Could a spacecraft spin so fast that it spontaneously deconstructs?

Question

I have to admit, first and foremost, that this was inspired by Kerbal Space Program. During a launch I decided to hold the "E" key and spin my rocket to the fastest rotation speed possible. At some point I feel that the rocket should have deconstructed, then thought again-- on what basis did I make that assumption?

Could a ~~reaction wheel~~ spacecraft spin so fast that it deconstructs itself? What force would cause this (if it would) and which parts would be most susceptible to this force?

Bonus points (Putting Quantities to the Discussion):

I'm also interested in the actual calculations behind this as well. I read in another question "How far can you fall on the moon without injury" that the average human femur takes 4,000 newtons to snap. Assuming the average human femur is 50 centimeters long, weighs 250g and is a uniform cylinder; how fast would the bone be required to spin in a vacuum before reaching the critical limit of 4,000 newtons? You may assume the point of failure in a specific location if you like.

~~My calculation:~~

sqrt(4.00kN*.25m/.25kg) = ~63.25 m/s from F = (M*v^2)/r

Is this remotely correct? The bone would break at approximately ~63.25 m/s of rotational velocity?

After reading more all of that was over-simplifying a complex problem, @TomSpiker did a great job of explaining this variation of the calculation.

This sounds a lot more like a physics SE question to me. You'll probably get = if they're welcoming - an answer along the lines of "you can spin anything fast enough to make it deconstruct, except black holes." Or something like that. On Physics there are some folks that will have the background to detail that answer. There are here, too, of course, but there are more of them collected over there. — Don Branson, Jul 17 '18 at 14:37
@DonBranson I agree to be honest, I was about to ask "what is the theoretical limit to how fast a spacecraft could spin." That's *definitely* a Physics SE question, if anyone wants to migrate this I wouldn't mind. Kind of curious how we quantify the spinning of a black hole now too. — Magic Octopus Urn, Jul 17 '18 at 14:39
@Magic Octopus Urn: More of an engineering question than a physics one, though, since it depends on what the spacecraft is made of, and how it's put together. — jamesqf, Jul 17 '18 at 16:42
Related info - You can destroy a CD by spinning it too fast: https://www.youtube.com/watch?v=zs7x1Hu29Wc ... if you can do that to a spinning CD, why couldn't you do it to a anything larger and more complicated? — WernerCD, Jul 17 '18 at 20:17
Regarding your femur calculation: 50 cm has two significant figures, so you should give your answer in two significant digits (the force is written as 4000N, which is a putative four significant digits, which is clearly absurd; it's probably only one). The F = (M*v^2)/r equation refers to a situation with a fixed r. Most of the femur is less than 25cm from the center, and so is not contributing as much to the force. As a result, the force will be one third of what would be calculated using r = 25cm. The 4kN figure is likely compression, while rotation causes a tension force. — Acccumulation, Jul 17 '18 at 19:04
Okay: excessive significant digits are all over the place, so good for trying to help tame that, Acc., but "4000N" has only 1 s.d., not 4. "4000.0" has 5. If you want 4 s.d. (and you really shouldn't for this), you'd need to do something like "4.000e3 N." or "4.000 kN". — The Nate, Jul 17 '18 at 20:56
OTOH, the astronaut will be squeezed against the side of the spacecraft so the force experienced by the bones will be stronger... and death due to high-g will be long before bones start breaking. — SF., Jul 17 '18 at 20:57
@SF. I'm talking about a literal spinning femur in space, no mention of life or death, nor anything but the femur. I chose a bone because I know the (approx) tensile strength of it (4000N). This is a femur spinning in a vaccum with no external forces upon it other than the ones making it spin. — Magic Octopus Urn, Jul 17 '18 at 21:00
@TheNate slightly better? Note: I am not a physicist (obviously). — Magic Octopus Urn, Jul 17 '18 at 21:03
It's cleaner, now, yeah. Retaining absurd numbers of significant digits is something that scientists and engineers should never do, because it overestimates what is known. It didn't really injure your point.since it was clear you were asking about that sort of thing, and so not deceiving anyone, even yourself... which is the worst danger this error represents to engineering and science. (Engineers need to know what they do and don't know, and that's just one of many disciplines that ought always be maintained.) — The Nate, Jul 17 '18 at 21:14
If you think on it a bit, it doesn't make sense to average the force that tiny little finger bones can carry with the loads that your mighty legs do. What engineers use instead of force to calculate material breaking is stress. That is the force distributed over some type of area of the material. How that area is defined is the kind of stress in question. The units end up the same as pressure, since the force gets distributed through that area. This allows you to calculate the effects of forces on various sized objects. — The Nate, Jul 17 '18 at 21:36
@MagicOctopusUrn re spinning black holes:I'm not sure that it's meaningful to think of actual objects rotating, when talking of singularities. However, we can confidently say that they have angular momentum, because that's a conserved property and stuff tends to spiral into them. IANAexpert though, and the correct answer for stuff going on inside an event horizon is probably horribly complicated, as well as highly theoretical ;-) — Flyto, Jul 18 '18 at 08:51
@Flyto There are four quantities conserved when an object collapses into a black hole: mass, charge, angular momentum, and linear momentum. The angular momentum distorts spacetime (which is considerably warped to begin with!), so calculations of dynamics in the vicinity of a black hole must consider the spin. You have to take the linear momentum conservation in its context: masses that don't wind up inside the event horizon can indeed affect the post-collapse momentum through gravitation. Asymmetries cause some supernovae to eject the resulting black holes at high velocities. — Tom Spilker, Jul 19 '18 at 16:17
Neil Armstrong's first spaceflight, Gemini 8 (alongside pilot David Scott), almost ran into this rotation-pocalypse scenario. Armstrong solved it, using three quarters of the reentry maneuvering fuel, and triggering a mission rule to immediately abort the mission. — Jacob Krall, Nov 26 '18 at 20:31

