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Calculating Centrifugal Force(8)

jamesxi (Types) (OP) 
1 Mar 02 18:22 
I am designing a unique gyroscopic/centrifugal machine, but know little about the calculations involved, so am hoping that someone here will be able to help. Assume I have a 10cm radius centrifuge, and a 1kg mass at the end of the centrifuge, with the weight's center exactly at the 10cm mark. The centrifuge is spinning at 1,000 RPM. How would I calculate the centrifugal force (in kg) that such a system generates. I don't really need to factor in friction yet. So, my variables are: r(radius) = 10cm m(mass) = 1kg RPM = 1000 Cf(centrifugal force) = ? Jimmy jamesxi@yahoo.com 

TD2K (Chemical) 
1 Mar 02 18:40 
First, force is never measured in kg. That's a unit of mass.
Secondly, find the equation for centrifugal force and plug the numbers in.
Can you please post more a little more information on this problem? 

Guest (visitor) 
1 Mar 02 20:03 

(3) cowski (Mechanical) 
1 Mar 02 21:46 
Hopefully this will get you off on the right foot.
1. convert your rpm to radians/sec (cycles per sec, frequency) 1000 rev/min * 1 min/60 sec * 2pi radians/rev 1000/60 * 2pi = 104.72 rads/sec
2. compute velocity using standard units (in this case kilograms, meters, seconds) v = 2pi * r * frequency v = 2pi * 0.01m * 104.72rad/sec = 6.58 meters/sec
3. compute force F = m * a for constant circular motion a = v^2/radius a = (6.58 m/s)^2 / 0.01 m = 4329.64 m/s^2 F = m * a = 1 kg * 4329.64 m/s^2 = 4329.64 kg m / s^2
Since Weight = mass * gravity, this would be equivalent to 4329.64/9.81 = 441.3 kg.
note: this assumes horizontal circular motion (like a merrygoround) and ignores the effects gravity would have; if it were vertical (ferris wheel) the force of gravity would play a bigger part in the calculations.
best of luck in all your inventing! 

jamesxi (Types) (OP) 
1 Mar 02 22:31 
Thanks for the equation, it will definitely help me out. Check back in a few days to see my 100% great invention(or absolutely useless one). 

bgkms (Mechanical) 
2 Mar 02 5:27 
centrifugal force = m*w*w*r
m=Kg.
w= 2*3.14*speed in rpm / 60 rad/sec.
r=radius (dist.) m
centrifugal force= Newtons.


(2) svanels (Petroleum) 
2 Mar 02 6:26 
To www.ex.ac.uk (visitor),some people belief anything that is on the internet or in a calculator. kg is a unit of mass, in the past the term kgf or kilogramForce was used to express Force so 1 kgf = 1 kg * 9.8 * m/s ^{2} = 9.8 Newton Steven van Els SAvanEls@cqlink.sr 

Dear james, You know radius, mass, and rpm. You need to calculate the centrifugal force, right; Its simple:: CF= mass * (rpm/2*PI)^2 * radius .... from this relation u can calculate centrifugal force. R u satisfied. with smile kumaravel. S. Kumaravel, Finite element Analyst, Hyderabad  500 033.


Sorry SK, you've made a mistake in that equation. There's a factor of 60 to convert rpm to rps that you've ignored. Cheers
Greg Locock 

jamesxi (Types) (OP) 
3 Mar 02 22:42 
Here is the tough part: What happens when you have a negative radius? To imagine how this is possible, start with a large fan, with one fan blade having three times the length of the other blades. All the fan blades have a large mass at their tips. Fold the extralong fan blade until it is the same length as the other blades. The mass will be past the center of the fan, but the force will be directed in the opposite direction. Should I just take the absolute value of the radius?
kumaravel, I would have to say that the 441kg cowski calculated sounds about right, but plugging in your numbers:
m: mass in kg r: radius in meters Cf: Centrifugal force in newtons?
Cf = m*(RPM/2*Pi)^2 * r Cf = 1*(1000/2*Pi)^2 * .01 = 24,674 newtons(?)
That would be 2.77 tons of centrifugal force. That sounds a bit high, don't you think?
I did finally get my USPTO number though its looking more and more like my machine does not work, at least not as efficiently as I would like. I'll post a link tomorrow so people can see my lump of scrap. 

