# Sum of oscillating frequency and angular velocity

Homework Statement:
.
Relevant Equations:
. The "egg" initially spun around axis 1 with at ##\omega_s##. After being disturbed, it has started to possess angular velocities along 2 and 3. The question is to find the rotational speed of ##\vec \omega=\vec\omega_1+\vec\omega_2+\vec\omega_3## to a fixed observer.

It is calculated that ##\omega_2\text{ & }\omega_3## fluctuates at a frequency of ##\lambda=\left|\frac{I_1-I_{\perp}}{I_{\perp}}\right|\omega_s##, where ##I_{\perp}## represents the inertia around 2 or 3. The equations of angular speed 2,3 are:
$$\begin{cases}\omega_2=\omega_\perp \cos\lambda t\\ \omega_3=\omega_\perp\sin\lambda t\end{cases}$$
And they are shown by the diagram below. Then the author makes the following argument:
As the drawing shows, ω2 and ω3 are the components of a vector ω⊥ that rotates in the 2–3 plane at rate γ. An observer fixed to the body would see ω⊥ rotate relative to the body about the 1 axis at angular frequency γ. Since the 1, 2, 3 axes are fixed to the body and the body is rotating about the 1 axis at rate ωs, the rotational speed of ω⊥ relative to an observer fixed in space is $$\lambda+\omega_s=\frac{I_1\omega_s}{I_\perp}$$
(γ=##\lambda##)
I am totally bewildered by this derivation. While it is dimensionally correct, how can one add angular frequency to angular velocity, and why does the sum give you the answer? Could someone explain the reasoning behind this?

Steve4Physics
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how can one add angular frequency to angular velocity, and why does the sum give you the answer? Could someone explain the reasoning behind this?
You can't add angular frequency (a scalar) to angular velocity (a vector). But I think I know what you mean. I don't think I can guide you in question/answer style but I’ll have a go explaining my (non-rigourous) interpretation.

We know:
##\vec{\omega_2}## and ##\vec{\omega_3}## are mutually perpendicular angular velocities. And both are perpendicular to axis-1
##\vec\omega_\perp = \vec{\omega_2}+ \vec{\omega_3}##
##\omega_2=\omega_\perp \cos\lambda t##
##\omega_3=\omega_\perp\sin\lambda t##
This means that ##\vec{\omega_\perp}## is a vector with its direction perpendicular to axis-1 and rotating about axis-1 with angular speed ##\omega_\perp## = ##\lambda##.

If the last step isn't clear, think about the angular speed of a point where x=cos(3t), y = sin(3t).

Total angular speed about axis-1 is ##\omega_\perp## + ##\omega_s## which is the same as ##\lambda + \omega_s##.

• Leo Liu
This means that ω⊥→ is a vector with its direction perpendicular to axis-1 and rotating about axis-1 with angular speed ω⊥ = λ.
So you are saying that angular frequency of axis 2 & 3 ##\frac {2\pi}{T_{\perp}}## is equal to ##\omega_{\perp}##. This makes sense. Thank you.

• Steve4Physics
This means that ω⊥→ is a vector with its direction perpendicular to axis-1 and rotating about axis-1 with angular speed ω⊥ = λ.
I am sorry for bothering you again. But I still don't understand why the direction of ##
\lambda## is in the direction of axis 1. Would you mind explaining it? Also, isn't the rate of change of the direction of ##\omega_\perp## its angular acceleration? Why do we still get an angular velocity?

Steve4Physics
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First, please note I was wrong to say ##\omega_\perp = \lambda## in my previous message. ##\omega_\perp## and ##\lambda## are independent values. Apologies if this has caused confusion. I can't correct it, but see expanded explanation below.
But I still don't understand why the direction of ##\lambda## is in the direction of axis 1.
Let's think of axes 1,2 and 3 as z, x and y axes respectively.
If we have a vector with components ##V_x=3cos(5t)##, ##V_y=3sin(5t)## and ##V_z=0## the vector has constant magnitude (3) and rotates about the z-axis with angular speed 5. Check up on your circular-motion notes if you are not familiar with this. '5' correponds to ##\lambda##, '3' correponds to ##\omega_\perp## and axis-1 corresponds to the z-axis.
Also, isn't the rate of change of the direction of ##\omega_\perp## its angular acceleration?
Direction is changing but it is unnecesary (and unhelpful) to start thinking in terms of (centripetal) acceleration. The magnitude of angular velocity is constant. Just think of ##\vec\omega_\perp## as an arrow of constant length rotating with some constant angular speed (which is ##\lambda##) about axis-1.

