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I Derivative of angular velocity of rotating co. system

  1. Jul 22, 2017 #1
    What is time derivative of angular velocity ( measured w.r.t. an inertial frame ) of a rotating co. system w.r.t. the same rotating co. system?

    I think a person sitting in a closed rotating box feels the an object at rest w.r.t. him as rest. He doesn't observe the angular velocity of the object changing.
    So, the answer to the above question is 0.

    Is this correct?
  2. jcsd
  3. Jul 22, 2017 #2


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    No. In the following I work with Cartesian vector and tensor components. The Einstein summation convention is used, i.e., you always have to sum over repeated indices from 1 to 3.

    Let ##\vec{e}_j## denote a Cartesian coordinate system fixed in an inertial frame and ##\vec{e}_j'## one fixed in the rotating system. Then you have
    $$\vec{e}_j'=D_{ij} \vec{e}_i,$$
    where ##\hat{D}=(D_{ij}) \in \mathrm{SO}(3)## is a time-dependent rotation matrix. For the position vector you have
    $$\vec{x}=x_j' \vec{e}_j'=D_{ij} x_j' \vec{e}_i \; \Rightarrow x_i=D_{ij} x_j'.$$
    Now you get
    $$v_i=\dot{x}_i=\dot{D}_{ij} x_{j}'+D_{ij} \dot{x}_j'=D_{ij} [\dot{x}_j+(D^{-1})_{jk} \dot{D}_{kl} x_l'].$$
    Now since ##\hat{D}^{-1}=\hat{D}^T## you have ##\hat{D} \hat{D}^T=1## and thus ##\dot{\hat{D}} \hat{D}^T+ \hat{D} \dot{\hat{D}}^T=0## or
    $$\hat{D} \dot{\hat{D}}^T=-(\hat{D} \dot{\hat{D}}^T)^{T}.$$
    That means that the matrix ##\hat{D} \dot{\hat{D}}^T## is antisymmetric and thus we can express it with help of the Levi-Civita symbol via an axial vector, i.e.,
    $$(D^{-1})_{jk} \dot{D}_{kl}=\epsilon_{jkl}\omega_k',$$
    and thus
    $$v_i=D_{ij} (\dot{x}_j' + \epsilon_{jkl} \omega_k' x_l').$$
    Or writing the components as column vectors ##\underline{v}## etc. you get
    $$\underline{v} = \hat{D} (\dot{\underline{x}}' + \underline{\omega}' \times \underline{x}').$$
    Now in the same way you can take another time derivative, leading to
    $$\underline{a}=\dot{\underline{v}} = \hat{D} [\ddot{\underline{x}}' + 2 \underline{\omega}' \times \dot{\underline{x}}' + \underline{\omega}' \times (\underline{\omega}' \times \underline{x}') + \dot{\underline{\omega}'} \times \underline{x}'].$$
    Multiplying by the mass of the particle you get finally the Newtonian equation of motion in the rotating frame as
    $$m \ddot{\underline{x}}'=\underline{F}' - 2m \underline{\omega}' \times \dot{\underline{x}}' -m \underline{\omega}' \times (\underline{\omega}' \times \underline{x}') -m \dot{\underline{\omega}'} \times \underline{x}'].$$
    So despite the "true force" represented by the components ##\underline{F}'## wrt. the rotating basis you get three sorts of "inertial forces" which, as the derivation shows, comes from the rotation of the rotating frame wrt. the inertial frame. Sometimes one thus calls them also "fictitious forces", because you can get rid of them by transforming back into an inertial reference frame.

    The inertial forces on the right-hand side of the final equation are called Coriolis force (2nd term), centrifugal force (third term). The fourth term has no special name, as far as I know, but it's due to the angular acceleration for the rotating reference frame wrt. the inertial frame. In the usual application of the motion of particles (or as important in meteorology fluids like the air) on Earth the latter force usually can be negected since the rotation of the Earth is with pretty constant angular velocity in first approximation. For motion close to the Earth also the centrifugal term has to be lumped into the effective gravitational force, approximated by ##m \underline{g}'##. So that you finally have for this special case
    $$m \ddot{\underline{x}}'=m \underline{g}'-2 m \underline{\omega}' \times \dot{\underline{x}}'.$$
    You can use this formula to calculate the free fall on the rotating Earth or the fascinating motion of the Fourcault pendulum.
  4. Jul 22, 2017 #3


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    The time derivative calculated in a coordinate system whose rotational rate is changing would not include (detect?) it's own rotational acceleration.
    A person would certainly feel it if he was in a box whose rate of rotation changed. He would have to explain the difference between the physical force he feels and any time derivatives he calculates in the box coordinates.

    The equations of motion that are used to understand the flight of an airplane have to take all the rotational accelerations into account in the calculations.
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