_0^t{\frac{\alpha(\xi)}{\sqrt{t-\xi}}d\xi}+M_c\frac{R_0^2+R^2}{2R_0}\alpha$$The angular acceleration $\alpha$ is kinematically related to the acceleration of the center of mass of the can "a" by: $$\alpha=\frac{a}{R_0}$$Substituting this into the previous equation yields: $$F=2\rho \frac{R^3}{R_0^2}L\sqrt{\pi \nu}. \int_0^t{\frac{a(\xi)}{\sqrt{t-\xi}}d\xi}+M_c\frac{1+(R/R_0)^2}{2}a(t)$$$$=M_L\kappa^2\left(\frac{2\sqrt{\pi \nu}.}{\pi R}\right)\int_0^t{\frac{a(\xi)}{\sqrt{t-\xi}}d\xi}+M_c\frac{(1+\kappa^2)}{2}a(t)$$where $M_L$ is the mass of liquid in the can and $\kappa=\left(\frac{R}{R_0}\right)$.

 

[Chestermiller]

Your report is lacking important information.

Mass of can
I assume the wall thickness is 0.75 cm
The radius you gave in your report was for the outside radius
Thickness of lids

Results for rolling an empty (air filled) can (this will give you the results for a fluid of zero viscosity and density)

Physical properties of liquids
water density = 1.0 gm/cc, viscosity = 0.01 Poise
sunflower oil density = 0.92 gm/cc, viscosity = 0.49 Poise
honey density = 1.45 gm/cc, viscosity = 4-230 Poise
transmission fluid = 0.87 gm/cc, viscosity = 0.7 - 1.6 Poise

You should be plotting the dimensionless time $t\sqrt{\frac{g\sin{\theta}}{2L}}$ vs fluid density and viscosity. Why are there no graphs in your report. You should show predicted dimensionless temperature for infinite viscosity and zero viscosity to illustrate that you data falls within these bounds.

If the volumes of all the fluids is the same, why aren't their masses proportional to their densities?

From the masses of liquid you report, it seems like the inside diameter is 7 cm, rather than 5.5 cm. What gives?

 

[mostafaelsan2005]

You're somewhat correct, but, in this case, the boundary layer thickness is the penetration depth of the velocity profile from a value of V at the rotating inner surface of the can to a value of zero in the bulk of the fluid away from the surface. Increasing the boundary layer thickness is the same as decreasing the velocity gradient in the fluid near the surface, which results in reduced drag on the surface as time progresses.

In Eqn. 1 of post #34, the velocity V is the tangential velocity of the inner surface of the can as reckoned by an observer who is traveling down the incline at the velocity of the center of mass of the can. As written, the equation assumes that the can rotation starts from rest and then stays constant for all subsequent time. But, in our system, the rotation rate of the can is not constant, but is changing as time progresses. The next issue to be addressed is how this equation can be modified to take this varying velocity history into consideration. Any ideas?

Well from what I researched and, correct me if I am wrong, the most accurate way to calculate the torque exerted by the fluid on the inner surface of the can is to use the Navier-Stokes equations and I know for water since we can assume that it is incompressible and Newtonian, as well as having the characteristic of a laminar flow then we can simplify it to the Poiseuille equation (which can be solved analytically as far as I know) to obtain an expression for the torque exerted by the fluid on the inner surface of the can. Also in terms of what you have asked, so you mean that the equation that we used for the boundary layer thickness is only for an instant tangential velocity which in this case cannot be applied because the rotation is not constant... how so? Isn't the rotation constant as I am keeping it at a fixed angle so the rotational and translational forces would act in a constant way? Anyhow, is it possible if we integrate the torque equation over time to determine the total torque exerted by the fluid on the inner surface of the can. We would have to assume that the rotation rate is changing linearly which I believe it does and then use the following equation:
τ = 2πηR^3[(omega_f - omega_0)/ln(R/r)]

where omega_0 and omega_f are the initial and final rotation rates, respectively, and r is the radial distance from the center of the can.

Therefore, if we integrate that equation over the time interval of however many seconds the can takes to reach the bottom of the ramp, then I guess we would be able to find the total torque exerted by the fluid on the inner surface of the can. Does this make sense or are my assumptions far-fetched?

 

[mostafaelsan2005]

Your report is lacking important information.

