Gyroscopic Precession

Torque is the answer! The rotational moment of inertia of the spinning mass prevents the weight of the spinning mass from becoming a torque at the persons wrist. The remaining torque at the wrist comes from just the mass of the about a meter long rod and the precession effect of the spinning mass.

Does that fill in the blanks?

Please provide a free body diagram and some cool videos too.

DUH! Nope!

Still don't "see" how the 40 pounds is transferred through his arm to the scale platform.

Can you simplify it even further for this old country bumpkin?

"I don't understand all I know about gravity..."HTRN

If he picks up the weight with one hand, or both hands, the weight is the same. If he picks it up with both hands, then lets go with one hand, the weight is the same even though it creates a torque on the hand still holding the weight.

With both hands holding the weight, the torque is the same on both hands. With just one hand holding the weight, his body has to provide a counter-torque to keep the weight from hitting the ground, and his body pushes against the ground while supplying this counter-torque -- but that just means his weight is now unequally distributed; i.e., he shifts more of his weight to one leg to help provide the counter-torque.

If you suspended this same flywheel on a mechanical arm,and put strain gauges at all joints,what would be the path of the weight transfer to the scale below?

What would the gauge at the shoulder register with the wheel spinning,versus stationary?

you have to keep in mind, that a flywheel is a storage device, as its spinning it is using the stored energy, and depending on it position, it will affect the scale.

That does not answer my question:

How does the weight get transferred to the scale?

Please draw a diagram showing how the 40 pounds is transferred to the scale without adding any apparent strain to his arm?

The force has to come through his hand,wrist,and shoulder,yet it does not appear to be causing any strain.

Even it he put each foot on a separate scale,what difference would it make to the force on his arm?

"I don't understand all I know about gravity." HTRN.

I don't have the time to draw a diagram myself but they're not hard to find. Particularly since so many sites do a great explanation of the moment of inertia for a gyroscope. MIT has a nice text that goes into all of the details and shows a nice diagram for you, too.

The text from the MIT link goes through the details.

One can read all about it, or have people tell you all about it.

But at times, the best thing is hands on to really understand it.

I had one when I was a kid, and I still have it.

there a negative side,..... try holding it/moving it incorrectly. ie,... move it where it doesn't want to go.

What if you had two flywheels same weight, same speed,rotating in opposite directions on same shaft?

What would be the result?

How about attaching another identical flywheel at 90 degree angle to the first one?

Could you suspend a 40 pound weight on an arm designed to hold 20 pounds maximum with no damage?

If so,this is a whole new branch of engineering allowing massive weights to be suspended with a small support.

Consider a crane,for instance

Can you say Pyramids,anyone?

In our physics lab in college, we had a swivel chair that sat (2) people opposed from each other (back to back). and each had what looked like a motorized bicycle wheel that had (2) grips on the axel where you could hole with your hands.

You could get that chair spinning pretty good when you had both gyroscopic bicycle wheels working with each other.

The other result.... ahhh, A 'StarGate' opened

Even simpler-ever notice how difficult it is to keep a bicycle upright when the wheels are not spinning?

Consider the amount of energy/work expended to get that wheel to spin at (3000?) rpm. As it is expended (if left on its own, it will eventually slow down to zero rpm). How long, left to itself, would it take to expend to stored energy? Next, as in the video, using/transferring some of that stored energy as it dissipates to help lift it, likely the time to spin down to zero will be less? Proving that the stored energy is fixed and the amount of time it takes to disappates depends on how it is used. You could apply the spinning wheel to another wheel, which would absorb the rotation and also begin to spin up. And the length of time for the system to release the stored energy would grow smaller the larger and heavier the other wheel is.

Consider the force at his arm. Without considering the spinning mass. I am not good at conversions but know 30 kg is heavy and so is 40 lb so to make units nice I will use 30 kg.

Lets say the weight is .5 meter from his hand. At his hand there is a moment of .5m × 30kg= 15kgm. there is also a downward force of 30kg.

I am not 100% on how to explain how precession works, but i will do my best. You have to consider the position of parts of the mass as it is moving. Consider the bicycle wheel. A spinning wheel not affected by any outside forces only has a force downward at the axle. The moving mass is just as balanced as if it were not spinning (with exception for friction at the bearing). The mass moving up on one side is countered by the mass moving down on the other side. Newton says that the mass wants to move in a straight line, but the structure of the wheel makes it move in a circle. This is ok because the other side balances it out, the sum of all forces are zero (with exception for friction at the bearing...and wind).