score 60 · Answer 1 · edited Jul 18 '18 at 06:26

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Did it really happen?

Yes.

The investigation of Japanese Hitomi spacecraft's failure found that it was spinning too fast due to attitude control error. As a result, the spacecraft spun so fast that several pieces of debris were registered.

But it was caused by thrusters, not reaction wheels.

edited Jul 18 '18 at 06:26

Kat

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answered Jul 17 '18 at 14:56

Heopps

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I apologize for not accepting this, based on the time posted this did answer the question first. However, I really appreciated the additional information posed by TomSpiker, and that is why his was accepted over yours. This is still a great addition to the question though, and I don't want to understate that fact. Thank you again. – Magic Octopus Urn Jul 19 '18 at 19:57

Tom Spilker · Accepted Answer · 2018-07-19T21:07:44.617

To parallel @Heopps answer:

Did it really happen?

Yes.

In spectacular fashion!

In 1965 NASA launched a boilerplate Apollo command module on a Little Joe II rocket to test the Launch Escape System (LES), and got more of a test than they'd bargained for. Due to an erroneous installation of gyros the control vanes on the fins went to full deflection upon launch and caused the rocket to spin up, to the point that centrifugal force broke the motor mounts for the solid motors and the rocket came apart, well below the altitude where the LES was supposed to initiate. The video of that flight is very cool!

Despite the unintended flight profile, the LES got a successful test under a real abort situation.

Note that this example does not truly involve a spacecraft, since this test never reached, and wasn't intended to reach, the Karman line, and used only a boilerplate command module. @Heopps answer steers you to a true spacecraft's unfortunate demise.

EDIT 2018 July 18

To address the spinning femur, a real femur is a lot more complicated than a cylinder.

But first, let me say that for someone who's only dipped a toe (so far!) in the vast ocean of physics, you made a good try! Many would get brain-freeze as soon as they saw an equation, but you didn't. Good for you!

Regarding the cylinder calculation, remember that this object is spinning around its center of mass. If you assume all the mass is concentrated at the ends of the cylinder (which it's not, of course, but assume that for now), then two 125-gram masses are rotating with a radius of gyration of 25 cm. For the gyrating mass and radius of gyration to be 250 gm and 50 cm, the rod would have to be rotating around one end, not the center, and all the mass would have to be at the other end. Instead we have the two 125-gm masses rotating with a radius of gyration of 25 cm, so the equation becomes V = SQRT(4kN*0.25m/0.125kg) = 89.44 m/s, or ~3400 RPM. (This assumes the 4 kN is an ultimate tensile strength) That's actually faster than your calculation indicated! This equation calculates the force applied by only one of the 125-gm masses: the other will apply the same force, just in the opposite direction, so calculating only one of them is sufficient.

