1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

About rotation and precession

  1. May 18, 2009 #1
    I was just reading about torque-free precession and having difficulty seeing how an object's angular velocity vector could at any moment point in a direction that is different than the direction of its angular momentum vector, since that is not possible for an isolated particle

    [tex]

    \vec \omega = \frac {1}{r^2} ( \vec r \times \vec v)

    [/tex]

    [tex]

    \vec L = m ( \vec r \times \vec v)

    [/tex]

    (where [tex] r = | \vec r | [/tex])

    and in both cases one is just adding up the vector contribution of all the object's constituent particles.

    However, the answer to the following question might resolve the matter for me: when an object spins freely in space -- that is, unattached to any other object and with no torque being applied to it -- does the angular velocity vector of each of its particles at any given moment point in the same direction? My intuition says yes, but it is often wrong.
     
  2. jcsd
  3. May 18, 2009 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    On object is not a point mass. The velocities of the particles that constitute a rigid body are constrained. The inertial frame velocity of the ith particle in a rigid body is

    [tex]
    \boldsymbol v_i =
    \boldsymbol v_{\text{com}} +
    \boldsymbol{\omega} \times (\boldsymbol r_i - \boldsymbol r_{\text{com}})
    [/tex]

    where the position and velocity vectors are with respect to some external reference frame (typically an inertial frame), [itex]\omega[/itex] is the angular body of the object with respect to that frame, and the com subscript indicates the position and velocity of the object's center of mass.

    After a lot of tedious grinding through math, the angular momentum of the object is, with a bit of abuse of notation,

    [tex]L_{\text{obj}} = m_{\text{obj}}\boldsymbol r_{\text{com}} \times \boldsymbol v_{\text{com}} +
    \mathbf I \boldsymbol \omega
    [/tex]

    The abuse of notation comes from a mixing of coordinate frames. The position and velocity are preferably with respect to and represented in some inertial reference frame. The inertia tensor is invariably represented in some frame fixed with respect to the object and with origin at the object's center of mass. Finally, the angular velocity is the object's angular velocity with respect to inertial but represented in rotating frame coordinates.
     
    Last edited: May 18, 2009
  4. May 18, 2009 #3
    Thanks, D H.

    I have a couple immediate questions about your response:

    1. In the first equation, should there be a cross-product symbol between the omega vector and the difference of the radii vectors? If not, how are the vectors combined?

    2. What does "com" mean?
     
  5. May 18, 2009 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Fixed. See the updated post.
     
  6. May 18, 2009 #5

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Now to answer the original question.

    Elementary physics often teach that the rotational analog of Newton's second law,

    [tex]\aligned
    \frac{d\boldsymbol p}{dt} &= \boldsymbol F \qquad\qquad\text{or} \\
    \boldsymbol F &= m\boldsymbol a
    \endaligned[/tex]

    are

    [tex]\aligned
    \frac{d\boldsymbol L}{dt} &= \boldsymbol {\tau} \qquad\qquad\text{or} \\
    \boldsymbol \tau &= I\frac{d\boldsymbol {\omega}}{dt}
    \endaligned[/tex]

    This is not true in general; it is true if various simplifying assumptions are met (which these introductory courses of course ensure are the case).

    Rather than write this up one more time, see [post=2175593]this post[/post], [post=2091802]this post[/post], and [post=2006454]this post[/post].
     
  7. May 18, 2009 #6
    Well, those last two threads are pretty deep. I would like to narrow the matter down if possible. My first question, then, is . . . if a rigid body moves freely in space - that is, without the influence of any external forces - and I view it in an inertial (non-accelerating, non-rotating) reference frame, is it true that at any given moment the object will be rotating about an axis that passes through its center of mass?
     
  8. May 18, 2009 #7

    diazona

    User Avatar
    Homework Helper

    Yep (if it's rotating at all), because if it were rotating about some other axis, the center of mass would be undergoing a centripetal acceleration, which would require some external force to maintain.
     
  9. May 18, 2009 #8
    OK, then. So at any given moment in this inertial frame we can define an angular velocity vector [itex] \vec \omega [/itex] for the object, based on this rotation?
     
  10. May 18, 2009 #9

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Correct.
     
  11. May 19, 2009 #10
    And for every [itex] \Delta m [/itex] that makes up the object, a [itex] \Delta \vec L [/itex] can be assigned relative to this axis and rate of rotation?
     
  12. May 24, 2009 #11
    OK, I've given this some more thought and I now have a couple follow-up questions which may clear this matter up for me:

    Suppose I have an object that is spinning freely in space and experiencing torque-free precession. Its momentary angular velocity is [itex] \vec \omega [/itex] and its (constant) angular momentum is [itex] \vec L [/itex] and these are not pointing in the same direction.

