Question on hoop stress tension in rotating objects

In summary, the conversation discusses the concept of hoop stress, which arises from internal forces resisting deformation caused by radial forces in a spinning object. There is a debate about whether a greater inward force opposing the radial centrifugal force would result in compression or tension in the hoop stress. The respondent believes that hoop stress is produced by radial forces, and if an opposing force is greater than the centrifugal force, there would be net compression instead of tension. The other person believes that hoop stress is produced through some other means and would still occur even with a greater inward force.
  • #1
paulsterx
12
1
TL;DR Summary
Is the hoop stress tangential force acting on a spinning solid object a direct result of the radial (centrifugal) force occurring due to its rotation?
Hi, I have a question about hoop stress or tangential force acting within a spinning object such as a solid flywheel. As described in a textbook I’ve seen, the hoop stress tension force acting as if across the diameter of the object, trying to pull it apart, is a resultant of forces acting within that object. My question is, is that resultant tangential force the resultant of radial forces (centrifugal force) acting within the object due to rotation? Or something else? Again, my understanding is that if you were to calculate the vector sum of all centrifugal forces acting on the object (calculating centrifugal force as if it’s a real force) in one specific direction, the resultant of these radial forces would equal the tangential tension force (in a simplistic ideal situation with no other forces or masses taken into account). Is my understanding correct?

As a follow up, if the radial forces acting away from the centre of rotation were opposed by a much greater force acting towards the axis, would there still be a tangential tension force acting on the object, or would the net result now be a compression force acting to push the object together?

This is to settle a debate with a colleague!

Many thanks in advance!

Paul
 
Engineering news on Phys.org
  • #2
paulsterx said:
My question is, is that resultant tangential force the resultant of radial forces (centrifugal force) acting within the object due to rotation?
Hoop stress arises from the internal forces, resisting the deformation caused by radial forces, keeping a body together.

1637167951622.png

In this diagram, if the internal pressure is very large, the circle will tend to expand, increasing its circumference. If we were to cut this circle, lay it flat, and pull it from both ends, we would see a similar increase in length. In the simple tension case, it's easily apparent that our stretching it is producing internal tensions which resist the stretching force. Hoop stress is similar in this regard.

paulsterx said:
As a follow up, if the radial forces acting away from the centre of rotation were opposed by a much greater force acting towards the axis, would there still be a tangential tension force acting on the object, or would the net result now be a compression force acting to push the object together?
No longer tangential tension, but net tangential compression. That said, there is a bit more to these complex stress states than simply tangential compression/tension.
 
  • Like
Likes paulsterx
  • #3
Thanks so much for your reply! Just to be sure can I ask what your thoughts are about what he thinks on this subject? In his words: “There is no radial force in an object that spins like a top, proven by the fact that there is no pulling force on the axis.” He quotes the Machinery Handbook: “if a body rotates around it’s own centre of mass, R equals zero and V equals zero. This means that the resultant of the centrifugal forces of all the elements of the body is equal to zero, in other words no centrifugal force is exerted on the axis of rotation.” Then he concludes from this “so, opposing a radial force does not in any way oppose or ‘prevent from occurring in the first place’ hoop stress, (which is due to a force that acts across the diameter) which requires molecular bonds and a spinning body to develop”

His view is that no matter whether a greater force may oppose the radial centrifugal force, hoop stress will still exist in the spinning body. Mine is that if a force greater than the outward radial force is acting towards the axis of rotation, hoop tensional stress (as a resultant of a net radial force away from the axis) cannot occur and instead there will be a net compression.

I think from your reply you agree with my take rather than his, would this be fair?
 
  • #4
Let's just talk about the case where you have a circular ring rotating within its plane around the axis of the ring. Do you think that there is a hoop stress in the ring, and if so, what produces this hoop stress?
 
  • Like
Likes Baluncore
  • #5
Hi Chester, my understanding is that hoop stress in the ring is produced by the radial centrifugal forces experienced by that ring and is equivalent, in anyone specific direction, to the vector sum of the radial force magnitudes in that direction of one half of the ring. Were there to be an opposing force greater than centrifugal force acting on the ring towards the axis (‘inwards’) then hoop stress tension would not occur and instead there would be a net compression force acting on the ring.

