B Artificial gravity rotating on two axes

[vanhees71]

No matter, how you write it, $\vec{\omega}$ has the dimension 1/time.

 

[erobz]

I get the mathematics, but I still don't "see it"

The rotation for the diagram I shown has a magnitude of the vector in the shaded plane, and is rotating about an axis through the origin in the direction of the vector. My brain is refusing to cooperate in understanding this.

 

[OmCheeto]

I'm not understanding something, is this following scenario not considered rotation about two axis at once?

View attachment 301451

I'm with you and Dave on this one. It sounds like everyone else is saying the two circular rotations add up to a new circular rotation for any given point on the object. But it's shown in the video you posted (#10) that the points are not following a simple circular paths.

 

[PeroK]

I'm with you and Dave on this one. It sounds like everyone else is saying the two circular rotations add up to a new circular rotation for any given point on the object. But it's shown in the video you posted (#10) that the points are not following a simple circular paths.

There is an external torque on the object in that video, so its angular momentum is not constant. As explained in post #18, rotation is the special case where the angular momentum is constant - or, at least, the axis of rotation is constant. The video shows an example of the more general motion where the instantaneous axis of rotation is changing with time.

In this case the applied torque changes its direction as the contact point moves around the circle.

In this respect, pure rotation is the equivalent of linear motion for Newton's first law. If you apply an external force, then an object may move in a curved path. And, if you apply a torque, then you may lose the pure rotational motion.

 

[vanhees71]

Let's see, how mathematically the angular velocity comes about. To this end think of a rigid rod with one end fixed and free to rotate around this point. Make $\vec{n}(t)$ the vector from the fixed end to the other end. Then it's clear that
$$\vec{n}(t) = \hat{D}(t) \vec{n}_0,$$
where $\vec{n}_0=\vec{n}(0)$ is the vector at time $T=0$, and $\hat{D}(t)$ is a rotation. In the following I consider $\vec{n}$ as the column vector of its components wrt. a Cartesian basis. Then $\hat{D}(t)$ is a orthogonal matrix, fulfilling $\hat{D}^{\mathrm{T}} \hat{D}=\hat{D} \hat{D}^{\mathrm{T}}=\hat{1}$ with $\mathrm{det} \hat{D}=1$.

Now let's take the time derivative:
$$\dot{\vec{n}}(t)=\dot{\hat{D}}(t) \vec{n}_0 = \dot{\hat{D}}(t) \hat{D}^{\text{T}}(t) \vec{n}(t). \qquad (*)$$
Now from the orthogonality of $\hat{D}$ we have
$$\dot{\hat{D}} \hat{D}^{\text{T}} = -\hat{D} \dot{\hat{D}}^T = - \left ( \dot{\hat{D}} \hat{D}^{\text{T}} \right)^{\text{T}},$$
i.e., $\hat{\Omega}=\dot{\hat{D}}(t) \hat{D}^{\text{T}}(t)$ is an antisymmetric matrix, and we thus can write
$$\Omega_{jk}=\epsilon_{jlk} \omega_l,$$
and thus from (*)
$$\dot{n}_j=\Omega_{jk} n_k = \epsilon_{jlk} \omega_l n_k = (\vec{\omega} \times \vec{n})_j,$$
or
$$\dot{\vec{n}}=\vec{\omega} \times \vec{n}.$$
One calls $\vec{\omega}$ the angular velocity. The change of the vector $n$ during a small time $\mathrm{d} t$ is
$$\mathrm{d} \vec{n} = \mathrm{d} t \vec{\omega} \times \vec{n}.$$
From the geometrical meaning of the dot product that means that the infinitesimal rotation is around an axis given by the direction $\vec{\omega}/|\vec{\omega}|$, implying the direction of rotation to be according to the right-hand rule, and the infinitesimal rotation angle is $|\vec{\omega}| \mathrm{d} t$.

Of course you can decompose $\vec{\omega}$ in terms of an arbitrary (Cartesian) basis, and the infinitesimal rotation can be seen as composed of several infinitesimal rotations around the corresponding axes, but still it's just a rotation around one given axis.