See if this revised explanation helps...

##\vec\omega=\vec\omega_1+\vec\omega_2+\vec\omega_3##

##\vec\omega## does not have a constant direction in this problem. ##\vec\omega## precesses. That means its direction rotates (about axis-1). Be clear that we are trying to find the angular speed of precession, i.e. the angular speed of ##\vec\omega##’s direction-vector about axis-1.

We do this in 2 steps.

Step 1: Find the angular speed of ##\vec\omega_\perp## (= ##\vec\omega_2+\vec\omega_3##) about axis-1, as measured by an observer spinning with angular speed ##\omega_s##.

Step 2: Note that to a stationary external observer, the whole egg has angular speed ##\omega_s## about axis-1. So we need to add ##\omega_s## to step 1’s value. This gives the total angular speed with which ##\vec\omega## precesses about axis-1 as measured by a stationary external observer.
_________________

##\omega_2## and ##\omega_3## oscillate with period T so that:
##\lambda = \frac{2\pi}{T}##
##\omega_2=\omega_\perp \cos\lambda t##
##\omega_3=\omega_\perp \sin\lambda t##
(It’s worth noting that ##\lambda## could take any value)

Step 1.
As seen by an observer with angular speed ##\omega_s##, ##\vec\omega_\perp## has magnitude ##\omega_\perp## and precesses around axis-1 with angular speed ##\lambda##.

Step 2.
To a stationary external observer the whole egg has angular speed ##\omega_s## about axis-1. So ##\vec\omega## is observed to precess around axis-1 with angular speed ##\lambda + \omega_s##

I'm not 100% convinced that's correct, but I think it's what the text-book solution is doing.

Last edited:
• Leo Liu
Steve4Physics
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I was wrong to use ##\omega_\perp = \lambda## in post #2. As I can't edit it, I'm adding this corrected version of the relevant parts of post #2.

##\vec{\omega_2}## and ##\vec{\omega_3}## are mutually perpendicular angular velocities. And both are perpendicular to axis-1
##\vec\omega_\perp = \vec{\omega_2}+ \vec{\omega_3}##
##\omega_2=\omega_\perp \cos\lambda t##
##\omega_3=\omega_\perp\sin\lambda t##
This means that ##\vec{\omega_\perp}## is a vector with its direction perpendicular to axis-1 and rotating about axis-1 with angular speed = ##\lambda##.

Total angular speed about axis-1 is ##\lambda + \omega_s##.

• Leo Liu
Thanks again for spending time answering my dumb question.
As seen by an observer with angular speed ωs, ω→⊥ has magnitude ω⊥ and precesses around axis-1 with angular speed λ.
Could you tell me what the point of calculating the precession angular velocity of another angular velocity is? I don't quite get it.
Is it the angular velocity of the rotation of the axes relative to the observer?

Steve4Physics
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Thanks again for spending time answering my dumb question.
Definitely not a dumb question. It's a tricky topic. And the original textbook material you attached is far from perfect. It doesn't make clear what the difference beween ##\vec\omega_s## and ##\vec\omega_1## is. Or why ##\gamma## is used when it is stated as being equal to ##\lambda##. Or the purpose of marking ##\gamma t## on the diagram. And probably other things.
Could you tell me what the point of calculating the precession angular velocity of another angular velocity is? I don't quite get it.
The system is basically a gyroscope and the question is about finding the frequency of precession. This important/useful both theoretically and practically.
Is it the angular velocity of the rotation of the axes relative to the observer?
Sort of. I can't give a simple yes/no answer. A simple explanation is: if you look at the system's angular momentum vector(L), it does not point in a fixed direction. To a stationary external observer, L's direction changes at the precessional frequency. In fact things are more complicated because something called nutation also occurs (look it up!).

You will need to read-up about gyroscopes and precession (and nutation) if interested. Plenty of information out there including YouTube videos. E.g. one I just found with a quick search is

• Leo Liu
Or why γ is used when it is stated as being equal to λ.
I wrote lambda because it was the variable I used, while the book used gamma instead, so it's not the book's problem really. But I do think it could explain the derivation better.
if you look at the system's angular momentum vector(L), it does not point in a fixed direction. To a stationary external observer, L's direction changes at the precessional frequency. In fact things are more complicated because something called nutation also occurs (look it up!).
I just read that part of the book again and realized that I misinterpreted "the rotational speed of ##\vec\omega##" as the magnitude of the spin angular velocity. What it genuinely means is the rate of change of the direction of ##\vec\omega##. So the frequency represents how fast the orientation of ##\vec\omega## changes relative to axis 1. Then we add ##\omega_s## to it because we want to obtain this infor at a point fixed in space. So the answer is ##\lambda+\omega_s##. Am I right?