The person I’m speaking to seems to think hoop stress is produced not from the radial forces, but through some other means, and no matter how great an inwards radial force was applied, hoop stress tension would still be present in the ring.

Thanks for your reply!
 
  • #6
paulsterx said:
He quotes the Machinery Handbook: “if a body rotates around it’s own centre of mass, R equals zero and V equals zero. This means that the resultant of the centrifugal forces of all the elements of the body is equal to zero, in other words no centrifugal force is exerted on the axis of rotation.”
The very next sentence in my copy of Machinery's Handbook is "The centrifugal force of any part or element of such a body is found by the equations given below, where R is the radius to the center of gravity of the part or element..."

So even though the resultant of the centrifugal forces of all the elements of the body is equal to zero, that doesn't mean that individual parts or elements of a body aren't subjected to radial forces which cause internal stresses.

There is probably some extreme case to be considered...
 
  • #7
Ok, so essentially hoop stress is basically the radial force but measured in a different way, tangentially rather than radially?
 
  • #8
paulsterx said:
Hi Chester, my understanding is that hoop stress in the ring is produced by the radial centrifugal forces experienced by that ring and is equivalent, in anyone specific direction, to the vector sum of the radial force magnitudes in that direction of one half of the ring. Were there to be an opposing force greater than centrifugal force acting on the ring towards the axis (‘inwards’) then hoop stress tension would not occur and instead there would be a net compression force acting on the ring.

The person I’m speaking to seems to think hoop stress is produced not from the radial forces, but through some other means, and no matter how great an inwards radial force was applied, hoop stress tension would still be present in the ring.

Thanks for your reply!
So you think that, if it were a rotating disc instead of a rotating ring, there would be no hoop stress in the disc?
 
  • #9
Chestermiller said:
So you think that, if it were a rotating disc instead of a rotating ring, there would be no hoop stress in the disc?
No, he thinks (I do not agree) that in a rotating disc the hoop stress is not caused by a radial force. My understanding is that hoop stress is basically the radial force (they are one and the same), measured tangentially rather than radially. He thinks that although there is hoop stress, that no radial force exists within the disk.
 
  • #10
paulsterx said:
No, he thinks (I do not agree) that in a rotating disc the hoop stress is not caused by a radial force. My understanding is that hoop stress is basically the radial force (they are one and the same), measured tangentially rather than radially. He thinks that although there is hoop stress, that no radial force exists within the disk.
Back to the ring problem. We can settle this all with a free body diagram. Let's see your free body diagram for a small section of the ring between angular location ##\theta## and angular location ##\theta+\Delta \theta##, with T representing the tension in the hoop.
 
  • #11
Chestermiller said:
So you think that, if it were a rotating disc instead of a rotating ring, there would be no hoop stress in the disc?
These are his words: “hoop stress is due to molecular bonds resisting a force that acts along the diameter created by spinning a solid body around a centre of mass…while outward radial force is due to orbiting said body about an axis away from its centre of mass…just because the equations for the former are BASED on the classic equation from mechanics does not mean they are ‘one and the same force’ they are not. The moment you start spinning the solid body about its centre of mass is the moment you experience hoop stress,”

I have to say I’m not 100% sure what he thinks here - he’s sort of right in some things but I think he believes there’s no radial force occurring within a solid spinning body, due to the force on the axis being zero, and so hoop stress in a solid object is NOT caused by a net outward radial force.
 
  • #12
This might be easier to understand if we look at the forces on an element in a spinning disk. This free body diagram is copied from the chapter on rotating disks in Advanced Strength of Materials by J.P. Den Hartog. The book is still in print: https://www.amazon.com/dp/0486654079/?tag=pfamazon01-20.
Flywheel.jpg

The element has a body force due to centrifugal force on the mass, radial stresses, and hoop (tangential) stresses. The book takes several pages to cover the integration of those forces and stresses, resulting in equations which you can find by searching flywheel stress. If you find different equations, look closely at the boundary conditions - hollow vs solid center, compressive stress in center hole, disk vs constant stress design vs rim and spoke design, etc.
 