Now consider rotating the wheel on a vertical axis. Newton said it wants to move in a straight line, I said it will be happy in a circular line due to the balance but now consider the mass at the extreme edge of the wheel, moving the fastest as it is also moved sideways, it is no longer balanced by the other side because it is moving in the opposite direction. This applies a twisting moment to the axle. If you hold a spinning top and try to turn it like I describe you will feel it twist.

I have never attempted the trick shown but if it is not a gimick and I understand it correctly this should explain it:

Now back to our guy holding the weight. With the mass not spinning, all forces are downward, plus the moment at his hand. when the mass is spinning, and he revolves it around himself it is causing that twisting force I described above. The direction of forces at his hand are not all directly downward and the moment is countered by the twist too. as long as the mass keeps spinning and he keeps rotating the forces will allow him to hold it.

As for how he holds it above his head, I think he tosses it up and holds it instead of just slowly lifting as he did when it wasn't spinning.

Drew K

AAHHHHHHH you just hit a nerve!!!!!!!!

The torque is not 15 kgm and there is not a force of 30 kg. Kilograms is a unit of mass.....not force.

The force of a 30 kg mass would be 30 kg * 9.81 m/s/s or 294.3 N (Newtons). Likewise the torque has units of distance times force thus Nm not kgm. At 0.5 m the 294.3 N force would provide 147.15 Nm of torque.

Rant off.

Sometimes rants are good.

Force and torque have different units because they are different.

A net force changes the position of an object. A net torque will change the rotation rate of an object.

I was just simplifying it. No need for a rant. It was the concept of the moment I was explaining.

Was the rest of my explanation disregarded just because I simplified the math?

Drew K

Was the rest of my explanation disregarded just because I simplified the math?

Oh no, I don't disregard the rest of it at all. Just hit a sore point with me. But as engineers it's important to be accurate with language and units. The fact is, you didn't simplify the math...your math (multiplying and adding and such) was fine. You incorrectly stated the problem, or incorrectly applied the math to the example you created.

In college, one of the claimed advantages of using the metric system is that it's less confusing as you don't have to deal with things like pounds representing mass and pounds representing force. I think that's quite true. But then years later working with a Japanese company that designed servo valves, they kept talking about applying so many kilograms of force. It took awhile before I was able to determine they were using kilograms as mass and as a force unit, too.

Another pet peeve of mine is the stating that the pressure is xxx pounds of pressure. I admit it's much easier to say than xxx pounds per square inch or even xxx psi....but for crying in a bucket let's not muck up communication because we're lazy.

Wait....am I ranting again. Oops. Sorry.

I'm not so anal that I can't tolerate such mis-speak from non-technical folks - (don't ask my kids though).

I didn't use newtons because I didn't want to bring time into the explanation, that per second squared would have made it more confusing.

And yes, in college we often used kg as a unit of force. I think it is ok especially when considering moments. You have a force times a distance; a torque wrench in the U.S. and older ones in the U.K. use ft lbs so why not kg m as a unit of moment?

What are your thoughts on my circular Newton's law?

Drew K

"older ones in the U.K. use ft lbs so why not kg m as a unit of moment?"

The reason that it's okay to use ft-lbs for torque is because pounds is a force unit and it's also mass unit. It's usually distinguished as an abbreviation lbf for pounds force and lbm for pounds mass.

When I was in college, our professors would not allow us to use kg as a unit of force as it is a mass unit. They were quite rigorous about it.

"I didn't use newtons because I didn't want to bring time into the explanation"

I don't see a need to introduce time into the equation. Just gravity. You could have easily said that the 30 kg mass would exert about 300 N of force (30 kg * 9.81 m/s/s = 294.3 N which is easily rounded up 300 N for keeping the math simple.

My take on this is - using lb (or kg) as unit of mass and force is OK for some applications, as long as everybody knows what's going on. But you have to be careful if it's a bit more scientific. E.g. if you have a body with moment of inertia MoI kg*m² subject to torque T N*m, angular acceleration a = T/MoI rad/s².

But if it's MoI lb*ft² and torque T ft*lb, T/MoI doesn't give the right answer. Don't know how you handle it over there, could use a = T*g/MoI where g = 32 ft/s², or convert the MoI to slug*ft² = lb*ft² /g (obviously answer is the same either way). Do you still use slugs? = mass which 1 lbf accelerates at 1 ft/s². My old copy of Perry uses lb for mass and force, but puts g in equations to put things right.

When I was at school (early 1960s UK) we were taught a system going the other way, with a force unit called poundal, which accelerates 1 lbm at 1 ft/s². At uni I did physics, all metric, but my mech eng friends used lbf and slugs in those days. Nowadays of course it's metric over here and a good thing too .

It's magic....

