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Suppose the angular acceleration velocity of a body as measured in a fixed inertial frame (with fixed basis $\big\{\hat X_1,\hat X_2,\hat X_3\big\}$) is $\vec\omega$ with direction $\hat \omega$.

Next, suppose we consider the body frame, with a rotating basis $\big\{\hat x_1, \hat x_2, \hat x_3\big\}$.

I initially thought that since angular velocity rotates with the body frame, then the direction of angular velocity will always be the same, i.e. $\hat\omega\cdot \hat x_i$ is fixed in time. That way, angular velocity simply represents the axis about which the body frame rotates, and the angles between $\hat x_i$ and $\hat \omega$ are constant in time.

However, this interpretation has a problem since $$\bigg(\dfrac{d\hat \omega}{dt}\bigg)_\mathrm{fixed} = \bigg( \dfrac{d\hat \omega}{dt}\bigg)_\mathrm{rot}+\vec\omega\times\hat \omega$$ implies that $\big(\frac{d\hat \omega}{dt}\big)_\mathrm{fixed} = \big( \frac{d\hat \omega}{dt}\big)_\mathrm{rot}$. So, that means $\hat\omega$ can't be fixed in the body frame, but I don't understand why that is so?

Why would angular velocity direction change with respect to the $\hat x_i$ axis? Don't we track the rotation of the $\hat x_i$ axis based on the orientation of $\vec\omega$?

Why can’t $\hat\omega$ be fixed in the body frame but change in the inertial one? Intuitively I can’t make sense of this.

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  • $\begingroup$ Could you elaborate on what's the difference between inertial frame and body fixed frame? I've never heard the second one being used so they sound the same to me. $\endgroup$ Oct 11, 2022 at 1:23
  • $\begingroup$ @ReetJaiswal The inertial frame is fixed in space. The body frame rotates with the body and its axes are usually aligned with the body's principal axes $\endgroup$
    – user256872
    Oct 15, 2022 at 9:08
  • $\begingroup$ 5. (Minor point) "angular velocity simply represents the axis about which the body frame rotates" is not exactly correct. The axis of rotation is a straight line that is characterized by its direction and a position relative to the coordinate origin. The angular velocity does not contain the latter. $\endgroup$
    – kricheli
    Oct 15, 2022 at 12:26
  • $\begingroup$ (I've deleted some of my earlier comments since they were obsolete after more carefully reading a sloppily written question.) Still remains: 3. Where do you take the formula from/what's it for? (If the formula implies that angular acceleration is the same in both frames, maybe it's only valid in frames that have zero angular acceleration with respect to each other and your using it under different conditions is causing your confusion...) $\endgroup$
    – kricheli
    Oct 15, 2022 at 12:38
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    $\begingroup$ If I understood correctly, a physical example to intuit is better than a page of calculations, as earthlings (observer O), we believed for centuries that the sun revolves around the earth (Ptolemy's system), but it is the reverse which is true (Copernic system, observer O' on the sun), all is relative, the angular speed is constant compared to the observer considers himself at rest (fixed). $\endgroup$
    – The Tiler
    Oct 15, 2022 at 15:41

4 Answers 4

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Regarding $$\bigg(\dfrac{d\hat \omega}{dt}\bigg)_\mathrm{fixed} = \bigg( \dfrac{d\hat \omega}{dt}\bigg)_\mathrm{rot}+\vec\omega\times\hat \omega$$ $\hat \omega$ is the same vector in both the fixed and rotating frames; it is the angular velocity of the rotating frame with respect to the fixed frame.

$\bigg( \dfrac{d\hat \omega}{dt}\bigg)_\mathrm{fixed}$ is the time rate of change of $\hat \omega$ expressed in fixed frame cooordinates.

$\bigg( \dfrac{d\hat \omega}{dt}\bigg)_\mathrm{rot}$ is the time rate of change of $\hat \omega$ expressed in rotating frame cooordinates. This is not the rate of change of the angular velocity IN the rotating frame; the angular velocity IN the rotating frame is a different vector, call it $\hat \Omega$ and is zero. $\hat \Omega$ is not $\hat \omega$.

