Period of oscillation of a ring on a cylinder in gravity field 3

[strokersix]

I'm having trouble calculating the oscillation period of a ring on a shaft. If I use the compound pendulum calculation it gives reasonable values if the shaft is very small compared to the ring inner diameter. Increasing the shaft size causes a shorter oscillation period experimentally (observing parts here at my desk) but my simple mathematical model doesn't correlate. I think the missing piece is related to slippage between the shaft and inner diameter of the ring. My math model assumes slippage as the ring moves about it's center of oscillation but the real parts do not slip. It appears that when the shaft and inner diameter of the ring are very close the period approaches that of a simple pendulum. Any help in calculating the period of this pendulum will be much appreciated.

Thank you to all in advance for your time and advice.

 

[EnglishMuffin]

Although anything is possible, I doubt if it's due to slippage - more likely due to the fact that the locus of the ring CG is not a circular arc as you probably assumed, but is an epicycloid (at least I think it is - off the top of my head. But what on earth is this ? I hope it's not one of those homework questions.

 

[EnglishMuffin]

Sorry - I'm wrong of course - forgot to engage brain - the CG does move in an arc but the rate of rotation about the CG is different than it would be for a normal pendulum because of the rolling action.

 

[EnglishMuffin]

To elaborate a little further, if the shaft diameter is d and the ring diameter is D, then the CG of the ring moves through an arc of radius (D-d)/2, centered on the shaft. If you calculated the period based on a normal compound pendulum, you would be solving the equation:

I*omega^2/2 - m*g*(D-d)/2*cos(theta)= constant

where I is the moment of inertia of the ring about an axis at the center of the shaft, m is the mass of the ring, and theta is the angle of swing of the "pendulum". For a normal compound pendulum, omega is d(theta)/dt.

However, for this pendulum, omega is (1-d/D)*d(theta)/dt, so the usual solution must be modified slightly unless d is very small compared with D.

Hope my math is right.

 

[EnglishMuffin]

And to elaborate still further, I think the new equation that you have to solve is actually:

m*((D-d)/2)^2*(d(theta)/dt)^2/2+I*(d(theta)/dt)^2*(1-d/D)^2/2 - m*g*(D-d)/2*cos(theta)= constant

where I is now the moment of inertia about the center of the ring.

The first term is the instantaneous translational KE of the ring as it pivots about the center of the shaft. The second term is the rotational KE of the ring about its mass center - which is less than it would be for a normal compound pendulum because of the rolling action - and a constant minus the third term is the gravitational potential energy.

Something like that anyway - I'm in too much of a hurry really. I expect someone else will correct me if I'm wrong

 

[strokersix]

Thank you Englishmuffin.

I'll study this for a bit and see if I can wrap my brain around it.

This is not a homework problem. I'm trying to construct an absorber to combat a torsional oscillation problem using a concept similar to pendulous absorbers in engine crankshafts. One difference here is the potential energy is stored in the height of the ring (against gravity) instead of the radius relative the the crank axis (against centrifugal force).

Any other thoughts? Thank you!

 

[GregLocock]

My first thought is that if this is a rotating sytem then speeds would have to be very low otherwise the ring will hula hoop.



Cheers

Greg Locock

 

[EnglishMuffin]

Well, if the whole thing rotates continuously, rather than just the ring oscillating, then I expect my whole analysis will be incorrect. And if it's going to operate as any kind of shaft "damper" (more correctly absorber?), then presumably the shaft must at least oscillate, even if it does not continuously rotate, which again renders my analysis invalid. However, I rather get the impression that strokersix is merely trying to correlate some physical measurements for the non-rotating case with theory, as a first step. (A journey of a thousand miles ..etc). Unless of course I've misunderstood, which wouldn't be the first time!

 

[strokersix]

The shaft speed is typically 30-40 rpm. Ring inner diameter will be around 3 inches. I expect the ring to remain rolling in a suspended position. There are velocity oscillations which cause undesireable function of the machine. These oscillations are very small and can just barely be felt when touching the shaft but are enough to disrupt the machine function. Attempts to stiffen the upstream drive have improved the situation but have not eliminated it. I believe the root cause of the angular velocity oscillation to be inconsistent friction (stick-slip only a continuous rather than step function) combined with a torsionally flexible upstream drive. I don't know if this concept is going to work or not but it's important enough that I must try. What pendulum period should I target relative to the machine oscillation period? What mass pendulum relative to machine mass? Any other thoughts? Please share!

 

[EnglishMuffin]

Well, before you get into designing something like this from scratch, have you ruled out using a Houdaille damper, for example ? You can buy these ready made - they contain silicone oil, and I have employed them on drive line applications with good results. Another possibility would be a tuned mass torsion damper as used on crankshafts.

 

[strokersix]

I have tried a friction damper but it was not sensitive enough.

Simply increasing the machine mass only increased the oscillation period.

What I like so much about the pendulous ring is it's low cost and easy retrofit on the existing machinery. I have some geometry constraints that the ring concept fits well within.

Tuned mass torsional absorber using spring elements is also on the agenda. I think I'll revisit the friction damper as well if I can get it to be more sensitive. A viscous fluid damper sounds very attractive. I think I'll give it a try also but at this point I don't see how it could be done cheaply enough. This will have to be done in the 1-2 dollar range to be economically feasible. However, experimentally trying other concepts is certainly valid to learn what does and does not work.

 

[EnglishMuffin]

Well, the purchased Houdailles are going to be more than $2. But all they consist of is an annular container with a free floating mass inside (on a teflon bushing) - the cavity is then filled with silicone oil. I don't see why it should necessarily cost much more than what you are attempting (just a minimum of three simple round parts have to be made), but probably it would cost more than $2.

 

[GregLocock]

I'd go with the ready made thingo (partly because it doesn't need much tuning), but your idea is pretty enough in its own right. For maximum effect you should match the frequency of the absorber exactly to the problem frequency. I'm assuming this is a steady state problem?

An important consideration is how exactly you think the energy is going to be absorbed. I suspect it is air damping and a bit of friction at the interface - not very much power with the operating speed and size you give. Oil is a much better damping fluid than air.

Cheers

Greg Locock

 

[ivymike]

sounds a bit like a rotating version of a frahm absorber to me (the springy thingies that cancel a single frequency in the ideal world). as such, excess energy from the shaft would be transferred to KE+PE of the ring, and then back to the shaft.

 

[ivymike]

might work better if slip was prevented - by adding splines, perhaps.

another possibility - use a collar-mounted assembly of springs and weights arranged circumferentially about the shaft (think of a crankshaft TV damper with springs instead of rubber - more flexible, in other words)

 

[strokersix]

Thank you to all!

T=2pi*sqrt[((R-r)/g)(1+(R2^2+R^2)/2R)]

r = hub or shaft diameter
R = inner ring diamter
R2 = outer ring diamter

I derived this based on a textbook example of a disk rolling inside a circular surface. The result is off by about 20 percent but the trend is correct. There is either an error in my derivation or in my desktop measurement. If I find the error I'll post. Also when I try the concept in reality I'll post the results.

Thank you again to all who took the time to help.