  • Like
Likes paulsterx and Astronuc
  • #13
paulsterx said:
These are his words: “hoop stress is due to molecular bonds resisting a force that acts along the diameter created by spinning a solid body around a centre of mass…while outward radial force is due to orbiting said body about an axis away from its centre of mass…just because the equations for the former are BASED on the classic equation from mechanics does not mean they are ‘one and the same force’ they are not. The moment you start spinning the solid body about its centre of mass is the moment you experience hoop stress,”

I have to say I’m not 100% sure what he thinks here - he’s sort of right in some things but I think he believes there’s no radial force occurring within a solid spinning body, due to the force on the axis being zero, and so hoop stress in a solid object is NOT caused by a net outward radial force.
Like I said, let's see your free body diagram of a small section of the ring.
 
  • #14
Chestermiller said:
Like I said, let's see your free body diagram of a small section of the ring.
Hi Chester, I’m sorry but I’m not trained in drawing free body diagrams, I’m not sure exactly what you’re after? I’m not an engineer and haven’t trained as one, which is why I’m asking this question here. It doesn’t need to get complex, take the simplest situation. Basically it’s 2 questions:

Is circumferential tension caused by radial (centrifugal) force?

If another force far greater than centrifugal force was acting on the body towards the axis of rotation, would hoop stress still be present in the body or would there instead be a compression force acting to push the body together, rather than trying to pull it apart?

Thanks!
 
  • #15
paulsterx said:
Hi Chester, I’m sorry but I’m not trained in drawing free body diagrams, I’m not sure exactly what you’re after? I’m not an engineer and haven’t trained as one, which is why I’m asking this question here. It doesn’t need to get complex, take the simplest situation. Basically it’s 2 questions:

Is circumferential tension caused by radial (centrifugal) force?
The circumferential tension is in balance with the radial centrifugal force. The direction of the circumferential tension varies with circumferential position around the ring, and this variation of tension direction results in a net force (per unit length around the circumference) necessarily to accelerate the particles of ring mass radially inward toward the center of the ring.
paulsterx said:
If another force far greater than centrifugal force was acting on the body towards the axis of rotation, would hoop stress still be present in the body or would there instead be a compression force acting to push the body together, rather than trying to pull it apart?

Thanks!
If another force far greater than the centrifugal force was acting on the ring towards the axis of rotation, a hoop stress would be present, but this hoop stress would be compressional in the circumferential direction (rather than tensile)..
 
  • #16
paulsterx said:
Ok, so essentially hoop stress is basically the radial force but measured in a different way, tangentially rather than radially?
Let me try my hand here. If I miss the mark, so be it:
The system I consider is a bicycle wheel with spokes and a thin flexible rim. As I increase the rotational speed of the wheel, centripetal forces must be applied to each part of the rim to keep it from disintegrating. In a real bicycle wheel these forces come from both circumferential tension in (along) the rim and radial tension in the spokes. The balance between them depends in detail upon the stretching of the component parts.
A solid disk is even more complicated because there are shear forces and moduli in addition to the tensile ones. As usual the answer is the forces are distributed in a complicated way and all contribute to the integrity of the wheel.
paulsterx said:
Summary:: Is the hoop stress tangential force acting on a spinning solid object a direct result of the radial (centrifugal) force occurring due to its rotation?

This is to settle a debate with a colleague!
The answer is "it depends". The forces come from everywhere: the net accelerations are centripetal (meaning towards the center)
 
  • #17
Chestermiller said:
The circumferential tension is in balance with the radial centrifugal force. The direction of the circumferential tension varies with circumferential position around the ring, and this variation of tension direction results in a net force (per unit length around the circumference) necessarily to accelerate the particles of ring mass radially inward toward the center of the ring.