See Goldstein, Classical Mechanics and my answer to The time derivatives of vectors in rotating frames on this exchange.

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Yes. With the conditions given the rotational velocity vector is fixed in all coordinate directions.

This is because if vector $\vec{A}$ is changing on the rotating frame then in the inertial frame its rate of change is

$$ \frac{\rm d}{{\rm d}t} \vec{A} = \frac{\partial}{\partial t} \vec{A} + \vec{\omega} \times \vec{A} $$

But in the case of rotational velocity $\vec{\omega} \times \vec{A} = \vec{\omega} \times \vec{\omega} = \vec{0}$, and so

$$ \frac{\rm d}{{\rm d}t} \vec{\omega} = \frac{\partial}{\partial t} \vec{\omega} $$

which means if the rate is zero in one frame, it is zero on the other frame also.

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  • $\begingroup$ I figured as much, but I am not assuming that $\dfrac{d\omega}{dt}$ is zero, instead that $\dfrac{d|\omega|}{dt}=0$. from that constraint I get that $\sum \omega_i \dot\omega_i = \sum \omega_i ' \dot \omega_i' = 0$ and $\sum \omega_i^2 = \sum \omega_i'^2 = |\omega|^2$ but I don't know where to go from there $\endgroup$
    – user256872
    Oct 10, 2022 at 23:16
  • $\begingroup$ If ${\rm R}$ is the 3×3 rotation matrix that transforms the basis vectors (directions) then $$\vec{\omega} = {\rm R} \vec{\omega}'$$ and the rotational transformation preserves vector lengths (by definition). $\endgroup$ Oct 11, 2022 at 0:01
  • $\begingroup$ "Why can’t ω^ be fixed in the body frame but change in the inertial one? Intuitively I can’t make sense of this." - Angular momentum is fixed in the space frame. Angular velocity might be fixed in the space frame if rotating about a principal axis. If it is rotating about a principal axis, then the rotation would be fixed in the body frame also as shown above. $\endgroup$ Oct 15, 2022 at 15:25
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the angular velocity components in body frame is

$$\vec\omega_B=\mathbf R\,\vec\omega_I$$

where $~\mathbf R~$ is the transformation matrix between inertial system and body system and $~\mathbf R^T\,\mathbf R=\mathbf I_3~$

from here $$\vec\omega_B^T\, \vec\omega_B=\vec\omega_I^T\,\mathbf R^T\,\mathbf R\,\vec\omega_I=\rm constant$$

hence the magnitude of the angular velocity in B system is also constant.

and with

$$\frac{d}{dt}\left(\vec\omega_B^T\, \vec\omega_B\right)=0\quad\Rightarrow\\ \vec\omega_B\cdot \vec{\dot{\omega}}_B=0$$

hence the magnitude of $~\vec{\dot{\omega}}_B~$ is not zero

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Short answer

For sure, the components of the angular velocity w.r.t. the body frame, moving with the body, can change in time during the motion of a rigid body. The evolution of the components of the angular velocity w.r.t. the body frame are governed by the Euler equations (nothing more than the components in th\dfrac{d \boldsymbol{\hat{\omega}}}{dt}e body frame of the dynamical equation for the rotational dynamics), that can be written as

$I_{G,XX} \dfrac{d \Omega_X}{dt} - ( I_{G,YY} - I_{G,ZZ} ) \Omega_Y \Omega_Z = M_{G,X}^{ext}$
$I_{G,YY} \dfrac{d \Omega_Y}{dt} - ( I_{G,ZZ} - I_{G,XX} ) \Omega_X \Omega_Z = M_{G,Y}^{ext}$
$I_{G,ZZ} \dfrac{d \Omega_Z}{dt} - ( I_{G,XX} - I_{G,YY} ) \Omega_X \Omega_Y = M_{G,Z}^{ext}$ ,

when referred to the center of gravity $G$ using the principal axis of inertia, so that the tensor of inertia is diagonal when written in the the body reference frame

$\mathbb{I}_G = I_{G,XX} \mathbf{\hat{X}}\otimes \mathbf{\hat{X}} + I_{G,YY} \mathbf{\hat{Y}}\otimes \mathbf{\hat{Y}} + I_{G,ZZ} \mathbf{\hat{Z}}\otimes \mathbf{\hat{Z}}$

and the angular velocity is written as $\mathbf{\omega} = \Omega_X \mathbf{\hat{X}} + \Omega_Y \mathbf{\hat{Y}} + \Omega_Z \mathbf{\hat{Z}} = \Omega_i \mathbf{\hat{X}}_i$.

