View Full Version : The weight of a spinning mass?

30th Jun 2001, 14:09
Here is a question that kept me awake last night and I would like to hear your thoughts.

Does a spinning mass (eg a helicopter main rotor) weigh less when it is revolving than when stationary?
My thinking goes like this:- does centrifugal force somehow oppose the force of gravity to in effect make the rotor lighter? Obviously the mass is unchanged but if I am right, could you in theory spin something so fast as to negate the effect of gravity?? Also, does it matter in what plane the rotor axle is lying(ie vertical for a helicopter rotor or horizontal like an aircraft propellor)?
Or am I just talking a load of rubbish? Do you know what I am trying to say?
Please put me out of my misery.

30th Jun 2001, 14:19
Mass = m
Weight = mg (g is 9.81 m/s/s on Planet Earth)
'Centrifugal force' = mass x radius of plane of rotation x square of angular velocity

'Weight' is unaffected by rotational spinning of any kind.

30th Jun 2001, 14:27
Ok, but doesn't the fact that stationary drooping blades when spun are no longer drooping mean anything? (not a facetious answer but a genuine query)

30th Jun 2001, 14:49
There could be two possible explanations for this:

1. (Most likely) Rotor blades act like wings, and when spinning, they start to produce aerodynamic lift which will support their own weight and stop them from drooping.

2. (Not particularly significant to this example) As a body is spun, the centripetal force acts in a plane to which the axis of rotation is perpendicular. This pulls the blades outwards, and to a certain extent, if they are drooping, this could appear to 'lift' them, even though the downward component of their weight remains the same.

As you should be able to see from this, simply spinning a heavy object around an axis does not result in any net transational force. If the object is out of balance, a net force will be produced, but the fact that the force is changing direction constantly means there is still no net acceleration.

The only known way to produce a force on an object is to push it. This invariably means it either interacts with an external body or medium (e.g. air) or simply ejects matter to produce a momentum change.


[This message has been edited by Pielander (edited 30 June 2001).]

Bally Heck
30th Jun 2001, 15:47

Weight is no longer an approved term. Mass is the new weight and is variously defined as the quantity of matter contained in a body or the resistance to inertia of a body or something like that.

The effect of gravity produces a force of 9.81 m/s/s x m giving a downward force of 9.81 newtons for a one kg mass

The rotation of the body exerts a force which is m x r x omega squared.

Where r is the radius of rotation in metres and omega is the rotational speed in radians per second.

Radians per second is revolutions per second multiplied by pi (3.141)

So a 1kg mass rotating at 1 rps at a radius of 1m will produce a force of about 9.9 newtons.

If this was a conker on a piece of string then the force of gravity would pull it down and the centripetal force would pull it in the plane of rotation. As both forces in this case are approximately equal, the string would be at an angle of approximately 45 degrees. As the rotational speed increases, the string will approach the plane of rotation...but never actually get there because gravity will always have it's 9.81 newtons downward effect (on the earths surface.)

Clear as mud?

Think this is about right but I'm dredging up knowledge and formula from twenty years ago so please forgive any minor or major errors.

Lu Zuckerman
1st Jul 2001, 03:51
I have to go back a few years to give you an answer. If I remember correctly on the CH-37 the centrifugal loading exerted by an individual blade on the attachment cuff and then to the head was somewhere around 72,000 pounds. On a smaller helicopter (S-58) it was around 24,000 pounds.

As an aside, I once sat in on a maintenance course for the H-37 and the instructor was asked what would happen if one of the blades came off. His reply was that the other blades would move accordingly and take up for the missing blade. These blades as massive as they were are balanced to inch ounce.

The Cat

1st Jul 2001, 05:15
The MASS of your helicopter blade is the same where ever it is. Earth, Moon, Mars, floating in water, rotating or stationary. Mass is to do with the elemental composition of an object.

Weight is a FORCE. And someone correctly stated that force is related to mass by the equation force=mass x acceleration. Earthly objects experience an acceleration due to gravity of around 32 ft/sec/sec. Most people (incorrectly) use the words force and mass interchangeably.

So on the moon, your helo blade will weigh less, while retaining its mass.

On the ground, the blade's weight will remain unchanged, stationary or rotating. Its apparent weight, measured relative to the aircraft, will vary dramatically as the forces of flight change.

But if the blade is rotating a few thousand feet agl in a hover, it WILL weigh a teensy bit less, as the force of gravity drops with altitude. And if you can climb to the moon, its weight will drop to about one sixth of its earthly weight.