If you assume a true thin cylinder with the mass distributed uniformly along its length, the parts of the cylinder closer to the rotational axis are moving at a speed slower than the ends, so they produce less centrifugal force than if they were out at the ends. The net tension produced at the midpoint is half as much (not one third as much!) as the case where the mass is concentrated at the ends, so you could actually rotate faster, by a factor of √2.

Real femurs are somewhere in the middle between the uniform cylinder and the situation above where the mass is concentrated entirely at the ends, but there is a complication.

The complication is that, as you can see from the image, the femur isn't cylindrically symmetrical. It isn't even homogeneous. There's a significant chunk of mass (the head, according to Gray's Anatomy) extending to one side at the upper ("proximal") end. And the shaft is often not straight, though that's not shown here, and the bone isn't homogeneous, so local mass densities aren't constant. When rotating, the asymmetry means that in addition to purely tensile stress (which @TheNate correctly identified stress as the primary metric of interest) there is a bending moment applied to the shaft. At a given cross-section of the shaft (say, for instance, at the center of mass) this bending moment decreases the stress on one side and increases it on the other. Assuming no pre-existing cracks in the bone, the first break would occur where that increased stress occurs and quickly spread as a Griffith crack across the entire shaft. This would happen at a rotation rate slower than those calculated above.

Great example! The roll looks really pretty gentle; I was surprised at how early it came apart. I guess they really hadn't expected that kind of centrifugal stress on the booster. — Russell Borogove, Jul 17 '18 at 21:18
@RussellBorogove Yep, the gyros on that flight were supposed to prevent roll, not cause it! So they probably had fairly beefy structures for axial loads, not for radial loads. If you look carefuly at the video, it's the unignited Algol motors that break free first: with their full fuel load still there, they have more mass and thus more centrifugal force. — Tom Spilker, Jul 17 '18 at 21:35
@TomSpilker wow! Awesome! I was honestly mostly curious about the point of failure and how it would shift due to weight distribution, this helps a lot with that. Pointing out that we have "two masses spinning" instead of one makes a lot of sense. The explanation of why it would break sooner provided the actual structure (due to masses being concentrated at the ends) was also very enlightening :). Thanks for taking the time to edit. — Magic Octopus Urn, Jul 19 '18 at 12:13

score 14 · Answer 3 · answered Jul 17 '18 at 14:56

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Reaction wheel? No, these are way overpowered in KSP, and don't exert nearly as high torque in real spacecrafts, never mind the issue of saturation (maximum speed of reaction wheel motor).

But asymmetric thrust is a definite danger. If some engines of a rocket fail to fire, or stop firing early, the remaining ones will send it into a spin. And that may end really badly. And not just on ascent - a stuck RCS thruster in orbit is equally bad news.

answered Jul 17 '18 at 14:56

SF.

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Thank you for the clarification, I've edited the question to rule-out reaction wheels and focus more on the spinning problem, but gave you +1 for the clarification :). – Magic Octopus Urn Jul 17 '18 at 16:25
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And KSP reaction wheels do not saturate. Real world ones do. – Loren Pechtel Jul 17 '18 at 21:44
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@LorenPechtel: Note though that KSP puts a hard limit on RPM of any in-game object. – SF. Jul 17 '18 at 21:51
It does? I’ve definitely seen rotations fast enough to appear aliased (i.e. > 180 degrees in 1/60 sec = 1800 RPM). – Russell Borogove Jul 18 '18 at 13:45
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@RussellBorogove: I think explosions/collisions are exempt; possibly propulsive rotation too. You can't reach arbitrary speeds with reaction wheels though. – SF. Jul 18 '18 at 13:48
@SF. that's a limit on the reaction wheel, not the rotation of in-game objects ;) – Baldrickk Jul 18 '18 at 15:32
I'd say it's even worse in orbit - there's no atmosphere to slow you down and maybe give you extra time to correct the problem. – corsiKa Jul 20 '18 at 04:47

Russell Borogove · Answer 4 · 2018-07-18T19:33:18.610

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Yes, absolutely.