I have a horrible suspicion I know what you are designing, but I'll leave that out, for the time being. Instantaneously the centrifugal force will ALWAYS act away from the centre of rotation, so when you fold your long blade over then the force will still tend to fling the mass away. The shape of the folded blade affects the stresses in that blade, but do not affect the forec on the mass. Build a model out of paper and tape, and coins, to check this. Cf with your initial figures is 1096.6 N per 1 kg mass. Cowski has included an extra factor of 2*pi in working his velocity out. This thread hasn't exactly been our finest hour, has it? Cheers
Greg Locock 

cowski (Mechanical) 
4 Mar 02 8:14 
Greg,
It is 2pi/60 to get from rpm to rads/sec and there is another 2pi in getting from a circular frequency to a linear velocity. I stand by my equations for now, but if you can provide corrections I will be glad to look at them. Thanks 

FINN (Mechanical) 
4 Mar 02 9:21 
I use this formula all the time. I design vibratory compactors using eccentric weights. Given: mass=1kg, radius=.01m, rpm=1000 and F=m*a a=r*w*w w=1000rev/min*2pi radians/rev*1min/60sec=104.72rad/s a=.01m*104.72rad/s*104.72rad/s=109.66m/sec^2
Therefore: F=m*a=1kg*109.66m/sec^2=109.66kg*m/sec^s=109.66 Newtons
I hope this clears up all this miscalculations. You must remember to carry your units along in order to keep from making a major error.


RobWard (Industrial) 
4 Mar 02 10:14 
What a fantastic thread from such a simple request! I can't wait to get the link to see what's been designed. 

I may be just an Aeorspace Engineer, but isn't 10 centimeters .1 meter? All of the calculations I saw on the thread said .01 meter. T 

metrication, yes
1 cm =0.01 m
hence
10 cm = 0.1 m 

OK, let's take it from the top. r=.1 m w=2*pi*1000/60= 104.7 rad/s (roughly) m=1 kg centrifugal force = m*r*(w^2) (which incidentally is the same as m*(v^2)/r, since v=r*w) =1*.1*104.7*104.7 N =1096.6 N, a number I have seen before. You don't need a 2pi to convert from rad/s to velocity, it is in there already. Cheers
Greg Locock 

jamesxi (Types) (OP) 
4 Mar 02 19:04 
Thanks for the calculation corrections guys. Here would be the complete allinone equation:
Cf: centrifugal force in newtons m: mass in kg r: radius in meters
Cf = (2*Pi*RPM/60)^2 * m* r (2*Pi*1000/60)^2 * 1 * .1 = 1096.6N This is equal to 247 pounds of force
Also, Greg, after thinking about its hard to see it any other way than what you were saying. ("Instantaneously the centrifugal force will ALWAYS act away from the centre of rotation") That would actually make my machine better than otherwise, so I'll raise my odds of it working to 25,000 to 1. I think I will have a link to it tomorrow evening.
One last question, why is w used as the symbol for radians/second? 

w is the nearest roman equivalent to the greek letter omega, which in its lowercase form looks like a curly w. As to why we almost always use lowercase omega for angular velocity  beats me! Cheers
Greg Locock 

*Beats Greg* Really interesting thread. A lot of formulas for a simple one at the end, I like this. When I read such stuffs, I know why I study engineering FrenchCAD Université Joseph FOURIER Département Génie Mécanique Grenoble  France cyril.guichard@wanadoo.fr 

Speedy (Mechanical) 
5 Mar 02 7:37 
Could someone explain the negative radius bit?


cowski (Mechanical) 
5 Mar 02 8:30 
Thanks to FINN and Greg for setting me straight, maybe I should go design satellites for NASA or something! :) 

svanels is correct. Think of a negative radius as a positive radius with a 180° phase shift. 

And force is allways a Newton or lbf. Using word "weight" can be misleading, unless you say if it means mass or force. "kg of force" was used long time ago, some countries called it "kilopond" (I believe Germany) but because of the confusion it was replaced by Newton (1kgf = 9.81N) gearguru


jamesxi (Types) (OP) 
6 Mar 02 22:00 
Its taking longer than I thought to get the full animation of my working machine, so I have a link to the nonworking version that is somewhat similar. In this animation I only displayed the two main parts of the invention, so you can't see the structure that keeps the "thrusting arm" at a bearing of 90 degrees. This machine is a centrifugal propulsion machine, which uses simple mechanical motion to generate centrifugal force meant to propel the machine. Keep in mind that this version does not work. It consists of a a centrifuge arm, a "thrusting" arm, and a "thrusting mass". The red line is only the path of the centrifuge. Looking at the invention I was at first thinking that it would be thrusted to the 90 degree mark using centrifugal force. But, I'm pretty sure that the center of the centrifuge has only been raised(correct me if I am wrong). http://www.ultimateinteractive.com/images/animation2.gifFortunately their is a similar version that I am quite sure that does work. Unfortunately I'm not so sure how well it works because the math in calculating its effectiveness is more complex. I'll post a link in a couple days. 

ivymike (Mechanical) 
6 Mar 02 22:43 
well if you can't get your desired thrust out of it, you can at least turn it over and let it walk, eh?
;)