..."

http://www.met.reading.ac.uk/pplato2/h-flap/phys2_8.html

http://www.real-world-physics-problems.com/gyroscope-physics.html

Bear with me,but I don't quite get the full picture.

It is in black and white,but not full color yet.

Ok,the math makes sense,but common sense says that the guy should feel the

entire weight of the flywheel if indeed,it shows up on the scales he is standing on.

If it was a 1 ton flywheel,would it make any difference if it was supported and he

grabbed it by the shaft from an elevated position and hefted out of it's hook?

Could he hold it?

Maybe this old dog is too old to learn a new trick.

Ok, answer two questions:

How much force or weight is transferred through the shoulder joint to the body?

His feet obviously feel the weight.

So let's increase the weight to something humanly unbearable,and the rpm accordingly.

Would it still appear to be very light?

I don't think it ever appears "very light". The gyroscopic action just makes it feel as though the center of gravity is moved laterally closer.

Instead of a man holding it, picture it suspended from a spring scale. As the flywheel slows, it will droop lower and lower eventually hanging vertically.

The reading on the scale will not change during the entire event.

He still feels and bears the entire weight of the device whether the wheel is spinning or not, what he no longer feels when the wheel is spinning is the downward turning moment that is removed by the gyroscopic action of the spin.

When the wheel is static, the weight of the device is controlled by an upward force on the hand closest to the wheel, and the downward turning moment due to the position of that hand not being at the centre of gravity of the device is controlled by a downward force on the more distant hand.

When the wheel is spinning, that downward moment is now converted to a rotational moment about his vertical axis and he is as unable to resist that as he was the downward one and hence he must turn with the device as it precesses.

If the weight of the wheel was not lift-able when stationary, then it would also not be when spinning.

So back to my question:

How much weight would be measured--- at the shoulder joint--- when it is spinning,

and not spinning when held at the same angle from the body?

If it is the same, why could he not lift the weight before it was spinning?

There is still the vertical force at his hand. But there is no torque, disc mass x rod length, when it is spinning. It has to be allowed to precess to achieve that, if the precession were stopped it would fall down. I know that doesn't explain how it happens, probably somebody here has given a good explanation.

If it helps, I did a calc of precession velocity, for disc thickness 40mm, dia 269mm, density (steel) 8000kg/m3, 1m long rod and disc spinning at 4000rpm. I make it 24.8rpm.

CM, do you have that formula handy? I know I can Google it, but I also know (from experience with the formulae for evaporation above a standing body of water) that it becomes necessary to do considerable sorting to find the formula which best approximates one's situation in detail.

If you'd be willing to supply the formula you used, it would save me considerable sorting.

Thank you.

I used ω_p =T/M_a where

ω_p = angular velocity of precession, rad/s

ω_d = disc angular velocity rad/s

m = mass of disc

d = dia of disc

I = disc MoI, kg*m2 = m*d²/8

L = arm length

T = torque due to weight = m*g*L

M_a = disc angular momentum = m*ω_d

I did it on Mathcad and I'm fairly sure it's correct, though you do have to be careful with angular velocities on Mathcad that you're not out by a factor 2*pi. As angles are dimensionless you don't an instant warning.

BTW has anybody noticed on CR4 when composing a comment you can't copy part of it. I have to go into Preview, copy something, and then go back via Edit. Might be my computer, but it didn't used to happen.

CM. Thank you. I'll use this.

As far as the copy-paste issue, the last time I did that was probably less than three days ago, replying to someone, and I had no problem copying it before I started editing, then pasting it into my comment, adding quotes, and (as you guys sometimes say) Bob's your Uncle.

I'm trying, but British English is somewhat difficult for me to get. Friends over there joke about two countries separated by a common language.

BTW, I use Win7-64bit, with Mozilla Firefox, for what it's worth.

I don't have a problem pasting into the reply screen, it's (or was) copying from reply, to paste lower down or into Word in case something happens and I lose it. Not sure which you meant. But by highlighting and right-clicking, I don't get a Copy option, but a box with Undo/Redo/ etc. But I just tried using Control C and that works OK. Should have thought of that before!

I use Mozilla Firefox, but I'm still on XP, I don't know whether that means it's Win7-64bit.

Ah. I usually use the Control key shortcuts (Ctrl-C to Copy, Ctrl-X to cut, Ctrl-V to paste) just by habit, so I never even looked for any kind of a drop down menu choice.

Win7-64bit is Windows 7, for 64-bit computers. Next gen after Vista, and avoided it's pitfalls. Short of Windows Eight, which I understand also has mixed reviews.

But, other than the 64-bit address and operations space width, Win7 is not a lot different than XP, at least in running it. Maybe faster, and MAYBE a few better features. But not much to learn.