1st Jul 2001, 07:20
If I may echo some other posts..

What seems to give people difficulty with these sorts of problems and concepts is the need for juggling a number of different forces (external and inertial) acting in different directions to end up with an instantaneous resultant force which relates to what the mass is having done to it at that time. Very much like the navigation solution's vectors or the usually seen resolution of the wing's aerodynamic force vector into (vertical) lift and (horizontal) drag componenents.

Unless you need quantitative answers (ie numbers), forget the sums as that is an exercise in systems of units confusion. (Typically this causes undergraduate students the dickens of a problem for several weeks until they finally get their heads around systems of units and conversions. For the slower ones, like me, read months for weeks). Realistically, unless you are in the design or maintenance game, qualitative (ie a feel for the matter) probably is of more use than quantitative (ie numbers).

If I interpret the intent of the original post reasonably, the following may be of use -

In essence you have a number of major things going on, including ..

(a) weight acting downwards on everything, but the rotor blades in particular. This is no different to the sort of "weight" you see on the bathroom scales (the force the scales need to exert to stop you falling onto the floor) in the morning and accounts for the blades drooping at rest. The hub mechanisms prevent the rotors falling down to the ground. A bit like trying to hold your arms out horizontally for a prolonged period.

The suggestion that "weight" is no longer PC is a new one for me. I suspect that that comment related more to the use of the term in typical weight control calculations where "mass" would be more appropriate from a mechanics viewpoint. However, as we all know more or less what it is that we are referring to it really is an inconsequential consideration. "Weight", in the present case, merely refers to the forces arising due to the influence of gravitational attraction.

(b) a measure of engine-provided torque to drive the rotor assembly against drag forces, except in an autorotative state.

(c) as the rotor disk winds up, centripetal accelerations and forces act radially inward to keep the rotor going around in circles. Centrifugal reactions act radially outward. These terms often are used interchangeably in common discussion. This is a bit like getting a length of string with a lump of something on the end and spinning it around your head. The effort you put in to keep the lump from hurtling off into the darkness is the centripetal force needed to keep the lump going in a circle. The lump, meanwhile, continually tries to hurtle off into the darkness. This reaction to the centripetal force is normally given the term centrifugal reaction (more commonly, if less inappropriately, called centrifugal force).

(d) as the rotor winds up, the individual surfaces generate forces as they whizz around in the air - what we normally resolve in lift and drag components. We can ignore blade pitching moments for the present discussion.

As the rotor RPM increases the radial forces will tend to make the blades rise above their rest position. A bit like attaching a couple of pieces of string to a lump of something resting on the kitchen table. If we pull the two bits of string in different directions (weight and centrigual reaction directions), the lump moves in a third direction depending on the geometry and size of the applied loads. This latter direction is where the blade elements will sit.

When the pilot applies collective, however, the blade lift forces will make the blades move considerable higher and above the horizontal. The radial loads prevent the blades from rising to a level which would be other than useful.

(e) gyroscopic loads will complicate the situation when the disk is tipped.

Either way, the blades will end up doing whatever is appropriate according to the net effect of what all the various loads are doing.

Comparison of rotor disk and prop disk is complicated by the orientation and the quite different nuts and bolts mechanisms involved. However, in a similar manner, the prop blades are subjected to loads which tend to cause deflections similar to what is seen in a rotor disk.

In respect of your base question, the spinning mass isn't going to "weigh" less as that relates only to what the bathroom scales would read - which is near enough to constant for the range of heights we might be talking about. However, if you were to define "weight" as the tension load in the string which you are whirling about your head, then that "weight" would be very much increased, to incorporate the centipetal load component. LZ's numbers for a particular rotorcraft puts that bit into big number perspective.

These things are much easier to understand with a few models and pictures. Hopefully my attempt here hasn't needlessly complicated the matter.

Lu Zuckerman
1st Jul 2001, 18:44
The analogy of a weight spinning on a string does not apply to a spinning rotor blade as the effects of gravity would be nil compared to the centrifugal forces involved. If on the Moon the mass of the blade would be the same but the effects of gravity, which is much less, would make it much easier to pick the blade up. However if you had the means to spin that blade up and rotate it at 250 rpm the centrifugal loading would be the same as that on earth.

On earth when the blades are being spun up the blades will assume a radial position and extend straight out from their attachment point on the rotorhead. They would also hang back from the radial line due to their inertia. Whether this same inertial effect would occur on the moon I couldn’t say because the math involved is way beyond my understanding.