 

[erobz]

Let's see, how mathematically the angular velocity comes about. To this end think of a rigid rod with one end fixed and free to rotate around this point. Make $\vec{n}(t)$ the vector from the fixed end to the other end. Then it's clear that
$$\vec{n}(t) = \hat{D}(t) \vec{n}_0,$$
where $\vec{n}_0=\vec{n}(0)$ is the vector at time $T=0$, and $\hat{D}(t)$ is a rotation. In the following I consider $\vec{n}$ as the column vector of its components wrt. a Cartesian basis. Then $\hat{D}(t)$ is a orthogonal matrix, fulfilling $\hat{D}^{\mathrm{T}} \hat{D}=\hat{D} \hat{D}^{\mathrm{T}}=\hat{1}$ with $\mathrm{det} \hat{D}=1$.

Now let's take the time derivative:
$$\dot{\vec{n}}(t)=\dot{\hat{D}}(t) \vec{n}_0 = \dot{\hat{D}}(t) \hat{D}^{\text{T}}(t) \vec{n}(t). \qquad (*)$$
Now from the orthogonality of $\hat{D}$ we have
$$\dot{\hat{D}} \hat{D}^{\text{T}} = -\hat{D} \dot{\hat{D}}^T = - \left ( \dot{\hat{D}} \hat{D}^{\text{T}} \right)^{\text{T}},$$
i.e., $\hat{\Omega}=\dot{\hat{D}}(t) \hat{D}^{\text{T}}(t)$ is an antisymmetric matrix, and we thus can write
$$\Omega_{jk}=\epsilon_{jlk} \omega_l,$$
and thus from (*)
$$\dot{n}_j=\Omega_{jk} n_k = \epsilon_{jlk} \omega_l n_k = (\vec{\omega} \times \vec{n})_j,$$
or
$$\dot{\vec{n}}=\vec{\omega} \times \vec{n}.$$
One calls $\vec{\omega}$ the angular velocity. The change of the vector $n$ during a small time $\mathrm{d} t$ is
$$\mathrm{d} \vec{n} = \mathrm{d} t \vec{\omega} \times \vec{n}.$$
From the geometrical meaning of the dot product that means that the infinitesimal rotation is around an axis given by the direction $\vec{\omega}/|\vec{\omega}|$, implying the direction of rotation to be according to the right-hand rule, and the infinitesimal rotation angle is $|\vec{\omega}| \mathrm{d} t$.

Of course you can decompose $\vec{\omega}$ in terms of an arbitrary (Cartesian) basis, and the infinitesimal rotation can be seen as composed of several infinitesimal rotations around the corresponding axes, but still it's just a rotation around one given axis.

I get the mathematics

I remember once taking a class in linear algebra (I do still have the textbook)! I retract my earlier statement...

 

[DaveC426913]

From the geometrical meaning of the dot product that means that the infinitesimal rotation is around an axis given by the direction $\vec{\omega}/|\vec{\omega}|$, implying the direction of rotation to be according to the right-hand rule, and the infinitesimal rotation angle is $|\vec{\omega}| \mathrm{d} t$.

Once more, in tiny words?

 

[vanhees71]

What do you mean? What's not clear?

 

[erobz]

Once more, in tiny words?

I think it's going to be an impossibility to explain in a simplified way. There is no 1D conceptualization I'm afraid. Hence, the multi-dimensional analysis is a minimum model.

 

[erobz]

What do you mean? What's not clear?

I think he is giving a nod to the complexity of the issue. For me, I can understand most of the bits and pieces individually (probably), but putting it together...I'm not sure how long that would take.

 

[DaveC426913]

Presumably, the calcs are describing a real world scenario, yes? That real world scenario must be fairly easy to describe. For example: to the question of does any solid rotating body always have a single axis of rotation, presumably there's a yes/no.

 

[OmCheeto]

I think he is giving a nod to the complexity of the issue. For me, I can understand most of the bits and pieces individually (probably), but putting it together...I'm not sure how long that would take.

I don't understand any of it. Perhaps PF needs a level below "B" as apparently even high school maths is WAY over my head.

Anyone seen @Janus lately? I'd like him to render the dual axis spinning cylinder as it would work on the ISS. Clear, with just a single point being traced out.
I found a youtube video of a sphere with dual rotation, but it was solid with flowers, and my brain couldn't handle whether or not each point was actually traveling in a circle.