If another force far greater than the centrifugal force was acting on the ring towards the axis of rotation, a hoop stress would be present, but this hoop stress would be compressional in the circumferential direction (rather than tensile)..
Hi Chester, many thanks for your time, it’s much appreciated. I have a confession to make as I’ve not been entirely honest with you. The person I’ve been speaking to is a flat earther who has tried for a long time to use this argument (that tensional hoop stress is always present in a rotating body no matter what other forces are present), to try and argue that the Earth should ‘tear itself apart’ due to hoop stress and thus cannot be a rotating globe. I’ve tried to explain again and again that hoop stress is a result of radial forces and that gravity, with at least 300x the inward radial force compared to centrifugal force at the equator, prevents hoop stresses from forming within the Earth and instead results in a net compression of the earth, as in any other planet sized object.

He ‘challenged’ me to get validation of this from a qualified engineer, these are his words:

“if an engineer will state that "gravity" can "prevent hoop stress from
occurring in the first place" AND enable a cracked
basalt water covered pearoid to spin over 1000 mph
Make sure he agrees that "gravity" opposes
"the force that acts along the diameter', AND he
explains how this force exists in a cracked body, He
must disagree with my textbook explanation: Hoop
stress is created by spinning a solid body and
molecular bonds resist the force that acts along any diameter. And prove that he’s qualified”

So here’s my take on that statement.

Gravity does prevent tensional hoop stress from occurring, by opposing radial centrifugal force and causing a net force towards the earth’s centre, resulting in compression, not tension.

By creating a net compression through the earth, gravity holds it together and enables a ‘cracked basalt covered pearoid’ to rotate at 15 degrees per hour (or 1000mph linear speed at the equator, as flat earther like to say as they like big impossible sounding numbers).

Gravity does not directly oppose a force acting across the diameter (hoop stress), rather it prevents tensional hoop stress forming in the first place.

I assume you’d agree with the ‘text book explanation’ as hoop stress is indeed caused by rotation and molecular bonds do resist this stress - if it exists in the first place, which is not the case within the Earth due to gravity.

Feel free to ignore this, but your views as a qualified engineer would be really interesting! As I said I’m not an engineer, but I do have some physics education - there may be some details I’m not quite correct on. Apologies for not telling you this sooner, but I didn’t want to be accused of causing bias by bringing the Earth and gravity into the equation from the start.

Thanks,

Paul
 
  • #18
Keep fighting the good fight. Luddites (and worse) are everywhere. Is your friend vaccinated?
 
  • Like
Likes paulsterx
  • #19
paulsterx said:
Hi Chester, many thanks for your time, it’s much appreciated. I have a confession to make as I’ve not been entirely honest with you. The person I’ve been speaking to is a flat earther who has tried for a long time to use this argument (that tensional hoop stress is always present in a rotating body no matter what other forces are present), to try and argue that the Earth should ‘tear itself apart’ due to hoop stress and thus cannot be a rotating globe. I’ve tried to explain again and again that hoop stress is a result of radial forces and that gravity, with at least 300x the inward radial force compared to centrifugal force at the equator, prevents hoop stresses from forming within the Earth and instead results in a net compression of the earth, as in any other planet sized object.

He ‘challenged’ me to get validation of this from a qualified engineer, these are his words:

“if an engineer will state that "gravity" can "prevent hoop stress from
occurring in the first place" AND enable a cracked
basalt water covered pearoid to spin over 1000 mph
Make sure he agrees that "gravity" opposes
"the force that acts along the diameter', AND he
explains how this force exists in a cracked body, He
must disagree with my textbook explanation: Hoop
stress is created by spinning a solid body and
molecular bonds resist the force that acts along any diameter. And prove that he’s qualified”

So here’s my take on that statement.

Gravity does prevent tensional hoop stress from occurring, by opposing radial centrifugal force and causing a net force towards the earth’s centre, resulting in compression, not tension.

By creating a net compression through the earth, gravity holds it together and enables a ‘cracked basalt covered pearoid’ to rotate at 15 degrees per hour (or 1000mph linear speed at the equator, as flat earther like to say as they like big impossible sounding numbers).

Gravity does not directly oppose a force acting across the diameter (hoop stress), rather it prevents tensional hoop stress forming in the first place.

I assume you’d agree with the ‘text book explanation’ as hoop stress is indeed caused by rotation and molecular bonds do resist this stress - if it exists in the first place, which is not the case within the Earth due to gravity.

