How Does Resolving Angular Velocity into Components Affect Its Magnitude?

[aspodkfpo]

Homework Statement

I'm pretty sure I'm tripping myself up.

Relevant Equations

n/a

Angular velocity is the degrees by which something rotates over a time period. If I have an angular velocity in one direction and I resolve it into its components, its components would obviously be of lesser value. Here's what I don't get. When I imagine this scenario, I see that the thing rotates the same amount of degrees per second in both the resolved component and the original vector. Help.

 

[wrobel]

Angular velocity is a quite subtle concept. Try to rest on definition and theorems only

 

[Delta2]

You have to tell us what exactly is the scenario you imagine. We can't read your mind :D:

 

[haruspex]

Homework Statement:: I'm pretty sure I'm tripping myself up.
Relevant Equations:: n/a

Angular velocity is the degrees by which something rotates over a time period. If I have an angular velocity in one direction and I resolve it into its components, its components would obviously be of lesser value. Here's what I don't get. When I imagine this scenario, I see that the thing rotates the same amount of degrees per second in both the resolved component and the original vector. Help.

If you are resolving angular velocity into components then I take it this is in 3D.
Unlike linear motion, angular displacements don't add, so looking at the angles of rotation over extended periods is not helpful. Instantaneous angular velocities can be added because the displacements are over infinitesimal periods.

Or have I misunderstood your question?

 

[aspodkfpo]

If you are resolving angular velocity into components then I take it this is in 3D.
Unlike linear motion, angular displacements don't add, so looking at the angles of rotation over extended periods is not helpful. Instantaneous angular velocities can be added because the displacements are over infinitesimal periods.

Or have I misunderstood your question?

You have to tell us what exactly is the scenario you imagine. We can't read your mind :D:

diagrams kind of sketchy tbh.

Also was wondering whether I could indicate a vector by just drawing a -- (bar) instead of an --> (arrow) to save time in exams.

 

[archaic]

a small thing you should keep in mind is that the angular velocity vector has no physical significance (only its magnitude does). it's just a good mathematical tool.

 

[etotheipi]

a small thing you should keep in mind is that the angular velocity vector has no physical significance (only its magnitude does). it's just a good mathematical tool.

Although angular velocity is technically a pseudovector, I think that is still a little harsh . For instance, the direction of the angular velocity vector coincides with the instantaneous axis of rotation.

 

[aspodkfpo]

So am I just meant to ignore the fact that 1 rev looks like it takes the same amount of time for the component and the original vector, and that makes it look like theta/t (angular velocity) is the same value?

 

[haruspex]

I think your notion of resolving an angular velocity vector into components in, say, orthogonal directions is not in general valid.
Consider a particle executing constant circular motion around some axis X at rate $\omega$. Considering its instantaneous velocity in respect of some axis X' at angle $\theta$ to X, its angular velocity about X' varies between $\omega\sin(\theta)$ and $\omega\csc(\theta)$. Do I have that right?

 

[aspodkfpo]

I think your notion of resolving an angular velocity vector into components in, say, orthogonal directions is not in general valid.
Consider a particle executing constant circular motion around some axis X at rate $\omega$. Considering its instantaneous velocity in respect of some axis X' at angle $\theta$ to X, its angular velocity about X' varies between $\omega\sin(\theta)$ and $\omega\csc(\theta)$. Do I have that right?

I have the same problem with the statement that you made. Say that I switch axes, I still see the particle moving the same theta/t, that is 1 rev in the same time period and thus angular velocity remains equal?

 

[haruspex]

I have the same problem with the statement that you made. Say that I switch axes, I still see the particle moving the same theta/t, that is 1 rev in the same time period and thus angular velocity remains equal?

It completes one orbit about the axis in the same time, but as I posted, the angular velocity keeps changing, so I think it is simply not valid to resolve an angular velocity into components in the same way as for a linear velocity.

However, it is ok to resolve angular momentum that way. In my scenario, the angular momentum comes out constantly as $mr^2\omega\sin(\theta)$.

 

[aspodkfpo]

It completes one orbit about the axis in the same time, but as I posted, the angular velocity keeps changing, so I think it is simply not valid to resolve an angular velocity into components in the same way as for a linear velocity.