 

[Filip Larsen]

Presumably, the calcs are describing a real world scenario, yes? That real world scenario must be fairly easy to describe. For example: to the question of does any solid rotating body always have a single axis of rotation, presumably there's a yes/no.

The answer is yes. There are no cases where this is not so.

The rotation of a rigid object relative to any other frame of reference can always be completely described with a single angular velocity vector which, in general, vary over time both in direction and magnitude. This include both cases where the object is rotating freely in space with net external torque being zero or when it is constrained to rotate following multiple axis as shown in some of the video and thus experience a non-zero torque.

For example, if an object rotates due to be put into a device that has two axis of rotation, one "inside" or "after" the other, where the rotational speed around each of those two axes is simply set constant, then even as this can be seen as a "rotation around two axes" it can completely equivalently be model as a rotation around a single axis. When using rotation matrices to describe "individual" rotation then the effective rotation is found simply by matrix multiplication. If we first rotate around, say, the x-axis $R_x(t)$ and inside this (i.e. intrinsically) rotate around what was originally the y-axis $R_{y'}(t)$ then this combine into the effective rotation of $R(t) = R_{y'}(t) R_x(t)$. Another example of "combining rotation around multiple axes" would be Euler angles, where a sequence of three angles each with an implied axis to rotate around is used to model any orientation of an object.

That said, rotations in 3 dimension are indeed a fair bit complex with a lot of detail rich formalism and no one should expect they can intuitively understand the complexities of such rotations based on, say, their experience with 2D rotations or linear translatory motion.

 

[vanhees71]

Presumably, the calcs are describing a real world scenario, yes? That real world scenario must be fairly easy to describe. For example: to the question of does any solid rotating body always have a single axis of rotation, presumably there's a yes/no.

No, usually the momentary axis of rotation changes. That's always the case when $\vec{\omega} \neq \text{const}.$ For an elementary theory of the spinning top see

https://itp.uni-frankfurt.de/~hees/pf-faq/spinning-top.pdf

 

[Janus]

I don't understand any of it. Perhaps PF needs a level below "B" as apparently even high school maths is WAY over my head.

Anyone seen @Janus lately? I'd like him to render the dual axis spinning cylinder as it would work on the ISS. Clear, with just a single point being traced out.
I found a youtube video of a sphere with dual rotation, but it was solid with flowers, and my brain couldn't handle whether or not each point was actually traveling in a circle.

Okay, it took me a bit to work out a good way to to do it, and here's what I came up with.
I did two animations, as there are different ways to treat "rotating on two axis"

Both start from here:

A transparent cylinder(so nothing gets hidden during the rotations. A red sphere on the wall of the cylinder near an end and a white rod connecting this sphere to the long axis of the cylinder.

The first animation has the cylinder rotate on the y-axis (green) while it also rotates on its own long axis. (here, the blue line follows this long axis. The rotations are at a one to one ratio. It leaves a trail of spheres to show the whole path.

With the next one, the z axis (blue) remains fixed, while the cylinder rotates around this axis and the y axis.

 

[Filip Larsen]

Anyone seen @Janus lately? I'd like him to render the dual axis spinning cylinder as it would work on the ISS.

Okay, it took me a bit to work out a good way to to do it, and here's what I came up with.
I did two animations, as there are different ways to treat "rotating on two axis"

Nice animations

Note however, if "as it would work on the ISS" refers to what should happen when an astronaut in free fall onboard ISS spins an object for the camera, i.e. when the object is put in torque free rotation, then note these animations do NOT show that. What they do appear to show is what the rotation would look like if a mechanical device via some bearings spins an object around two axis at once, e.g. using an intrinsic rotation sequence like in a gyroscope or an extrinsic sequence like when rotating a spherical shape using to orthogonal driver slip-wheels (think old mechanical mouse ball, but in reverse), then torques will in general be present at the bearings.

I also like to stress again, than when we in mechanics model the rotational state for a rigid object as rotation around a single vector as any given time this does not in general imply the rotation vector will be fixed over time in inertial space.