Feel free to ignore this, but your views as a qualified engineer would be really interesting! As I said I’m not an engineer, but I do have some physics education - there may be some details I’m not quite correct on. Apologies for not telling you this sooner, but I didn’t want to be accused of causing bias by bringing the Earth and gravity into the equation from the start.

Thanks,

Paul
If the Earth were not rotating, would he agree that, near the surface, the hoop stress is compressional. If so, then just argue that the centrifugal force, even at 1000 mph tangential rotation speed, would not be enough to overcome gravity, and gravity would win out. Otherwise people at the equator would be rising into space. So the hoop stress, even at the equator, would still have to be compression.
 
  • #20
Chestermiller said:
If the Earth were not rotating, would he agree that, near the surface, the hoop stress is compressional. If so, then just argue that the centrifugal force, even at 1000 mph tangential rotation speed, would not be enough to overcome gravity, and gravity would win out. Otherwise people at the equator would be rising into space. So the hoop stress, even at the equator, would still have to be compression.
I don’t think he would, no. Don’t forget that flerfs don’t believe gravity even exists - they think objects fall due to density differential alone - yes, I know, density is a scaler and has neither magnitude nor direction. He’s read that hoop stress forces act ‘as if across the diameter’ and run with that, thinking that the force creating hoop stress is something other than radial force. His conclusion is that radial forces only act on ‘objects at the surface’ and that within a solid body the force relationships are completely different, and he’s not entirely wrong, but it is still the case (unless you correct me), that hoop stress is essentially caused by radial centrifugal force and if the net radial force acts towards the axis of rotation, there will be no tensional force in the object.

Would you agree with my take on his statement in the previous comment?
 
  • #21
hutchphd said:
Keep fighting the good fight. Luddites (and worse) are everywhere. Is your friend vaccinated?
Of course not! Flat Earth and anti Vax go together like peas and carrots.
 
  • Like
Likes hutchphd
  • #22
Chestermiller said:
Gravity does not directly oppose a force acting across the diameter (hoop stress), rather it prevents tensional hoop stress forming in the first place.
No. Gravity does exactly create a force directly across the diameter. And we know exactly how fast the Earth's surface would need to spin to counteract that force, for this is the tangent speed at which satellites in low Earth orbit need to attain: 18,000 mph. So if the Earth spun at 18 times its present rate, pieces would spontaneously start going into orbit because gravity would be insufficient.
I hope your friend does not die from his idiocy, I know two people who have recently done so.
EDIT: I meant to reference the original quote from @paulsterx
 
  • #23
paulsterx said:
I don’t think he would, no. Don’t forget that flerfs don’t believe gravity even exists - they think objects fall due to density differential alone - yes, I know, density is a scaler and has neither magnitude nor direction. He’s read that hoop stress forces act ‘as if across the diameter’ and run with that, thinking that the force creating hoop stress is something other than radial force. His conclusion is that radial forces only act on ‘objects at the surface’ and that within a solid body the force relationships are completely different, and he’s not entirely wrong, but it is still the case (unless you correct me), that hoop stress is essentially caused by radial centrifugal force and if the net radial force acts towards the axis of rotation, there will be no tensional force in the object.

Would you agree with my take on his statement in the previous comment?
It doesn't pay to waste your valuable time arguing with someone like that.
 
  • #24
hutchphd said:
A solid disk is even more complicated because there are shear forces and moduli in addition to the tensile ones. As usual the answer is the forces are distributed in a complicated way and all contribute to the integrity of the wheel.
The principal stresses for a rotating disc are in the radial direction and in the hoop direction. There are no shear stresses between these directions.
 
  • Like
Likes hutchphd
  • #25
hutchphd said:
No. Gravity does exactly create a force directly across the diameter. And we know exactly how fast the Earth's surface would need to spin to counteract that force, for this is the tangent speed at which satellites in low Earth orbit need to attain: 18,000 mph. So if the Earth spun at 18 times its present rate, pieces would spontaneously start going into orbit because gravity would be insufficient.
I hope your friend does not die from his idiocy, I know two people who have recently done so.
EDIT: I meant to reference the original quote from @paulsterx
Yes agreed, his diatribe is that because hoop stress is a tangential force and gravity is a radial force, there’s no way that gravity could oppose hoop stress. My point is that for (tensional) hoop stress to exist, there must be a net force acting away from the axis of rotation and that due to gravity, the net force is towards the axis, therefore there is no tension stress, rather compression.