However, it is ok to resolve angular momentum that way. In my scenario, the angular momentum comes out constantly as $mr^2\omega\sin(\theta)$.

How exactly is it changing? I don't see why theta/t would be changing.

 

[wrobel]

Let a coordinate frame $Oxyz$ be fixed. And let a frame $O'x'y'z'$ has an angular velocity $\boldsymbol \omega_1$ relative the frame $Oxyz$. A rigid body $B$ has an angular velocity $\boldsymbol \omega_2$ relative the frame $O'x'y'z'.$

Theorem. The angular velocity of the body $B$ relative $Oxyz$ is equal to
$\boldsymbol \omega_1+\boldsymbol \omega_2.$

 

[aspodkfpo]

Let a coordinate frame $Oxyz$ be fixed. And let a frame $O'x'y'z'$ has an angular velocity $\boldsymbol \omega_1$ relative the frame $Oxyz$. A rigid body $B$ has an angular velocity $\boldsymbol \omega_2$ relative the frame $O'x'y'z'.$

Theorem. The angular velocity of the body $B$ relative $Oxyz$ is equal to
$\boldsymbol \omega_1+\boldsymbol \omega_2.$

w1+w2. Where is the angular velocity changing? What statement are you referring to?

 

[wrobel]

I formulated the statement which explains kinematics meaning of decomposing of the angular velocity vector by several directions. It was an answer to your original post as I understood it

 

[Lnewqban]

Angular velocity is the degrees by which something rotates over a time period. If I have an angular velocity in one direction and I resolve it into its components, its components would obviously be of lesser value...

Paraphrasing your description:
Angular velocity is the degrees by which something rotates over a time period around an axis and on a plane of rotation.
Any imaginary component that deviates out of that plane of rotation will be a trigonometric projection of the real angular or tangential velocity; therefore, its value will be less than real (multiplied by the sine or cosine of the angle respect to the real plane of rotation).

 

[haruspex]

Let a coordinate frame $Oxyz$ be fixed. And let a frame $O'x'y'z'$ has an angular velocity $\boldsymbol \omega_1$ relative the frame $Oxyz$. A rigid body $B$ has an angular velocity $\boldsymbol \omega_2$ relative the frame $O'x'y'z'.$

Theorem. The angular velocity of the body $B$ relative $Oxyz$ is equal to
$\boldsymbol \omega_1+\boldsymbol \omega_2.$

Yes, angular velocity vectors can be added and subtracted to find relative motion, but the OP's question is specifically about resolving a vector into components, as in sin and cos of an angle.
Consider the model in post #9. If a point is moving around the X axis at constant rate $\omega$ and radius r then the component of its instantaneous velocity normal to the plane containing it and the X' axis varies between a speed $r\omega\sin(\theta)$ at radius r and a speed $r\omega$ at radius $r\sin(\theta)$.
Thus, its angular velocity around X' varies between $\omega\sin(\theta)$ and $\omega\csc(\theta)$.

The OP worries that the time for the particle to complete a rotation about X and X' is the same, so in that sense the angular velocities are the same. My position is that resolving angular velocities in this way is invalid.

 

[etotheipi]

Yes, angular velocity vectors can be added and subtracted to find relative motion, but the OP's question is specifically about resolving a vector into components, as in sin and cos of an angle.
Consider the model in post #9. If a point is moving around the X axis at constant rate $\omega$ and radius r then the component of its instantaneous velocity normal to the plane containing it and the X' axis varies between a speed $r\omega\sin(\theta)$ at radius r and a speed $r\omega$ at radius $r\sin(\theta)$.
Thus, its angular velocity around X' varies between $\omega\sin(\theta)$ and $\omega\csc(\theta)$.

There are some interesting points here. If the origin is specified, then the angular momentum vector is uniquely determined: $\vec{\omega} = \frac{1}{r^2} \vec{r} \times \vec{v}$. So for the class of coordinate systems related by rotations about the common origin, they will all measure the same angular velocity vector. We should be able to resolve this into any of those coordinate systems just as we would any other vector.