Also, all this does not mean we cannot make a simpler geometric model of rotation that involves a rotation sequence, like when the object is constrained to rotate around one or more axes by being rigidly attached to bearings. If we are only interested in the kinematics of the object in this case (i.e. don't care about forces and torques) then surely a "two axis rotation" model can be much easier to describe. This correspond to constraining the linear motion of an object to move in a circle without caring about the actual forces that is needed for this to happen.

 

[DaveC426913]

OK, so it looks like, based on animation #2, that - notwithstanding all the geometry theory showing that an object seems to rotate about a single axis - the answer to the primary question is: a passenger inside such a vessel would indeed experience apparent gravity that changes in both direction and magnitude.

Does anyone disagree?I rode the zipper carnival ride once. I suspect it would feel much like that.

 

[Janus]

Nice animations

Note however, if "as it would work on the ISS" refers to what should happen when an astronaut in free fall onboard ISS spins an object for the camera, i.e. when the object is put in torque free rotation, then note these animations do NOT show that.

A camera spun on the ISS would tend to maintain its rotational axis relative to the stars. The ISS would also maintain its orientation relative to the stars if it wasn't for the fact that it has a system of gyroscopes designed to force it to keep one face pointing towards the Earth. Minis air drag, the camera would maintain its axis of rotation, while the ISS rotates around it. Introducing drag would (depending on the axis of rotation for the camera) cause a precession.

 

[DaveC426913]

The ISS would also maintain its orientation relative to the stars if...

Nitpick: Would it? In reality, I believe such oblong orbiting objects are subject to micro-tidal forces and will, in the longer term, align themselves with their long axis pointed Earthward, yes?

 

[Filip Larsen]

Nitpick: Would it? In reality, I believe such oblong orbiting objects are subject to micro-tidal forces and will, in the longer term, align themselves with their long axis pointed Earthward, yes?

That is correct. Gravity gradient stabilization is useful for maintaining attitude in low Earth orbit and (if I recall correctly) was often used by the Space Shuttle. For objects with (significant) three-axial moments of inertia, maintaining a fixed attitude wrt. the Earth (aka local vertical, local horizon) or the stars that is not aligned with the principal axes will require constant torquing by some means. For completely passive attitude control (i.e. no control) one need to have the minor inertial axis (often the geometrically longest axis) in the local vertical.

 

[Filip Larsen]

A camera spun on the ISS would tend to maintain its rotational axis relative to the stars.

Indeed. But I understand your animation to be the result of a visualization of a sequence of two geometrical rotations and not the result of a numerical simulation of a torque free rotation. If that is correct, then the cylinder in your animation will not rotate as if it would if spun freely on the ISS. The reason I mentioned this is just that you included the bit about ISS in the quote.

 

[Janus]

Indeed. But I understand your animation to be the result of a visualization of a sequence of two geometrical rotations and not the result of a numerical simulation of a torque free rotation. If that is correct, then the cylinder in your animation will not rotate as if it would if spun freely on the ISS. The reason I mentioned this is just that you included the bit about ISS in the quote.

I actually missed the mention of the ISS in that quote somehow. I probably was too focused on the animation he had in the post.

 

[wrobel]

Once more, in tiny words?

There is a unique (pseudo)vector $\boldsymbol \omega=\boldsymbol \omega(t)$ such that for any points $A,B$ of a rigid body one has $$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol \omega\times \boldsymbol {BA}$$

 

[erobz]

There is a unique (pseudo)vector $\boldsymbol \omega=\boldsymbol \omega(t)$ such that for any points $A,B$ of a rigid body one has $$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol \omega\times \boldsymbol {BA}$$

I'm assuming that $\boldsymbol{BA}$ is a position vector from $B \to A$; I haven't saw that notation ( but I don't see much )

 

[DaveC426913]

There is a unique (pseudo)vector $\boldsymbol \omega=\boldsymbol \omega(t)$ such that for any points $A,B$ of a rigid body one has $$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol \omega\times \boldsymbol {BA}$$

Note the "B" in the edumakashun level for this thread.

