Question on hoop stress tension in rotating objects

[Chestermiller]

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 $d\theta$

 

[LT Judd]

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?

 

[Chestermiller]

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?

 

[pbuk]

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.

 

[LT Judd]

$i_r$ 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 .

 

[Chestermiller]

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 are aware that you can cause a ring to rotate about its axis as fast as you want, and this can create centripetal accelerations far exceeding g, with associated huge hoop tensions, right?