Consequently, the components in two such rotated coordinate systems must be related like $\vec{\omega} = \sum_i \omega_i \vec{e}_i = \omega'_i \vec{e}'_i$.

A consequence of this would be that the angular velocity about a certain axis only equals $\frac{2\pi}{T}$ (where $T$ is the period of rotation about that axis) if the motion is completely within the plane defined by the other two axes.

If I've made a mistake, please do correct me. I am still reading this to try and get a better understanding: https://galileoandeinstein.phys.virginia.edu/7010/CM_26_Euler_Angles.html

 

[aspodkfpo]

I formulated the statement which explains kinematics meaning of decomposing of the angular velocity vector by several directions. It was an answer to your original post as I understood it

Yes, angular velocity vectors can be added and subtracted to find relative motion, but the OP's question is specifically about resolving a vector into components, as in sin and cos of an angle.
Consider the model in post #9. If a point is moving around the X axis at constant rate $\omega$ and radius r then the component of its instantaneous velocity normal to the plane containing it and the X' axis varies between a speed $r\omega\sin(\theta)$ at radius r and a speed $r\omega$ at radius $r\sin(\theta)$.
Thus, its angular velocity around X' varies between $\omega\sin(\theta)$ and $\omega\csc(\theta)$.

The OP worries that the time for the particle to complete a rotation about X and X' is the same, so in that sense the angular velocities are the same. My position is that resolving angular velocities in this way is invalid.

There are some interesting points here. If the origin is specified, then the angular momentum vector is uniquely determined: $\vec{\omega} = \frac{1}{r^2} \vec{r} \times \vec{v}$. So for the class of coordinate systems related by rotations about the common origin, they will all measure the same angular velocity vector. We should be able to resolve this into any of those coordinate systems just as we would any other vector.

Consequently, the components in two such rotated coordinate systems must be related like $\vec{\omega} = \sum_i \omega_i \vec{e}_i = \omega'_i \vec{e}'_i$.

A consequence of this would be that the angular velocity about a certain axis only equals $\frac{2\pi}{T}$ (where $T$ is the period of rotation about that axis) if the motion is completely within the plane defined by the other two axes.

If I've made a mistake, please do correct me. I am still reading this to try and get a better understanding: https://galileoandeinstein.phys.virginia.edu/7010/CM_26_Euler_Angles.html

I might have failed to convey the scenario that I was thinking about specifically.

Here's where I started to think about it, for reference.

10:10 - 11:00, where z = omega cos theta is stated.

 

[haruspex]

I might have failed to convey the scenario that I was thinking about specifically.

Here's where I started to think about it, for reference.

10:10 - 11:00, where z = omega cos theta is stated.

Yes, I see your issue with the method at that point.
But my issues start sooner. How did she pick that tilted axis for $\Omega$ in the first place?
I would have considered the motion of the system as the sum of a constant rotation about the z axis and a rotation about the axis of the rod, where that second axis varies over time.
At a given instant, yes, the angular momenta probably add vectorially to one aligned with her $\Omega$ axis, but that is not the same as saying the rotation vectors do.
As far as I can see, in her method, she could just as easily have calculated the angular velocity of the centre of the small disc about the z axis directly as its linear velocity divided by its distance from the z axis, getting a larger value than she got for the angular velocity about her tilted axis - instead of a smaller one.
This is exactly the problem I modeled in post #9, and as I wrote, you are much safer working with angular momentum vectors.

 

[aspodkfpo]

Yes, I see your issue with the method at that point.
But my issues start sooner. How did she pick that tilted axis for $\Omega$ in the first place?
I would have considered the motion of the system as the sum of a constant rotation about the z axis and a rotation about the axis of the rod, where that second axis varies over time.
At a given instant, yes, the angular momenta probably add vectorially to one aligned with her $\Omega$ axis, but that is not the same as saying the rotation vectors do.
As far as I can see, in her method, she could just as easily have calculated the angular velocity of the centre of the small disc about the z axis directly as its linear velocity divided by its distance from the z axis, getting a larger value than she got for the angular velocity about her tilted axis - instead of a smaller one.
This is exactly the problem I modeled in post #9, and as I wrote, you are much safer working with angular momentum vectors.

