Force generated by rotating mass
Force generated by rotating mass
(OP)
Hello
Could someone be kind enough to confirm or correct my calculation of the force generated by a mass of 80 pounds rotating at a radius of 7.5 inches at 2200 rpm. My calculation is 81,666 pounds of force... am I close?
Thank you for your time
Could someone be kind enough to confirm or correct my calculation of the force generated by a mass of 80 pounds rotating at a radius of 7.5 inches at 2200 rpm. My calculation is 81,666 pounds of force... am I close?
Thank you for your time





RE: Force generated by rotating mass
That is a weight of 80lb!
I am getting 82,500lb. Don't do it while I am in the area, okay?
RE: Force generated by rotating mass
F = m*r*ω2
http://fil
TTFN
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RE: Force generated by rotating mass
That sounds about right. I rounded mine off.
RE: Force generated by rotating mass
This is from a machine used to clear brush and small trees, the environmentally correct way to clear seismic survey trails, among other uses. A 8 foot drum attached to the front of an all terrain vehicle. There are various designs but this one has 6 rows equally spaced with 8 carbide tipped teeth on each row. Each tooth is about 10 pounds.
The issue is that the customer wants to balance the rotors without the teeth, his argument is that each row has 8 teeth and is directly opposed by another row.
My argument is that because the teeth are offset there will be considerably dynamic unbalance generated once the teeth are installed.
I wanted to try and quantify just how much of a force is generated at each bearing. I have not yet completed my calculations, perhaps... it will be low enough to be acceptable and his request will work... but I doubt it
Ralph
RE: Force generated by rotating mass
So, if you assume a worst case of a 2lb imbalance per row, AND a worst case of all 6 rows having the same imbalance in the same orientation, you'd have something like 12,372 lbf. Still, a rather large number, but clearly, the most likely imbalance contributor is going to be the teeth, particularly since they're located at the maximum radius.
TTFN
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RE: Force generated by rotating mass
Ted
RE: Force generated by rotating mass
This rotor has 6 rows with 6 evenly spaced teeth. there is a combined force due to adding the teeth of 15312 pounds acting at the center of each row of teeth. All references are from the right hand side bearing in inches.
The center of row 1 (where the cumulative force of the teeth react) is 36.375. The center of the row opposite is at 38.25 (1.875 inch offset).
Row 2 is at 34.5, and opposite at 40 (5.5 inch offset)
Row 3 is at 32.625 and opposite at 41.75 (9.125 offset)
From here I calculate what the force at the bearing is, eg. row 3 15312*9.125/32.625=4282. If I do this for the other 2 sets and combine the values (@ 120 degree intervals 789; 2441; 4282) I get 3,026 pounds.
This seems high, to the point I feel I am overlooking some basic math.
One of the issues is that the center of the force of each set is not directly in the center of the rotor. Set 3 for example is offset by 0.437 inches, how to include this is giving me a headache.
The rotor is 73.5 inches on bearing centers. Picture of rotor included (I hope)
Any advice / thoughts / lessons in math appreciated
Ralph
RE: Force generated by rotating mass
The whole thing weighs, what, several hundred pounds? If the shaft were light enough to be affected by the moment created by the tooth offset, it would be too light to maintain its rotational momentum when the teeth actually bite into something. The tearing force generated by this structure has to be high, and the rotational inertia has to be even higher, for this system to work correctly in the first place. Therefore, it cannot be affected lateral displacement of teeth.
TTFN
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RE: Force generated by rotating mass
Worry more about the design of tooth attachment and getting good welds so that teeth don't fly off the rotor.
Ted
RE: Force generated by rotating mass
This rotor sans teeth weighs 900 ish pounds and rotates at 2200 rpm. We are guessing that each tooth weighs 2.5 pounds with a center of gravity at about 7.5 inches.
Unfortunately, at the moment I have no feedback on how this is going to work out. Ideally I would hope the customer mounts the rotor and runs it with no teeth, makes a judgment on the level of vibration. Then install the teeth and makes a new assessment.
We will see, more work for us if it comes back. These are a bugger to balance, as one welds on weights the rotor distorts, creating runout on the bearing stubs. This then offsets the mass and requires yet more weight. To keep it "straight" I weld short passes on the opposite side. All very time consuming it takes at least an 8 hour day to do one
RE: Force generated by rotating mass
I have worked on some reciprocating devices, where I thought the back and forwards forces were rather high. Vibration turned out to not be a problem, because the device was rigidly coupled to a relatively heavy structure. The forces, however high, were exerted for a short period of time, on a substantial mass, and the resulting acceleration was not that high. The results were not as noticeable as I had anticipated.
The forces I was looking at were nowhere near your ballpark, but the principal works. Your 80klb imbalance would be manageable if the system were attached rigidly to a D9 bulldozer or an Abrams tank. Do you have a heavy vehicle to attach this thing to?
If you balance thing perfectly, you still are going to have your cutting forces to contend with.
RE: Force generated by rotating mass
RE: Force generated by rotating mass
your calculations for bearing reations to each resolved couple are incorrect. You should divide each resolved couple by the bearing to bearing distance, 73.500 inches. The couples act anywhere along the length of the shaft, not at any particular point of application. The bearing force reacting to each couple is acting through the length of the shaft, sum moments around each bearing. Reaction force at a bearing multiplied by the bearing separation equals the couple due to imbalance in the plane of the couple. Then the couple divided by the length of bearing separation equals the resultant bearing reaction in the plane of the couple.