Just read my #41 again and noticed a mistake.

M_a = disc angular momentum = m*ω_d should have said

M_a = disc angular momentum = I*ω_d

Hope it didn't cause a problem!

Didn't cause a problem yet, thanks. It's been too cold for my classes here (outdoors or unheated space), so I haven't used it yet. I'll see if I can get it wrong the next warm(er) day we have, though.

I am sure the explanation is there somewhere in all of the excellent feedback I have gotten.

It is staring me in the face.

I am too close to focus,so I will give it some distance and come back later.

I will have to digest it and sleep on it a while,perhaps consult with my friend Dewar.

I seem to do my best thinking with a nightly toddy.

To paraphrase Einstein:

"Reality is an illusion, but a very damn good one."

When you lift something, you not only have to lift it's weight but if you are not lifting at the center of gravity, you have to apply a torque to keep it balanced. The torque makes the object seem much heavier because the smaller muscles controlling your hand are involved.

A spinning gyroscope makes all the weight appear at the point it is being lifted, just as if it were being lifted from the center of gravity. The torque that you would supply if the gyroscope were not spinning is being supplied by the change in the angular momentum of the spinning disk.

Only torque is being supplied by the gyroscope, not lifting force. You're still lifting the whole weight, just not applying a twist to keep it balanced.

GA from me on this. It is the cantilevered (Torqued) portion of the weight which makes the 40 pounds seem quickly like 100. I can carry 100 lb feed sacks on my shoulders, two if I can get help getting them up there. But there is no way on this earth I could even carry a single 50 lb sack out in front of me. Getting them over my center of gravity is the key.

But I didn't understand the transfer before HTRN asked about this, either. Thanks to all of you.

This might help with a missing piece of the puzzle. Let's change the conditions a bit; instead of standing solidly on the floor, you're now standing over the center of rotation of a platform that is free to turn.

When you are handed the rotating flywheel the torque will cause you to rotate (precess) at a rate dependent upon how far your arms are extended and your own moment of inertia, in a direction that depends upon the orientation and which way the flywheel is spinning.

Now attach a scale horizontal and tangent to the platform and fix the other end to the floor. Instead of rotating the platform will act against the scale and cause it to register a force/weight. That measurement represents the "missing" force from the rotating flywheel; it's missing because it acts at right angles to the gravitational force.

If you were standing on a scale you would measure your own weight; if you are now handed the stationary flywheel you would measure the combined weight, and the horizontal scale would read zero. When the flywheel is spinning, the scale you are standing on records the same weight, but the resultant force is the vector sum of the combined weight plus the resisting force being transmitted by the horizontal scale. Bottom line, it's a three dimensional problem.

In the original experiment, your body is flexible and incrementally spreads that "missing" rotational force through your wrists, elbows, shoulders, spinal column, hips, knees, ankles and feet, to the soles of your shoes and their interface with the floor. You can check this by increasing the weight of your flywheel in 10 pound increments until the precessional force becomes so great that your body twists because it can no longer resist that turning moment.

A GA from me. You have explained what i too could not 'see'. Also by your explanation i can see that the type of scale is important. A scale platform that was free to rotate would indeed show LESS weight.

I have visited a kids interactive science centre and they have a suitcase that is fitted with a gyroscope inside it. When stationary no child can lift the suitcase. Once placed on the contacts and allowed to spin up to speed a child can easily lift this suitcase. An empirical example that eliminates the torque element. HTRN and myself agree that they should NOT be the same 'weight' on a scale. I tried this suitcase for myself and experienced the ability to lift with a single finger. Then i tried walking with it!!!!! Going in a right turning circle was fine. Significantly, the suitcase lifted up as i turned. Going straight made the case 'heavier' and it pressed hard against my leg. Turning left was going to break something and the case got much heavier and went down to the floor.

So to explain what i experienced and using your good explanation as a springboard i would say that a large flywheel rotating fast would actually lift a person off the scales if they went in a right ( in this example ) turning circle. The mass of the person would all be in a vector.

Feel free to correct me if you can see i am wrong. ( probably no need to say that )

Jim

compare with ufos,satellites and planets in constant motion

I have read all the answers and I am equally mystified. In the video the guy has a difficult time holding the gyroscope up over his head with one hand when it is not spinning. No torque is involved. It weighs 42 pounds.

When it is spinning, he has a very easy time raising it up over his head, and yet the scale shows his weight as roughly the same as when the gyroscope is not spinning, even though it fluctuates. Roughly 42 pounds.

On the surface, it appears that when the gyroscope is spinning he is stronger, because the scale shows that the gyroscope is not lighter.