Why is working with angular momentum vectors safer? There is still the angular velocity within it?

 

[haruspex]

Why is working with angular momentum vectors safer? There is still the angular velocity within it?

As I noted in post #11, in my one part, two axis model, the angular momentum around the second axis comes at as the expected constant, i.e. resolving in the usual (cos, sin) manner works.

 

[aspodkfpo]

If a point is moving around the X axis at constant rate ω and radius r then the component of its instantaneous velocity normal to the plane containing it and the X' axis varies between a speed rωsin⁡(θ) at radius r and a speed rω at radius rsin⁡(θ).
Thus, its angular velocity around X' varies between ωsin⁡(θ) and ωcsc⁡(θ).

It's this varying part that I don't get. I don't get how you arrive at any of the statements. This stuff is quite new to me.

 

[archaic]

@aspodkfpo I answered that JEE question on stackexchange before, I'll try and copy/paste the question.

Interesting question, though I think that the questions could have been asked in a better way.

First, let met explain the angular velocity with respect to the $z$-axis. If you look from atop, then you'd see something like this.

The smaller circle represents the projection of the position of the centre of mass of the disk onto the $xy$-plane, while the bigger circle represents the actual path taken by the point of the disk which is in contact with the aforementioned plane. (I have only drawn the little circle, but it doesn't matter since they both roll with the same angular velocity).

Call the angle that is described on the $xy$-plane $\varphi$, then the angular displacement is given by $\Delta s=\sqrt{l^2+a^2}\Delta\varphi$.

The crux of this problem lies in recognising that this angular displacement is equal to the angular displacement inside the disk! i.e $\Delta s=a\Delta\phi$ where I represent with $\phi$ the angle described as the disk rotates about its center of mass.
$$\begin{align*}
\sqrt{l^2+a^2}\Delta\varphi=a\Delta\phi
&\implies\Delta\varphi=\frac{a}{\sqrt{l^2+a^2}}\Delta\phi\\
&\implies\Delta\varphi=\frac{a}{\sqrt{25a^2}}\Delta\phi\\
&\implies\Delta\varphi=\frac{\omega t}{5}\\
&\implies\dot\varphi=\frac{\omega}{5}
\end{align*}$$

Now I'll explain the angular velocity about the origin. By this they meant you to imagine that the configuration is perfectly straight.

Do you see the line I have drawn? You need to imagine that it is as if the disks are rolling on that plane (we see it as a line from this point of view).

$\varphi'$ is the angle of rotation on the imaginary plane.

The angular displacement is (the same arguments as before holds, but in this case the smaller disk doesn't touch the ground, so we operate on the bigger one)$$\Delta s=2l\Delta\varphi'=2a\Delta\phi\implies\dot\varphi'=\frac{a\omega}{l}=\frac{\omega}{\sqrt{24}}$$
To go from this back to first case, then if you wish to project$^*$ $2l$ onto the $xy$-plane you will do $\left[2l/(\cos\theta)\right]\Delta\varphi=2a\Delta\phi$, hence $\omega_z=\dot\varphi'\cos\theta$, which is why projecting the angular velocity pseudo-vector about the point $O$ in the imaginary plane onto the $z$-axis works.

Possible confusion with regards to ($^*$):

Suppose that we have two referentials, $(x,y,z)$ and $(x',y',z')$, which are on top of each other, i.e the same. Imagine that there is a point particle rotating about the origin in the $x'y'$-plane such that its direction vector is $\vec u=r\cos\phi\,\hat x'+r\sin\phi\,\hat y'$.

Now let us say that $(x',y',z')$ will rotate about the $x$-axis such that $\angle(\hat z,\hat z')=\angle(\hat y,\hat y')=\theta$, while $\hat x=\hat x'$, then we get that $\hat y'=\cos\theta\,\hat y+\sin\theta\,\hat z$, hence $\vec u =r\cos\phi\,\hat x+r\sin\phi\cos\theta\,\hat y+r\sin\phi\sin\theta\,\hat z$.

If we project $\vec u$ onto the $xy$-plane, we get $\vec v=r\cos\phi\,\hat x+r\sin\phi\cos\theta\,\hat y$ which definitely doesn't describe a circle!