They’re not called Skiba and Focka by any chance are they?
 
  • #26
No but they are certainly deceased. Good Luck.
 
  • #27
hutchphd said:
No but they are certainly deceased. Good Luck.
Oh ok sorry, I thought you might be referring to 2 well known (within the flat Earth community anyway) flat earthers and Covid deniers who both recently died, from covid.
 
  • #28
Missed the reference. I try not to think Darwinian thoughts at times like these...not always successfully!
 
  • #29
Chestermiller said:
Like I said, let's see your free body diagram of a small section of the ring.
Good Day Chester,
I thought I would have a crack at the FBD. see attached- does that look right?
Being a ring rather than a disk I assumed that there is no inward radial force, not sure if that's correct.
So based on the FBD and the arising formula, it appears that the hoop forces could be huge.
Note this is based on a general thin ring, not for the "earth case", in that case you would also include gravity, as an inward force, I guess.
so given that there is a sineis also very small, it, i i small,
FBD thin ring.jpg
 
  • #30
This is close, but not quite right. If the magnitude of the tension is T, the tension vectorially is ##Ti_{\theta}## where ##i_{\theta}(\theta)## is the unit vector in the ##\theta## direction. So the force balance should read: $$(Ti_{\theta})_{\theta+\Delta \theta/2}-(Ti_{\theta})_{\theta-\Delta \theta/2}=-\rho(r\Delta \theta)r\omega^2i_r$$Differentially, this becomes $$T\frac{di_{\theta}}{d\theta}=-\rho r^2\omega^2i_r$$
 
  • #31
LT Judd said:
Good Day Chester,
I thought I would have a crack at the FBD. see attached- does that look right?
Being a ring rather than a disk I assumed that there is no inward radial force, not sure if that's correct.
So based on the FBD and the arising formula, it appears that the hoop forces could be huge.
Note this is based on a general thin ring, not for the "earth case", in that case you would also include gravity, as an inward force, I guess.
so given that there is a sineis also very small, it, i i small, View attachment 297052
I guess the only mistake here is that there should have been a 2##d\theta## in the numerator instead of just ##d\theta##
 
Last edited:
  • Like
Likes hutchphd
  • #32
Sorry , totally lost me there.
Is the units of T Newtons ( or kN) ?
and does ## i_r ## means the unit tension vector in the radial direction?
I also don't understand the notation ##i_\theta (\theta)##.
Why is ## \(theta) ## written twice and which one refers to the direction and what does the other refer to?
 
  • #33
LT Judd said:
Sorry , totally lost me there.
Is the units of T Newtons ( or kN) ?
and does ## i_r ## means the unit tension vector in the radial direction?
I also don't understand the notation ##i_\theta (\theta)##.
Why is ## \(theta) ## written twice and which one refers to the direction and what does the other refer to?
##i_r## is the unit vector (radial direction) in cylindrical polar coordinates and ##i_{\theta}## is the unit vector (angular direction ##\theta##). Geometrically, both these unit vectors depend on the polar angle ##\theta##: $$\frac{di_r}{d\theta}=i_{\theta}$$ and $$\frac{di_{\theta}}{d\theta}=-i_r$$

Are you not familiar with polar coordinates and use of unit vectors which change in spatial position?
 
  • #34
paulsterx said:
Yes agreed, his diatribe is that because hoop stress is a tangential force and gravity is a radial force, there’s no way that gravity could oppose hoop stress. My point is that for (tensional) hoop stress to exist, there must be a net force acting away from the axis of rotation and that due to gravity, the net force is towards the axis, therefore there is no tension stress, rather compression.
You don't need to calculate the tangential force to see that it is compressive: if it were not then the oceans would part.

There is no point arguing with a flat Earther with facts as they have already opted to ignore certain facts that do not support their delusion.
 