Mass of can
I assume the wall thickness is 0.75 cm
The radius you gave in your report was for the outside radius
Thickness of lids

Results for rolling an empty (air filled) can (this will give you the results for a fluid of zero viscosity and density)

Physical properties of liquids
water density = 1.0 gm/cc, viscosity = 0.01 Poise
sunflower oil density = 0.92 gm/cc, viscosity = 0.49 Poise
honey density = 1.45 gm/cc, viscosity = 4-230 Poise
transmission fluid = 0.87 gm/cc, viscosity = 0.7 - 1.6 Poise

You should be plotting the dimensionless time $t\sqrt{\frac{g\sin{\theta}}{2L}}$ vs fluid density and viscosity. Why are there no graphs in your report. You should show predicted dimensionless temperature for infinite viscosity and zero viscosity to illustrate that you data falls within these bounds.

If the volumes of all the fluids is the same, why aren't their masses proportional to their densities?

From the masses of liquid you report, it seems like the inside diameter is 7 cm, rather than 5.5 cm. What gives?

My report is in a work-in-progress (I have until October 2nd to finalize it), I still am aiming to do so and change the theoretical framework which is why I haven't added any graphs. Also I measured the outside diameter to be 7 cm from the bottom of the can which was flat, and quite honestly, did not measure the inner radius just yet but I can do that tomorrow or the day after. Also the volumes of the fluid should be equal but there is a chance that the scale was not precise since it was quite old but I can remeasure the masses after tomorrow as well just to make sure, as well as rolling the hollow cylinder. Apologies for the missing information and I appreciate that you are helping me still by the way

 

[mostafaelsan2005]

Because the torque imposed by the shearing fluid on the inside surface of the can is a linear function of the tangential inside surface velocity of the can, the form of Eqn. 1 in post #34 for the case where the rotation rate of the can is variable in time can be obtained by linear superposition in terms of a so-called "convolution integral:" $$\tau(t)=2\rho R^2L\sqrt{\pi \nu}. \int_0^t{\frac{V'(\xi)}{\sqrt{t-\xi}}d\xi}$$where $V'=\frac{dV}{d\xi}$ and $\xi$ is a dummy (time) variable of integration. Note that we also have that $$V'(xi)=R\alpha(\xi)$$where $\alpha$ is the angular acceleration of the can. So, combining these equations, we get: $$\tau(t)=2\rho R^3L\sqrt{\pi \nu}. \int_0^t{\frac{\alpha(\xi)}{\sqrt{t-\xi}}d\xi}$$
MOMENT BALANCE ON CAN:$$FR_0-\tau=I_c\alpha$$where $R_0$ is the outside radius of the can, F is the frictional force, $I_c$ is the moment inertia of the can: $$I_c=M_c\frac{R_0^2+R^2}{2}$$with $M_c$ representing the mass of the can. Substituting and solving for the friction force, we obtain: $$F=2\rho \frac{R^3}{R_0}L\sqrt{\pi \nu}. \int_0^t{\frac{\alpha(\xi)}{\sqrt{t-\xi}}d\xi}+M_c\frac{R_0^2+R^2}{2R_0}\alpha$$The angular acceleration $\alpha$ is kinematically related to the acceleration of the center of mass of the can "a" by: $$\alpha=\frac{a}{R_0}$$Substituting this into the previous equation yields: $$F=2\rho \frac{R^3}{R_0^2}L\sqrt{\pi \nu}. \int_0^t{\frac{a(\xi)}{\sqrt{t-\xi}}d\xi}+M_c\frac{1+(R/R_0)^2}{2}a(t)$$$$=M_L\kappa^2\left(\frac{2\sqrt{\pi \nu}.}{\pi R}\right)\int_0^t{\frac{a(\xi)}{\sqrt{t-\xi}}d\xi}+M_c\frac{(1+\kappa^2)}{2}a(t)$$where $M_L$ is the mass of liquid in the can and $\kappa=\left(\frac{R}{R_0}\right)$.

I understand the process but the idea of this type of integration confuses me (as in why do we use it?), was I close with what I proposed or did something similar because I think I had the same idea but did not know about convoluted integrals

 

[Chestermiller]

Well from what I researched and, correct me if I am wrong, the most accurate way to calculate the torque exerted by the fluid on the inner surface of the can is to use the Navier-Stokes equations and I know for water since we can assume that it is incompressible and Newtonian, as well as having the characteristic of a laminar flow

Correct

then we can simplify it to the Poiseuille equation (which can be solved analytically as far as I know) to obtain an expression for the torque exerted by the fluid on the inner surface of the can.

Incorrect. The solution I presented is the solution to the transient Navier Stokes equations (since the flow is changing with time), and the solution for the velocity is a function of time, unlike the Poiseulle equation, which is a steady flow solution (independent of time.)