However, this is not what we have in this situation, the "projection" here is in the process of bringing the whole apparatus from the imaginary position to what we see on the problem statement's image, thus only changing the length of the radius of rotation, but keeping the whole system moving in a circle.

Thus the projection is taking the new radius of rotation of the point of contact of either of the disks with the ground, which is painting a circle!

Conclusion: (I might be incorrect here, some time have passed since I have thought about this)

If you have a mobile point (here we considered the point with which one of the disks, doesn't matter which one, touches the floor) naturally rotating about a point $O$ with a radius of rotation $r$, then, if it is brought to another level by rotation such that it still rotates in a circular fashion but about an axis that makes an angle $\theta$ with the original one, we can find that its angular velocity about this axis is equal to its angular velocity about the original axis scaled by $\cos\theta$.

Addendum about the question on the point P:
The op there asked this:

b) I still don't quite realize why only chosing Origin led to the correct answer. For instance , Check out the point P i mention in the question. Why can't we take the cosin e of the angular velocity vector relative to point P?

Let's consider the point $P$, and let's also place ourselves on the plane that contains $P$ and that is parallel to the $xy$-plane in order to find an expression of our angle.

You can see that that $z$ height corresponds with the $z$ height of the center of mass of the little disk, hence our rotation radius is $l\cosθ$ and thus $Δs=l\cosθΔφ$. We need to find a way to relate this to $Δϕ$ as I have done.

The point of contact is painting an angular dispalcement of $Δs_1=aΔϕ$, while the center of mass in painting an angular displacement $Δs_2=l\cosθΔφ$, but are both part of the same circle.

In general, two points of a rotating radius in a circle have this relation $$\frac{r_1\Delta\phi=\Delta s_1}{r_2\Delta\phi=\Delta s_2}=\frac{r_1}{r_2}$$ thus $$\frac{l\cos\theta\Delta\varphi}{a\Delta\phi}=\frac{l\cos\theta}{\sqrt{l^2+a^2}}$$ which means that $$\Delta\varphi=\frac{a\Delta\phi}{5a}=\frac{wt}{5}\implies\dot\varphi=\frac{\omega}{5}$$

 

[aspodkfpo]

To go from this back to first case, then if you wish to project∗ 2l onto the xy-plane you will do [2l/(cos⁡θ)]Δφ=2aΔϕ, hence ωz=φ˙cos⁡θ,

Explain this part a bit more please.

Here's what I'm concluding. Viewing the circle like I was before doesn't work because some of it is in an imaginary plane. Taking costheta of angular velocity works for solving problems.

 

[archaic]

Explain this part a bit more please.

Here's what I'm concluding. Viewing the circle like I was before doesn't work because some of it is in an imaginary plane. Taking costheta of angular velocity works for solving problems.

First of all, I am sorry for some sloppiness on my part. I used $\Delta\varphi$ to talk about the angle of rotation in both the $xy$-plane and the imaginary one.
I'll change it as $\Delta\varphi'$ for the latter.

In the imaginary plane, the arc that big circle describes is given by $2l\Delta\varphi'$. This arc is also equal to $2a\Delta\phi$. Think of a disk in real life that has on it wet paint. The distance painted on the floor is equal to the distance rolled.

Now imagine that we bring up this apparatus to the actual plane of the problem, the relationship between the distance painted and the distance rolled still holds, only the radius of the big rotation changes and becomes $\left[2l/(\cos\theta)\right]\Delta\varphi$.
$$\left[2l/(\cos\theta)\right]\Delta\varphi=2a\Delta\phi\implies\Delta\varphi=\frac al\omega\Delta t\cos\theta$$
We rearrange a bit to get:
$$\frac{\Delta\varphi}{\Delta t}=\dot\varphi=\omega_z=\left(\frac al\omega\right)\cos\theta$$
And I have shown that $\frac al\omega$ is the angular velocity about the imaginary axis.
We can see geometrically that $\cos\theta=\frac{l}{\sqrt{l^2+a^2}}$.

 

[etotheipi]

As I noted in post #11, in my one part, two axis model, the angular momentum around the second axis comes at as the expected constant, i.e. resolving in the usual (cos, sin) manner works.