You have to remember that this was done at the University of Sydney, and everything down here happens differently to what it does in the normal world...Up is down and down is up, left is right and right is left.. Water goes down the drain in the opposite direction.. Gravity behaves very differently here.

Yanks think that we drive on the incorrect side of the road, but few are aware that we actually drive upside down as well.

I bet that makes for some awkward newlywed honeymoons for Yanks if it is their first trip "down under".

"Now I see." said the blind carpenter,as he picked up his hammer and saw.

It finally dawned on me this morning.

Thanks to everyone for their help;(The voices in my head finally reached a consensus.)

Perhaps you should ask yourself why it is easier to stay upright on a moving bicycle than a stationary one. And then ask yourself why there is no twisting torque on your hand holding the Gyroscope. If you answer these two questions in sequence, you should know the answer to your question without the math.

As for the "easy lift", swinging the axle appropriately allows some of the angular momentum of the disk to be transferred to rotating the whole assembly over the operator's head. (Loosely analogous to "pumping up" a swing.)

I agree that the angular momentum would help rotate the assembly, but it does not explain why the guy is able to lift it far more easily while the whole assembly remains the same weight (according to the scale).

Because the rotation is helping top lift it, paid for by decrease of angular momentum of the disk. It is a bit like swinging a bucket to a higher position, getting started low and coasting up.

It seems like the rotation cannot be helping to lift it, since the scale shows no decrease in weight. Can it be helping lift it without showing an effect on the scale?

Here's an explanation of how a gyroscope works that's a bit more intuitive than angular momentum vectors.

Assume you are holding a gyroscope in front of you with a handle at each end. Imagine 4 points on the disk: point A at the top, point B farthest in front of you, point C at the bottom and point D at the closest point.

If the gyroscope is not spinning and you decide to lower your right hand and raise the left. A force is applied to point A and C which start to move. Points B and D, being on the axis about which you are applying torque, do not move. This is no different from any other object.

Now assume the gyroscope is spinning with point A moving away from you, point B moving down, point C moving toward you, and point D moving up. The difference now is that points B and D are forced to change direction. It is the reaction to this change of direction that causes rotation or precession about the vertical axis.

This explains it better than I can:

https://www.youtube.com/watch?v=_g3KSAqJ7Hs

why a top or bicycle or coin don't fall when in operation

They do.

Look carefully at the front wheel as the bicycle goes out of control.

While I couldn't slow down the video, it appears that the front wheel may have struck a pothole or blown a tire causing the rider to turn the wheel perpendicular to the direction of travel.

Observe a motocross rider as he clears a jump.

The front wheel always turns to the right due to gyroscopic precession, requiring the

rider to correct before landing.

With the exception of my first solo after my dad removed the training wheels, I never experienced what is shown on the video.

Actually, unless the rider is riding hands free, a bicycle is primarily kept in balance not by gyroscopic action but by steering to keep the tires underneath the center of gravity.

I saw a few comments on how it seemed easy to lift overhead while spinning, but I don't think I saw anyone say anything about the fact that he is throwing it up with both hands. He struggled to lift it slowly overhead with one hand, but would he struggle if he used both hands to smoothly throw it up with both hands then guide / lift with the one like he did when it was spinning?

The way I explain and understand gyroscopic forces are with newtons laws. The mass of the wheel are trying to go in a straight line but held to a circle by the wheel. The circular forces are dealt with by the strength of the wheel and not noticed by the outside environment because they are balanced. But if you try to turn the wheel you take the path of the mass outside of the circular line and you can feel the force of the wheel trying to continue in it's straight (but circular) line.

Drew K

Lots of posts here, not much time to read them all, so I'll just give a 'blind response.'

The aparatus weighs 40lbs. that is the same wether the flywheel is spinning or not.

With the flywheel stationary, you have most of the weight at the far end of a three foot rod, and that acts as a lever, with the wrist as a fulcrum. Don't know the formulas offhand, fur the longer the lever arm, the more effect the mass at the end has.

Once the flywheel is spinning, the gyroscopic forces want to keep the axis of the flywheel static (barring 'gyroscopic procession,' which I believe the video is showing with the apparatus wanting to turn around the holder), the forcing of the axis into a static plane provides a 'balance,' so the holder is 'feeling' most/all of the weight at the fulcrum (his palm) instead of out at the end of the lever.

Here is a link to a very good mechanical linkage page with videos of them in action.

Some are very basic,some more elaborate.

If you are thinking of reinventing the wheel,check here first;you may find what you are looking for.

http://makezine.com/2015/04/20/understand-1700-mechanical-linkages-helpful-animations/