Details

Vector equations. Equations of dynamics are vector equations, and thus it's convenient to write them as vector/tensor equations using abstract notation. Namely, if we trust in classical mechanics, the dynamical equation for the rotation of a rigid body can be written as

$\dfrac{d \mathbf{\Gamma}_H}{dt} = -\dot{\mathbf{x}}_H \times \mathbf{Q} + \mathbf{M}_H^{ext}$,

being $H$ the pole w.r.t. we evaluate the external moments $\mathbf{M}^{ext}_H$ and the angular momentum $\mathbf{\Gamma}_H$ that can be written as $\mathbf{\Gamma}_H = \mathbb{I}_G \cdot \mathbf{\omega} + (\mathbf{r}_G - \mathbf{r}_H) \times \mathbf{Q}$, and $\mathbf{Q}$ is the momentum of the body.

The form of this equation becomes easier if we use the center of gravity $G$ as the pole $H \equiv G$, namely

$\dfrac{d \mathbf{\Gamma}_G}{dt} = \mathbf{M}_G^{ext} \qquad $ with $\qquad \mathbf{\Gamma}_G = \mathbb{I_G} \cdot \mathbf{\omega}$.

Fixed (inertial) reference frame and body reference frame. Now we can introduce two reference frames with unit-length orthogonal vectors of the basis, usually used in the study of rigid body dynamics:

  • the fixed inertial reference frame as $\{\mathbf{x}_i\}_{i=1:3}$ that we assume fixed in space so that their time derivative is identically zero;
  • the body reference frame, moving in space with the rigid body, $\{\mathbf{X}_j\}_{j=1:3}$, useful because many properties of a rigid body (as an example, the components of the tensor of inertia) are constant in time when referred to the body reference frame. The angular velocity of the reference frame, is thus the same angular velocity $\mathbf{\omega}$ of the rigid body.

We can use these two reference frames to write every vector and tensor quantity; as an example, a vector $\mathbf{v}$ can be written as

$\mathbf{v} = v_i \mathbf{\hat{x}}_i = V_j \mathbf{\hat{X}}_j$

while its time derivative reads

$\dfrac{d\mathbf{v}}{dt} \quad = \quad \dfrac{d v_i}{dt} \mathbf{\hat{x}}_i + v_i \underbrace{\dfrac{d\mathbf{\hat{x}}_i}{dt}}_{=0} \quad = \quad \dfrac{d V_j}{dt} \mathbf{\hat{X}}_j + V_j \underbrace{\dfrac{d\mathbf{\hat{X}}_j }{dt}}_{=\mathbf{\omega} \times \mathbf{\hat{X}}_j \\ \text{(Poinsot)}} $
$\qquad \quad = \quad \dfrac{d v_i}{dt} \mathbf{\hat{x}}_i \qquad \qquad \quad = \quad \dfrac{d V_j}{dt} \mathbf{\hat{X}}_j + \mathbf{\omega} \times \mathbf{v}$

The angular velocity behaves in a "special way" when its time derivative is evaluated, since $\mathbf{\omega} \times \mathbf{\omega} = \mathbf{0}$. The angular acceleration $\mathbf{\alpha} = \frac{d \mathbf{\omega}}{dt}$ becomes

$\dfrac{d\mathbf{\omega}}{dt} = \dfrac{d \omega_i}{dt} \mathbf{\hat{x}}_i = \dfrac{d \Omega_j}{dt} \mathbf{\hat{X}}_j$$\qquad \qquad$$\mathbf{\alpha} = \alpha_i \mathbf{\hat{x}}_i = A_j \mathbf{\hat{X}}_j$

Change in direction of the angular velocity and orthogonal projectors. If we write the angular velocity $\boldsymbol{\omega} = \omega \boldsymbol{\hat{\omega}}$, its time derivative reads

$\dfrac{d \boldsymbol{\omega}}{dt} = \dfrac{d}{dt} (\omega \boldsymbol{\hat{\omega}}) = \dfrac{d \omega}{dt} \boldsymbol{\hat{\omega}} + \omega \dfrac{d \boldsymbol{\hat{\omega}}}{dt}$.