Here is one more note about the "negative radius". In the most generic form of the formula for the centrifugal force "cf" the radius is expressed as a vector. It makes sense, because also the force is always represented as a vector; cf has not just size but also the direction. The remaining variables in the equation (mass "m" and angular velocity "omega") are scalars, they have size only. When the object rotates, the direction of the radius vector pointing to the center of gravity of the mass is changing. The cf vector is "heading" in the same direction as radius vector. In the generic formula the radius vector defines also the direction of the cf vector. To calculate only the size of the force, the equation was simplified: the radius now is actually represented by the length of the radius vector. The length of a vector is defined as the "absolute value" of the radius vector. The absolute value is always positive. So  if we use the radius as a vector "r1", we can also use a vector "r2" negative to it (r2 = r1); r2 has the same size (length), but opposite direction than r1. r2 only changes the cf direction (reverses it in this case), not size. Again: r1 and r2 and cf here are vectors!!! Boy, wasn't it a simple question on the very top of this thread? gearguru 

jamesxi (Types) (OP) 
9 Mar 03 18:38 
I think that the equation is wrong: Cf: centrifugal force in newtons m: mass in kg r: radius in meters
Cf = (2*Pi*RPM/60)^2 * m * r (2*Pi*1000/60)^2 * 1 * .1 = 1096.6N This is equal to 247 pounds of force
After figuring this out all over again, I came to the conclusion that r is in the wrong place so this is the right formula: Cf = m * (pi * (RPM/60) * r)^2
Here is how I got the formula f  force in newtons a  acceleration in m/s^2 v  speed in m/s pi  3.14159 RPM  rotations per minute r  radius in meters
f = m * a a = v^2 v = (Pi * (RPM/60) * r)
f = m * (pi * (RPM/60) * r)^2 f = 1 * (pi * (1000/60) * .1)^2 f = 27.4 newtons f = 6.16 pounds 

no, Cf = m * (pi * (RPM/60) * r)^2 is WRONG because a = v^2 is WRONG. Cheers
Greg Locock 

jamesxi (Types) (OP) 
11 Mar 03 4:13 
Well with wrong in caps lock I must be wrong! I can't believe that I messed that up. Should read: a = (v^2)/r But using that equation I'm pretty sure its wrong too. The problem is that using the formula you gave me the numbers seem a bit high, but maybe the power of centripetal force is stronger than I think.
I would think that a 1kg weight spinning 1,000RPM on a 10cm radius(about 17 times a second) would generate no more than 50 pounds of force, so do you think that the actual number could be 250? 

quark (Mechanical) 
11 Mar 03 4:35 
The equation for centrifugal force is F = m x r x (2 x Pi x N/60) ^{2} F= force in Newtons m = mass in Kg r = radius in meters N = revolutions per minute If you calculate the force it comes out to be approximately 111.8 Newtons. One error in your post is you considered 1Kg to be mass of the object in the calculation, where as you say it is weight. To convert weight to mass devide it by 9.81 because W= m x g Regards, 

quark (Mechanical) 
11 Mar 03 4:59 
oops! It is said as weight elsewhere. What is it exactly? Also v = r x w(omega) Where w = 2 x Pi x N/60 because there are 2 x Pi radians for one revolution. and a = rw ^{2} 

Heyner (Mechanical) 
11 Mar 03 7:35 
Hi everybody!!!!
Did you see? we are still learning... every second. I was smiling all the time while reading this thread!!! "37 replies for a simple question" we say, but look where we are!
Let me try to help or make it worse:
1. Don't worry about weigth & mass of 1 kg (both valid). All you need is to have clear where and how to use the number.
2. F = m * a OK? so all we need is to find a, cause we have m = 1 kg.
3. Our a is the centrifugal aceleration, it permits the body to continue in a circular path, since it's trying to fly away all the time. Its direction: center of circle. Right?
4. so a = (w^2)*r where w is the omega Greglocock said. Of course we need it in radians per second, so the right conversion is: remember 1 revolution is 2*pi radians, ok? in some equation the "2" was missing, we all know the minute has (yet) 60 secs. No problem with that, conversion is like Quark said.
5. Ready. Multiply m by a. You got it. Astroclone let us know sometimes we pass over the error one time and another. Be careful. 10cm=0.1m oh, sounds so easy!!!!
Correct answer is 1096.6 Newtons. Remember this force has the same direction as aceleration vector, it is: center. Everything is consistent.
I'm not here to show you the answer, you already got it when I arrived. I just wanted to recognize your work: good thread, GOOD. 

Get it right unit of force in metric is expressed in Newtons not Kg. If I were your physics teacher which I suspect is part of an assignment or homework that you are doing, I would flunk you. 

IRstuff (Aerospace) 
11 Mar 03 13:08 
While on that subject, force direction inwards is tne centripetal force. TTFN 

Heyner (Mechanical) 
11 Mar 03 15:02 
IRstuff
You're right, I got a problem with centrifugal/centripetal.
Sorry. 

canox23 (Mechanical) 
11 Mar 03 21:13 

IRstuff (Aerospace) 
11 Mar 03 21:26 
Heyner, It's OK, centripetal force can be very nonintuitive TTFN 