Every part of a spinning, rigid structure is undergoing acceleration towards the center, proportional to the radius from the axis of rotation and to the square of the angular velocity. This is technically centripetal force, but in the local neighborhood of a body that could potentially become disconnected from the structure, it can be treated as an outward force in a rotating frame, so it's commonly called centrifugal force.

Because the force increases with the radius from the rotation axis, the outermost parts of the spacecraft are subject to the most force; if everything is equally securely fastened, the outermost bits will fall off first.

When parts fall off, they will depart with the instantaneous velocity they had at disconnection, i.e. on a line tangent to their radius of rotation. In the rotating frame of reference such a part was in before it fell off, this straight path appears approximately parabolic; the motion is very much like acceleration due to gravity.

The effect of this force can in principle be used to simulate gravity in a rotating spacecraft for the benefit of passengers, but for this purpose it's preferable to use a larger radius and a slower rotation in order to minimize Coriolis effect.

Kerbal Space Program's structural attachments are generally stronger than those of real-life spacecraft, but I believe you can make things fall off a spinning craft within the game.

(SF. is correct that you probably can't spin a craft fast enough with reaction wheels to cause damage to anything but the reaction wheels.)

edited Jul 18 '18 at 19:33

answered Jul 17 '18 at 14:54

Russell Borogove

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Every part of a spinning, rigid structure is undergoing acceleration towards the center, proportional to the radius from the axis of rotation and to the square of the angular velocity. - This, thank you. I'll have to revisit angular momentum and such concepts as I never actually got too far into that theory in my Physics courses. – Magic Octopus Urn Jul 17 '18 at 15:59
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@MagicOctopusUrn You can get some understanding from experiments with everyday objects. Tie a tennis ball to a string, and spin it over your head in a circle. If you spin fast enough, the string will break or the tennis ball will become unattached and fly off. You could do the same thing with a metal rod - you'd have to spin it a lot faster to break it, but it's the exact same principle. – Nuclear Hoagie Jul 17 '18 at 16:19
@NuclearWang interesting, I'd love to have a ballpark on how fast something like a tungsten rod of X length and Y diameter would have to spin to actually rupture. I added to the question a quantitative problem to grasp the scale of how fast you'd need to rotate to break a tensile object (average human femur). – Magic Octopus Urn Jul 17 '18 at 16:23
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FWIW, Nuclear Wang's comment about spinning metal pieces--sometimes jet engines throw a blade. That's how fast you have to spin pieces of metal before pre-weakened bits fly off. – Jul 17 '18 at 17:20
Thanks again for the link to the Coriolis force, I'm halfway through and feel like I understand a lot more about orbital mechanics, weather patterns, tidal patterns and various other topics. This is extremely cool stuff :). – Magic Octopus Urn Jul 17 '18 at 17:40
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@MagicOctopusUrn The topic you're asking about is called "engineering mechanics" "Dynamics" and "material science" are the areas of particular relevance, here. Specifically, centrifugal force, centripetal acceleration/force, strain, stress (particularly sheer and axial strain), Young's modulus, and trusses should be pretty much everything you'd need to run the calculations. If you're serious about reading up, that list of terms should get you sorted. – The Nate Jul 17 '18 at 21:07
@TheNate thank you! I never went much past Hooke's law in my curriculum, I'm looking into Young's modulus and the difference between sheer/axial after I finish up reading the page on Coriolis Force :). – Magic Octopus Urn Jul 17 '18 at 21:13
@MagicOctopusUrn: In a rotating reference frame, centrifugal force and centripetal force are the "equal and opposite forces" of Newton's 3rd law. In an inertial reference frame, centrifugal force doesn't exist. (Some people like to claim centrifugal force doesn't exist and centripetal is the only real one, but they're ignoring reference frames that move with the edge of the merry-go-round or whatever. https://xkcd.com/123/) – Peter Cordes Jul 18 '18 at 12:10
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The second paragraph of this answer seems to imply that centripetal force and centrifugal force are the same thing, just in different reference frames. That's not true, and they're not necessarily equal: i.e. they're not when the structure fails and deconstruction happens. Centripetal force is provided by the structure; centrifugal force is from the inertia of a mass under rotation. I think that needs to be clarified. – Erin Anne Jul 18 '18 at 19:32
I've slightly changed the wording in the hope of more accuracy, but I'm mainly relying on the linked WP pages to get it right. My understanding stops at XKCD 123... – Russell Borogove Jul 18 '18 at 19:37

score 4 · Answer 5 · answered Jul 17 '18 at 19:29

Could a spacecraft spin so fast that it deconstructs itself? What force would cause this (if it would) and which parts would be most susceptible to this force?

