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

[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.

 

[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'.