The bearing forces are much lower than what you calculated.
Do a search on two plane balancing for a methodology on balancing multiple imbalances on a stiff rotor.
Ted
RE: Force generated by rotating mass
So in my worst case set I have a (15312*9.125/32.625)=4282 pounds at the bearing. You are now saying this number should be divided by the distance between bearings? = 58.25 pounds. This is a much more realistic value.
I appreciate your analysis / help
Thank you
RE: Force generated by rotating mass
Thanks all
RE: Force generated by rotating mass
No, I'm saying 15312*9.125/32.625 = 4282lbs. should be 15312*9.125/73.5 = 1900lbs.
15132*9.125 in-lb is the resolved couple in the row 3-6 plane and 73.5 in. is the bearing spread. 1900lbs is the bearing reaction in that plane. Resolve this reaction with the other two reactions to the other two couples to find the resultant bearing reaction.
Ted
RE: Force generated by rotating mass
I now have bearing reactions in rows 1/4 =390.62; rows 2/5 =1145.83; rows 3/6 =1901.047. Adding these up give a total of 1308.06 pounds of force at each bearing.. ouch!
RE: Force generated by rotating mass
What will the cutting load be on the bearings?
Vibratory shakers apply 22,000 lbs to bearings.
Ted
RE: Force generated by rotating mass
RE: Force generated by rotating mass
The original question / discussion arose.. Because the customer wanted it balanced SANS teeth and... while in the process of balancing this rotor I wondered what the actual value (force) of the unbalance would be at the bearings.
RE: Force generated by rotating mass
Ted
RE: Force generated by rotating mass
Around one end I will have F(15321.54)@ D(73.5-32.625)/73.5=8515.648 Around the other end I will have 15312.54*41.75/73.5=8697.941
I do this for all 3 sets and add the results at each end giving me a net value of 675.59 and 631.48. A realistic value and is roughly equivalent to an ISO 1940 value of 60.
Have I gone wrong?? my calculations (spreadsheet) are attached
RE: Force generated by rotating mass
For your row set 3/6 example you did not take into account the moment caused by row 3 about the rh bearing, 15312*32.625/73.5 = 6797, in order to determine the lh bearing reaction 8698 - 6797 = 1901. Calculate 15312*(41.75-32.625)/73.5 for the bearing reaction of 1901 lbs. There are two moments around each bearing. One caused by each row of teeth. Each row generates a moment countered unequally by its opposing row resulting in a twisting couple which is the product of the force 15312 and the row center offset from each other. Divide this couple by the bearing spacing 73.5 to get the bearing reaction to that row's imbalance couple.
Combine each bearing reaction to arrive at the resultant reaction. Keep in mind the direction of all row set reactions when you calculate the resulting bearing reaction.
Ted
RE: Force generated by rotating mass
G60 would be on the order of 0.01 inch eccentricity, or 0.02 inch peak-to-peak or TIR.
See page 3 or 4 here -
http://w
RE: Force generated by rotating mass
RE: Force generated by rotating mass
Set 1/4, the right side bearing would see a force of (37.125*15312.54=35.25*15312.54+73.5*X) =390.626 pounds. The left side bearing would see the same force (36.375*15312.54=38.25*15312.54+73.5*X) but as a negative value of -390.626
Set 2/5, RS (39*15312.54=33.5*15312.54+73.5*X) =1145.836 LS (34.5*15312.54=40*15312.54+73.5*X)
Set 3/6 RS, (40.875*15312.54=31.75*15312.54+73.5*X) = 1308.064 LS (32.625*15312.54=41.75*15312.54+73.5*X)
When these forces are combined the net result is a pull of 1308.064 pounds on one bearing with an equal pull in the opposite direction on the other bearing.
This works out roughly to an ISO value of G120.. I am pretty sure I will see this rotor back soon to balance with the teeth installed.
RE: Force generated by rotating mass
Ted
RE: Force generated by rotating mass
RE: Force generated by rotating mass
My resultant force at the right hand bearing is 2640 lbs. acting nearly 90 degrees clockwise from row 1. I assumed rows numbered 1 thru 6 in a counter-clockwise direction looking end on at the left hand bearing.
There are some strange keystrokes in your equations. Typos?
Ted
RE: Force generated by rotating mass
You are quite right, I copied and pasted something that is different than the spreadsheet I was using. I do have 1901.047 as a result of set 3/6. From this computer I can not add the resulting forces so will take your word for the resulting 2640 pounds.
RE: Force generated by rotating mass
I may have to retire my DOS program or at least be skeptical of the results.. Thanks for your help
RE: Force generated by rotating mass
Ted
RE: Force generated by rotating mass
F = mrw^2
m = 80lb/32.17 ft/s^2 = 2.49 slug
r = 7.5 in = 0.625 feet
w = 2200 RPM/60 = 36.67 RPS
F = 2092 lbf
I may be way off base.
RE: Force generated by rotating mass
Ted
RE: Force generated by rotating mass
How accurate the number is..is what led to the original question. I don't recall from where I got this (perhaps some balancing course). I believe if it gives an approximate value but only over a certain speed range... For example if one uses the speed of 10 RPM the value is only 3 pounds.. obviously not true