I appreciate your help. Have a good night/day.

• Steve4Physics
Steve4Physics
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I just read that part of the book again and realized that I misinterpreted "the rotational speed of ##\vec\omega##" as the magnitude of the spin angular velocity. What it genuinely means is the rate of change of the direction of ##\vec\omega##. So the frequency represents how fast the orientation of ##\vec\omega## changes relative to axis 1. Then we add ##\omega_s## to it because we want to obtain this infor at a point fixed in space. So the answer is ##\lambda+\omega_s##. Am I right?
Yes, I agree. I'm assuming when you say:
"So the frequency represents how fast the orientation of ##\vec\omega## changes relative to axis 1"
that you mean:
"So ##\lambda## represents how fast the orientation of ##\vec\omega## rotates about axis-1 as measured by an observer with spin ##\omega_1"##.

(I'm assuming ##\omega_1## and ##\omega_s## are the same.)

Also I'm, assuming when you say:
"for at a point fixed in space"
that you mean:
"for a stationary observer"
because frequency of precession (which is what we are calculating) doesn't depend at what point an external observer is, only that they are in a non-rotating frame of reference.

A happy and safe 2021, from Devon, England!

• Leo Liu
etotheipi
This can be confusing, so I hope this helps.

It's useful to remember that the ##\{1,2,3\}## axes (along basis vectors ##\{\boldsymbol{e}_1, \boldsymbol{e}_2, \boldsymbol{e}_3 \}##) are body-fixed, i.e. rotating with the body. In this rotating frame fixed to the body, the angular velocity ##\boldsymbol{\omega}## vector of the rigid body is actually rotating around the body-fixed ##1## axis, as you described, at what appears to be angular velocity ##\lambda \boldsymbol{e}_1## with respect to the body-fixed frame.

Since the body itself rotates at ##\boldsymbol{\omega}## with respect to the lab frame, the angular velocity at which the vector ##\boldsymbol{\omega}## rotates w.r.t. the lab frame is thus ##\boldsymbol{\omega} + \lambda \boldsymbol{e}_1##, by the law of addition of angular velocities of frames in relative motion. By definition in the text, the decomposition of ##\boldsymbol{\omega}## with respect to the ##\{1,2,3\}## axes is$$\boldsymbol{\omega} = \omega_s \boldsymbol{e}_1 + \omega_2 \boldsymbol{e}_2 + \omega_3 \boldsymbol{e}_3$$so the angular velocity at which ##\boldsymbol{\omega}## rotates w.r.t. the lab frame can be simplified to$$\boldsymbol{\omega} + \lambda \boldsymbol{e}_1 = (\omega_s + \lambda) \boldsymbol{e}_1 + \omega_2 \boldsymbol{e}_2 + \omega_3 \boldsymbol{e}_3$$And this angular velocity has a component about the body-fixed ##1## axis of ##\omega_s + \lambda##.

N.B. this can be exceptionally confusing, given that at the end here we're measuring this particular angular velocity with respect to the lab frame, but expressing the angular velocity with respect to the body-fixed axes/basis vectors! As a rule of thumb, when we talk about some measured rotational quantity "about some axis" we mean "take the scalar product of that measured quantity with the basis vector along that axis". So the rate at which ##\boldsymbol{\omega}## rotates around the ##1## axis as measured by the lab frame is ##[\boldsymbol{\omega} + \lambda \boldsymbol{e}_1] \cdot \boldsymbol{e}_1##. Unfortunately, this is a quirk of rotational mechanics that one just needs to spend many nights crying over Furthermore, assuming this is torque free precession, the vector ##\boldsymbol{\omega}## will precess about the angular momentum ##\boldsymbol{L}## at some angular velocity ##\boldsymbol{\Omega}##. If the angle between the angular momentum and the body-fixed ##1## axis is ##\theta##, you can finally deduce that with respect to the lab frame, ##\boldsymbol{\omega}## precesses around the ##1## axis at ##\boldsymbol{\Omega} \cdot \boldsymbol{e}_1 = \Omega \cos{\theta}##, and now you can solve for the precession velocity by equating our two equivalent expressions.

For more details, Landau & Lifshitz, Course of Theoretical Physics Vol 1, Mechanics, is very good, the relevant section being chapter 6.

• • Leo Liu and Steve4Physics
Unfortunately, this is a quirk of rotational mechanics that one just needs to spend many nights crying over   • • Steve4Physics and etotheipi