  • #35
Chestermiller said:
##i_r## is the unit vector (radial direction) in cylindrical polar coordinates and ##i_{\theta}## is the unit vector (angular direction ##\theta##). Geometrically, both these unit vectors depend on the polar angle ##\theta##: $$\frac{di_r}{d\theta}=i_{\theta}$$ and $$\frac{di_{\theta}}{d\theta}=-i_r$$

Are you not familiar with polar coordinates and use of unit vectors which change in spatial position?
Hi Chester,
Sorry for the late reply, I wanted to have a long think about it. Yes I am familiar with polar co-ordinates and unit vectors , but I dare say not as much as you are. I just struggled with some of your notations and sign conventions.

Anyhow , I decided to look at this another way however and realized that I should have accounted for gravity in my FBD as it turn out that has a far greater effect that an any centripetal force or hoop stress.

You can deduce a formula for centripetal acceleration on a spinning sphere with some elementary physics: a=V^2/R, where V = linear speed at equator and R is the radius
For earth, V is 465 m/s ( circumeference of Earth divided by length of day), and the radius of the Earth is 6.4 million metres.
This gives a centripetal acceleration of 0.03 m/s^2 . Considering gravity at 9.8 m/s^2 is about 300 times greater and in the opposite direction, there is obviously no risk of anything flying outward and I think I can also conclude that there are no net hoop stresses on the Earth's crust. Rather the Earth's crust (and oceans) would be overall in compression.

As the previous poster noted , simple observation can tell you that , but it's been a good mental excersize to look into the physics behind it .
 
<h2>1. What is hoop stress tension in rotating objects?</h2><p>Hoop stress tension is a type of stress that occurs in objects that are subjected to rotational forces. It is caused by the centrifugal force, which pulls the object outward from its center, creating tension along the circumference of the object.</p><h2>2. How is hoop stress tension calculated?</h2><p>Hoop stress tension can be calculated using the formula σ = ρω²r, where σ is the hoop stress tension, ρ is the density of the object, ω is the angular velocity, and r is the distance from the center of rotation to the outer edge of the object.</p><h2>3. What are the effects of hoop stress tension on rotating objects?</h2><p>Hoop stress tension can cause deformation or failure in rotating objects if the stress exceeds the object's strength. It can also cause cracks or fractures along the circumference of the object.</p><h2>4. How can hoop stress tension be reduced?</h2><p>Hoop stress tension can be reduced by increasing the object's strength, decreasing the rotational speed, or increasing the distance from the center of rotation to the outer edge of the object.</p><h2>5. What are some real-world examples of hoop stress tension?</h2><p>Hoop stress tension can be observed in various rotating objects, such as wheels, gears, turbines, and flywheels. It is also a critical factor in the design of structures such as bridges and wind turbines.</p>

1. What is hoop stress tension in rotating objects?

Hoop stress tension is a type of stress that occurs in objects that are subjected to rotational forces. It is caused by the centrifugal force, which pulls the object outward from its center, creating tension along the circumference of the object.

2. How is hoop stress tension calculated?

Hoop stress tension can be calculated using the formula σ = ρω²r, where σ is the hoop stress tension, ρ is the density of the object, ω is the angular velocity, and r is the distance from the center of rotation to the outer edge of the object.

3. What are the effects of hoop stress tension on rotating objects?

Hoop stress tension can cause deformation or failure in rotating objects if the stress exceeds the object's strength. It can also cause cracks or fractures along the circumference of the object.

4. How can hoop stress tension be reduced?

Hoop stress tension can be reduced by increasing the object's strength, decreasing the rotational speed, or increasing the distance from the center of rotation to the outer edge of the object.

5. What are some real-world examples of hoop stress tension?

Hoop stress tension can be observed in various rotating objects, such as wheels, gears, turbines, and flywheels. It is also a critical factor in the design of structures such as bridges and wind turbines.

Similar threads

Replies
5
Views
1K
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
636
  • Introductory Physics Homework Help
Replies
19
Views
786
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
966
  • Introductory Physics Homework Help
Replies
4
Views
153
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
733
Replies
9
Views
1K
Back
Top