The Cat

Self Loading Freight
1st Jul 2001, 19:38
Things get heavier in motion due to relativistic effects. You'll have to get the rotors spinning at some appreciable fraction of C, which might raise some interesting engineering issues...


2nd Jul 2001, 03:25

The vertical weight component remains, whether it is a large or small fraction of the radial loading. Until you bring blade lift into consideration, the blade will not be able to achieve the horizontal due this vertical force component.

The blade, viewed simplistically, is no different to a mass on the end of a piece of string .. the mass is any blade element, and the string is the blade structure at reduced radii.

I suspect that the reference to Einstein is facetious ?

[This message has been edited by john_tullamarine (edited 01 July 2001).]

2nd Jul 2001, 05:03

The blades would only 'hang back due to inertia' while the blades were accelerating. After they had reached a constant rotational velocity, the tangential effect of inertia would be nil. They may well hang back due to air resistance, and indeed, they are pivoted at the base to allow them to do that.

I don't thing the reference to the moon really helps, because the motion and forces experienced in the plane of rotation, assuming this is horizontal, are entirely independent of gravity. This is really the answer to the original question about 'weight'. Out-of-plane accelerations open up a whole new can of gyroscopically precessing worms.


Lu Zuckerman
2nd Jul 2001, 05:45
To: John T

Are you saying that a blade that weighs about 262 pounds when spun up to about 250 rpm and has a centrifugal loading of 72,000 pounds will not rotate in a pure radial plane relative to the root attachment due to the gravitational effect of the blade weight?

To: Pielander

You stated,” The blades would only 'hang back due to inertia' while the blades were accelerating. After they had reached a constant rotational velocity, the tangential effect of inertia would be nil. They may well hang back due to air resistance, and indeed, they are pivoted at the base to allow them to do that”.

Boiling down three pages of the Sikorsky Helicopter theory of flight Handbook it says that you are right. It states that the only time that the blades would rotate in a pure radial position is if the blades were rotating in a vacuum, which covers my comment about the moon. The blades when rotating in an air mass will hang back behind the radial line due to drag of various types. The leading and lagging takes place behind the radial line and the only time the blades move forward of the radial line is during autorotation or when the rotor brake is applied which I mentioned in a previous post

The Cat

2nd Jul 2001, 08:23
If we are considering rotational dynamics in the absence of the influence of aerodynamic loads then indeed I am.

Due to the relative magnitudes of the vectors, the net result is trivially different from the disk's being horizontal - a fraction of a degree - but, in the context of the original question, it is both relevant and important.

[This message has been edited by john_tullamarine (edited 02 July 2001).]

2nd Jul 2001, 12:57

To answer your original question the weight (a vector quantity) due to effects of gravity on the mass of the blades remains unchanged.


What JT & Pie have been saying is true, a few other ways to convince yourself …

Take a wheel off your push bike, spin it up and see if it gets any lighter....


[This message has been edited by Zeke (edited 04 July 2001).]

3rd Jul 2001, 01:42
I'm with John Tullamarine - more power to you I say!

Other contributors: if you wish to bring in relativistic effects of motion and talk about the moon's gravity, you should also should take into account the effect of gravity of the sun and various nearby stars, the coriolis effect of the earth's rotation, the gravitational force of the atmosphere above you... etc etc etc.

Keep It Simple eh?

Getting back to the original question, I once read a treatise that suggested you could avoid Newton's laws of motion by rotating objects really fast. Unfortunately I don't remember the source.

In any case, it was rubbish from a conventional physics point of view.

[This message has been edited by Oktas8 (edited 02 July 2001).]

Steven JC
4th Jul 2001, 19:22
HI for what it's worth this is my explanation - An object having mass has a force exerted on it by another (larger) object also having mass- ie gravity. In the case of the earth this force will act directly toward it's centre ie centre of gravity. The only way that this force (weight) could be changed is if due to the rotory motion of the blades an additional component was generated. It is simply a vector addition. Therefore the centripetal force (which generates the equal and opposite centrifugal force)which acts in a different geometric plane will generally not affect the downward force (weight) of the blades unless an additional component is generated.

Think of the forces acting in two entirely different planes which don't bear any relationship ( as long as they act with right angles between them - that is the main point- as soon as a right angle is lost between the vectors then we have a dynamic mass since there will be addional components)

I hope this makes more sense to you than me - it's been 6 yrs since I studied it.


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