It's not a matter of a force causing it, so much as a lack of a force. According to Newton's First Law, an object remains at constant velocity unless acted on by some force. Since velocity includes both speed and direction, an object moving at constant velocity is either stationary or moving in a straight line.

In order for an object (such as part of a spaceship) to travel in a circular path, it must be continually accelerated toward the center of that path. Newton's First Law tells us that this is not possible unless some force is acting on it. Newton's Second Law tells us that this force is equal to the mass of the object multiplied by its acceleration.

In the example of a rotating object, whatever part is holding the outside parts onto the spacecraft is exerting a force onto those outside parts that is directed toward the center of the spacecraft. When the rotational speed increases to the point that the acceleration required to hold those outside parts in a circular path rises above the maximum force whatever is holding the outside object on can provide divided by the mass of the outside object, the part hold the outside object on will fail and the outside object will fly off in a straight line.

As Nuclear Wang mentioned in a comment, you can observe this same behavior with a ball on a string. When you spin the ball on the string, the string is exerting a force (called tension) on the ball toward the center of the rotation. When you spin the ball fast enough that the acceleration required to hold the ball in this circular path reaches the maximum tension force the string can provide divided by the mass of the ball, the string will break and the ball will fly off. If it weren't for air friction and Earth's gravity (which are also forces acting on the ball,) the ball would fly off in a straight line.

score 0 · Answer 6 · answered Jul 18 '18 at 19:25

Possibly related is a problem examined by Randall Monroe of xkcd fame in his What If article about the fastest way to complete a NASCAR race:

https://what-if.xkcd.com/116/

In the latter half of the article, he describes a system that is rotating around a central point. The limit of the speed of this spin is the breaking strength of the material doing the spinning. To quote the relevant entry (though I'd recommend reading the whole thing -- and the rest of the archive. It's good stuff!):

Imagine a "vehicle" anchored with Kevlar straps to a pivot in the center, reinforced with a counterweight on the other side. In effect, this is a giant centrifuge. This lets us apply one of my favorite weird equations, which says that the edge of a spinning disc can't go faster than the square root of the specific strength (tensile strength divided by density) of the material it's made of. For strong materials like Kevlar, this speed is 1-2 km/s. At those speeds, a capsule could conceivably finish the race in about 10 minutes—although definitely not with a living driver inside.

So in your case, the question becomes: what structural material in a spaceship has the lowest specific strength? Once it fails due to the rotational stress, the whole thing will presumably fail.

score 0 · Answer 7 · answered Jul 19 '18 at 22:33

I see the issue as being about the spin axis. End-over-end is how a long cylinder 'wants' to spin. They had issues with this with early satelites which wouldn't stay spinning around their length, because any flex in the material would transfer momemtum to the end-over-end tumbling rotation. As the rotational momentum increases, there is more momentum to transfer, and any deviation from perfect stiffness of the material would be amplified. The top and bottom would start to wobble.

And once it switches from twisting to tumbling, the force and temperature differentials would quickly rip it apart.

score 0 · Answer 8 · answered Jul 21 '18 at 21:44

That's sort of the rule. You need to take active measures against it. Any continuing acceleration/thrust from a point fixed to the solid frame of a craft that isn't exactly pointed at its center of gravity will cause a rotation around the center of gravity and continuing in the acceleration will increase the amount of rotation. There is no inherent limit or braking in space. A mitigating factor is that stable modes of rotation are only about the eigenvectors of the inertial tensor corresponding to the largest and smallest eigenvalue: those modes of rotation cannot decay to other axes. But any continuing contribution to rotation around the two stable axes of rotation will accumulate, and the resulting centripetal forces keeping the solid together will eventually break any material.

Could a spacecraft spin so fast that it spontaneously deconstructs?

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