 

[wrobel]

I'm assuming that BA is a position vector from B→A; I

yes

Note the "B" in the edumakashun level for this thread.

but the initial question is not of high school level at all. The concept of angular velocity of a rigid body is a tricky enough stuff. The inertia forces in an arbitrary motion is also a pretty subtle thing for high school

 

[DaveC426913]

but the initial question is not of high school level at all.

Sure it is. It can be answered simply, without having the math as the answer.

The inertia forces in an arbitrary motion is also a pretty subtle thing for high school

But the end result is not. At their simplest, the OP question(s) could be answered with a yes/no.

 

[wrobel]

then why do you ask questions when you know answers?

 

[vanhees71]

Note the "B" in the edumakashun level for this thread.

This is perfectly B-level.The proof is not that difficult:

To describe a rigid body, take a model that it consists of a discrete set of point particles connected by massless rigid rods. Let's describe the rigid body with respect to an inertial refrence frame. To describe the body's particles take an arbitrary point fixed in the body. The position vector of this body-fixed origin is denoted with $\vec{R}$, and any point of the body is described by
$$\vec{x}_i=\vec{R}+\vec{r}_i.$$
Since the $\vec{r}_i$ are vectors connecting the body-fixed origin with another body-fixed point, the rigidity of the body means that $r_i=|\vec{r}_i|=\text{const}$. It's pretty obvious without calculation that thus the $\vec{r}_i$ can only rotate. To see this formally, we write
$$r_i^2=\vec{r}_i^2=(\vec{x}_i-\vec{R})^2.$$
Since this is constant, taking the derivative with respect to time yields
$$\dot{\vec{r}}_i \cdot \vec{r}_i=0,$$
which means that the velocity of the body-fixed point relative to the body-fixed reference point is always perpendicular to the position vector $\vec{r}_i$ relative to the body-fixed position. That means that there must be a vector $\vec{\omega}$ such that
$$\dot{\vec{r}}_i=\vec{\omega} \times \vec{r}_i.$$
This implies that the velocity of this point in the body relative to the inertial frame,
$$\dot{\vec{x}}_i=\dot{\vec{R}}+\dot{\vec{r}_i}=\dot{\vec{R}} + \vec{\omega} \times \vec{r}_i.$$
The change of the relative vector in an infinitesimal step of time is
$$\mathrm{d} \vec{r}_i=\mathrm{d} t \vec{\omega} \times \vec{r}_i,$$
and this describes an infinitesimal rotation around a momentary rotation axis in direction of $\vec{\omega}$ by an angle $|\vec{\omega}|\mathrm{d} t$. Thus $\vec{\omega}$ is the angular velocity, describing the rotation of the body around the body-fixed origin, $\vec{R}$.

Finally we show that this relative angular velocity is independent of the choice of this body-fixed origin. So let $\vec{R}'$ be another choice. Then
$$\vec{x}_i=\vec{R}+\vec{r}_i=\vec{R}'+\vec{r}_i'.$$
Now as before we have
$$\dot{\vec{r}}'=\vec{\omega}' \times \vec{r}_i',$$
and we have to show that $\vec{\omega}'=\vec{\omega}$.

To see this we note that
$$\vec{R}'-\vec{R}=\vec{r}=\vec{r}_i-\vec{r}_i'$$
is also a body-fixed vector, because it connects two body fixed points. Thus there must be a vector $\vec{\omega}''$ such that
$$\dot{\vec{R}}'-\dot{\vec{R}}=\vec{\omega}'' \times (\vec{R}'-\vec{R}) = \vec{\omega}'' \times (\vec{r}_i-\vec{r}_i')=\vec{\omega} \times \vec{r}_i-\vec{\omega}' \times \vec{r}_i,$$
and this can hold true for all points $\vec{x}_i$ making up the body, if
$$\vec{\omega}''=\vec{\omega}=\vec{\omega}'.$$
So finally we also have
$$\dot{\vec{r}}_i'=\vec{\omega} \times \vec{r}_i,$$
and that was to be shown.

 

[vanhees71]

Sure it is. It can be answered simply, without having the math as the answer.But the end result is not. At their simplest, the OP question(s) could be answered with a yes/no.

Physics cannot be formulated with the adequate math. Euclidean vectors are still taught at high schools!