I don't agree that you need to treat angular velocities any differently in terms of resolving; after all, it's still just a (pseudo)vector. Consider a particle undergoing uniform circular motion in a plane that is the $x$-$z$ plane tilted by $\theta$ about the $z$ axis.

Let us resolve the components of $\vec{\omega}$ in this coordinate system, which are $\vec{\omega} = \omega_x \hat{x} + \omega_y \hat{y} + \omega_z \hat{z} = \omega \sin{\theta} \hat{x} + \omega \cos{\theta} \hat{y}$. Now, for purposes of an example, let us consider the $y$ component of the angular momentum. This is given by the expansion involving the middle row of the MoI tensor: $$L_y = I_{yx}\omega_x + I_{yy} \omega_y + I_{yz} \omega_z$$Now let us consider two different points along the motion, to show that we calculate the same quantities of motion.

At $x,y=0, z=r$, we have $I_{yx} = 0, I_{yy} = mz^2, I_{yz} = 0$. This means that $L_y = mr^2 \omega \cos{\theta}$.

At $x=-r\cos{\theta}, y=r\sin{\theta},z=0$, we have $I_{yx} = -xy, I_{yy} = x^2, I_{yz} = 0$. This means that $L_y = mr^2\omega \cos{\theta}[1-\cos^2{\theta}] + mr^2 \omega \cos^{3}{\theta} = mr^2 \omega \cos{\theta}$.

And indeed, if we were to resolve the original angular momentum vector into this rotated coordinate system we would also obtain $L_y = L\cos{\theta} = mr^2 \omega \cos{\theta}$.

 

[etotheipi]

Yes, angular velocity vectors can be added and subtracted to find relative motion, but the OP's question is specifically about resolving a vector into components, as in sin and cos of an angle.
Consider the model in post #9. If a point is moving around the X axis at constant rate $\omega$ and radius r then the component of its instantaneous velocity normal to the plane containing it and the X' axis varies between a speed $r\omega\sin(\theta)$ at radius r and a speed $r\omega$ at radius $r\sin(\theta)$.
Thus, its angular velocity around X' varies between $\omega\sin(\theta)$ and $\omega\csc(\theta)$.

The OP worries that the time for the particle to complete a rotation about X and X' is the same, so in that sense the angular velocities are the same. My position is that resolving angular velocities in this way is invalid.

The velocity for such a particle is described as$$\begin{pmatrix}v_x\\v_y\\v_z\end{pmatrix} = \begin{pmatrix}\omega_x\\\omega_y\\\omega_z\end{pmatrix} \times \begin{pmatrix}x\\y\\z\end{pmatrix}$$I did my working using a slightly different angle convention to you (specifically my $\theta$ is $\frac{\pi}{2}$ minus your $\theta$), however if we use the same decomposition of the angular velocity vector w.r.t. this new rotated coordinate system$$\begin{pmatrix}v_x\\v_y\\v_z\end{pmatrix} = \begin{pmatrix}\omega \sin{\theta}\\\omega\cos{\theta}\\0\end{pmatrix} \times \begin{pmatrix}x\\y\\z\end{pmatrix}$$Now you considered the positions on the circle where $z=0$, $x=r\cos{\theta}$, $y = -r\sin{\theta}$, in which case we only have a $z$ component of velocity which equals $$v_z = y\omega \sin{\theta} - x\omega \cos{\theta} = - \omega r$$and you also considered the point where $x,y = 0$ and $z = -r$, in which case $$v_x = -\omega r \cos{\theta}$$ $$v_y =\omega r \sin{\theta}$$ which are to be expected, as they are in your words the instantaneous velocities normal to the plane containing it and the axis of rotation.

The key thing to notice is that the linear velocity perpendicular to an axis divided by the perpendicular distance to that axis does not generally give the angular velocity about that axis. That only holds if the motion is confined to around one axis (e.g. in the 2D case).

More generally, the component of the velocity vector perpendicular to a given axis $\hat{z}$ is$$\vec{\omega} \times \vec{r} - (\vec{\omega} \times \vec{r} \cdot \hat{z})\hat{z}$$