Since $|\boldsymbol{\hat{\omega}}| = 1 = \text{const.}$, we get that $\frac{d \boldsymbol{\hat{\omega}}}{dt }\cdot \boldsymbol{\hat{\omega}} = 0$, and multiplying the time derivative of $\boldsymbol{\omega}$ by the $\boldsymbol{\hat{\omega}}$,

$\boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt} = \dfrac{d \omega}{dt} \underbrace{\boldsymbol{\hat{\omega}} \cdot \boldsymbol{\hat{\omega}}}_{=1} + \omega \underbrace{\boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\hat{\omega}}}{dt}}_{=0}$$\qquad \rightarrow \qquad$$\dfrac{d \omega}{dt} = \boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$

and using this expression of the time derivative of the magnitude of the angular velocity, we can get the expression of the time derivative of the unit-vector $\boldsymbol{\hat{\omega}}$

$\dfrac{d \boldsymbol{\hat{\omega}}}{dt} = \dfrac{1}{\omega} \left[ \dfrac{d\boldsymbol{\omega}}{dt} - \boldsymbol{\hat{\omega}}\boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}\right] = \dfrac{1}{\omega} \left[ \mathbb{I} - \boldsymbol{\hat{\omega}} \otimes\boldsymbol{\hat{\omega}} \right] \cdot \dfrac{d \boldsymbol{\omega}}{dt} = \dfrac{1}{\omega}\mathbb{P}_{\perp,\boldsymbol{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$

having introduced the definition of the normal projector $\mathbb{P}_{\perp,\boldsymbol{\omega}} = \mathbb{I} - \boldsymbol{\hat{\omega}} \otimes\boldsymbol{\hat{\omega}}$ in the direction orthogonal to $\boldsymbol{\hat{\omega}}$. Thus, we can summarize that:

  • time derivative of the magnitude of the angular velocity is the projection of the time derivative of the angular velocity along the angular velocity direction:

    $\dfrac{d \omega}{dt} = \boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$

  • time derivative of the unit vector indicating the direction of the angular velocity comes from the projection of the time derivative of the angular velocity in the directions orthogonal w.r.t. its direction $\dfrac{d \boldsymbol{\hat{\omega}}}{dt} = \dfrac{1}{\omega}\mathbb{P}_{\perp,\boldsymbol{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$

and thus write

$\dfrac{d \boldsymbol{\omega}}{dt} = \underbrace{\dfrac{d \omega}{dt} \boldsymbol{\hat{\omega}}}_{=\mathbb{P}_{//,\boldsymbol{\omega}} \cdot \frac{d \boldsymbol{\omega}}{dt}} + \underbrace{\omega \dfrac{d \boldsymbol{\hat{\omega}}}{dt}}_{=\mathbb{P}_{\perp,\boldsymbol{\omega}} \cdot \frac{d \boldsymbol{\omega}}{dt}} = \mathbb{P}_{//,\boldsymbol{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt} + \mathbb{P}_{\perp,\boldsymbol{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$

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  • $\begingroup$ Thanks for your answer. However I understand the Euler equations, how we choose our axes, etc. what I need help visualizing is how it is possible for the direction of angular velocity to change wrt to the body frame and why a constant angular velocity direction in the body frame corresponds to a fixed angular velocity in the fixed frame. The math makes sense, what I seek is the intuition. $\endgroup$
    – user256872
    Oct 15, 2022 at 11:22
  • $\begingroup$ I'm sorry but I really can't get your point $\endgroup$
    – basics
    Oct 15, 2022 at 12:29
  • $\begingroup$ I'm updating my answer with the time derivative of a vector written as $\mathbf{v} = v \mathbf{\hat{v}}$ $\endgroup$
    – basics
    Oct 15, 2022 at 13:23

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