Also in terms of what you have asked, so you mean that the equation that we used for the boundary layer thickness is only for an instant tangential velocity which in this case cannot be applied because the rotation is not constant... how so?

Well, first I solved for the time-dependent solution for the flow in a suddenly started can rotation, in which the can rotation is held constant; however, the velocity of the fluid in this solution is not constant except at the can surface. Away from the can surface it is changing with time. Next, I took this solution and, based on my fluid mechanics training, I knew how to modify it for the case of a can rotation which is changing with time. That is where the convolution integral came into play.

Isn't the rotation constant as I am keeping it at a fixed angle so the rotational and translational forces would act in a constant way?

No. The can rotation propagates into the fluid in the can analogous to a wave propagation. At short times, only the fluid near the can surface is moving. Even though the can surface is rotating, the fluid in the interior doesn't even know that anything is happening at the surface yet, and it is still at rest rotationally.

Anyhow, is it possible if we integrate the torque equation over time to determine the total torque exerted by the fluid on the inner surface of the can. We would have to assume that the rotation rate is changing linearly which I believe it does and then use the following equation:

The rotation rate does not change nearly with time, nor do the velocities and radial locations closer to the center of the can.

τ = 2πηR^3[(omega_f - omega_0)/ln(R/r)]

This equation is not correct.

where omega_0 and omega_f are the initial and final rotation rates, respectively, and r is the radial distance from the center of the can.

Therefore, if we integrate that equation over the time interval of however many seconds the can takes to reach the bottom of the ramp, then I guess we would be able to find the total torque exerted by the fluid on the inner surface of the can. Does this make sense or are my assumptions far-fetched?

It all makes sense only if we use the equation that I (correctly) derived.

 

[Chestermiller]

I understand the process but the idea of this type of integration confuses me (as in why do we use it?), was I close with what I proposed or did something similar because I think I had the same idea but did not know about convoluted integrals

Your concepts are partially correct, but I'm confident that the equation I derived is correct, at least if the product of kinematic viscosity and time is small (as in the case of water, for example). For larger values of this product, we would have to use the results in the Bessel functions reference I provided, which you were having to avoid.

Later, I can show you how this integral equation I presented can be used to solve for the distance traveled down the ramp as a function of time (which would then give the amount of time to reach the bottom of the ramp).

 

[Chestermiller]

My report is in a work-in-progress (I have until October 2nd to finalize it), I still am aiming to do so and change the theoretical framework which is why I haven't added any graphs. Also I measured the outside diameter to be 7 cm from the bottom of the can which was flat, and quite honestly, did not measure the inner radius just yet but I can do that tomorrow or the day after. Also the volumes of the fluid should be equal but there is a chance that the scale was not precise since it was quite old but I can remeasure the masses after tomorrow as well just to make sure, as well as rolling the hollow cylinder. Apologies for the missing information and I appreciate that you are helping me still by the way

Do you have any idea how much the can weights compared to the weight of the liquid? I'm thinking that the can weighs about 25-50 grams.

 

[Chestermiller]

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.

 

[mostafaelsan2005]

Do you have any idea how much the can weights compared to the weight of the liquid? I'm thinking that the can weighs about 25-50 grams.

I used this can: https://www.carrefouruae.com/mafuae...s/american-g-foul-medammes-eoe-400g/p/1153843 and I did not measure its weight but I expect it to be around the same as well based on how it felt when it was empty

 

[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: $$v=g\sin{\alpha}\left(1-\frac{4}{3\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}\right)t$$Integrating again to get the distance then give us $$L\approx g\sin{\alpha}\left(1-\frac{16}{15\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}\right)t^2/2$$or, to the same level of approximation, $$L\left(1+\frac{16}{15\sqrt{\pi}}\frac{\sqrt{ \nu t}}{ R}\right)\approx (g\sin{\alpha})t^2/2$$

What is the reason for solving for distance and exactly is it the distance of? Does it have to do with the boundary layer thickness or is this purely for the frictional forces involved?

 

[Chestermiller]

What is the reason for solving for distance and exactly is it the distance of? Does it have to do with the boundary layer thickness or is this purely for the frictional forces involved?

In this equation, L is the length of the ramp (145 cm) and t is the time required for the can to roll down the length of the ramp. See the solution to the equation for t that I have added to post # 46.

 

[Chestermiller]

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.

 

[Tom.G]

Please run a test where you have something totally rigid in the can in place of the fluid, like concrete or jello.

Or easier to find, damp Earth or damp sand, packed tightly.