    If I could grab the rotational axis of the object so that its rate of rotation did not change but its precession suddenly stopped, would [itex] \vec L [/itex] still be pointing in its original direction or would it become parallel to [itex] \vec \omega [/itex]?

    If I release the object, would precession resume?
     
  13. May 24, 2009 #12

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    What you are doing by grabbing the rotational axis and holding it steady is applying a torque to the object. Instead of the angular velocity vector rotating around the constant angular momentum vector, you are making the angular momentum vector rotate around the constant angular velocity vector. The torque will vanish when you release your hold on the rotational axis. The angular momentum vector will stop rotating (no torque) and the angular velocity vector will once again start precessing.
     
  14. May 25, 2009 #13
    So in this case when I hold on to the object's rotational axis, the orientation of the spinning is fixed, the revolutions per second is constant - but the angular momentum is changing?

    I can understand this if the time it takes for the angular momentum arrow to make one revolution about this axis is the same as the period of the spinning object (since the moment of inertia tensor is a function only of the distribution of the object's mass) but otherwise I don't see how this is possible.

    If one takes a look at the object at time [itex] t [/itex] and then again at time [itex] t + p [/itex] where [itex] p [/itex] is the period of its rotation, at both moments the positions and velocities of all the its constituent particles will be the same. What else could influence a mechanical quantity like angular momentum besides the mass, position and velocity of every particle in the system?
     
  15. May 27, 2009 #14
    I guess at this point it would be helpful for me to learn Euler's equations. I've read a little about them and there's something fundamental that I don't understand. The sources always mention that they are not for an inertial frame of reference, but for one that is "fixed in the rotating body" or "is rotating along with the body" or "whose axes are aligned along the principal axes of rotation of the body", etc.

    My question is: if the reference frame moves with the object, shouldn't [itex] \vec \omega [/itex] and its time derivative always be zero?
     
  16. May 27, 2009 #15

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Let's start with Euler's equations:

    [tex]\aligned
    I_1\frac{d{\omega}_1}{dt} + (I_3-I_2)\omega_2\omega_3 &= \tau_1 \\
    I_2\frac{d{\omega}_2}{dt} + (I_1-I_3)\omega_3\omega_1 &= \tau_1 \\
    I_3\frac{d{\omega}_3}{dt} + (I_2-I_1)\omega_1\omega_2 &= \tau_3
    \endaligned[/tex]

    where
    • The Ii are the moments of inertia of some body in one of that body's principal axis frames,
    • The ωi are the components of the angular velocity expressed in that frame, and
    • The τi are the components of the external torque expressed in that same frame.

    My advice: Forget Euler's equations. Euler died more than 100 years before the development of vectors. A much more general form for Euler's equations written in modern notation is

    [tex]\mathbf I \frac{d\boldsymbol{\omega}}{dt} = \boldsymbol{\tau} - \boldsymbol{\omega}\times(\mathbf I \boldsymbol{\omega})[/tex]

    where
    • I is the moment of inertia tensor expressed in some (not necessarily principal axis) body frame,
    • ω is the angular velocity vector expressed in that frame, and
    • τ is the torque vector expressed in that same frame.

    So, how does this expression arise and what does it mean?

    In introductory physics (up to and including freshman college physics), one is often taught that the rotational equivalent of F=ma is τ=Iα. Just as F=ma is not true in general (think varying mass), the rotational equivalent also is not true in general. The more general versions are

    [tex]\aligned
    \boldsymbol F &= \frac{d\boldsymbol p}{dt} \\
    \boldsymbol{\tau} &= \frac{d \boldsymbol L}{dt}
    \endaligned[/tex]

    where p and L are the linear and angular momenta as expressed in inertial coordinates. The simple expression for the linear case arises any time the mass is constant. That isn't true for the rotational problem. For a rigid body, the inertia tensor is constant in some body referenced frame, but it is not constant in an inertial frame. The inertia tensor in some inertial frame is

    [tex]\mathbf I_I = \mathbf T_{B\to I} \mathbf I_B \mathbf T_{B\to I}^T[/tex]

    Differentiating this with respect to time results in a bit of a mess. Since the inertia tensor for a rigid body is constant in the rotating body frame, a better way to look at things is from the perspective of a rotating frame. A general expression for the relation between the time derivative of any vector quantity q[/] from the perspectives of an inertial frame versus a rotating frame (taught with more than a bit of handwaving in the ophomore/junior classical mechanics class) is

    [tex]\boldsymbol q_I = \boldsymbol q_R + \boldsymbol{\omega}\times \boldsymbol q[/tex]

    Now let the vector q be the angular momentum vector L=Iω. Tada! The rotational equivalent of Newton's second law results:

    [tex]\mathbf I \frac{d\boldsymbol{\omega}}{dt} = \boldsymbol{\tau} - \boldsymbol{\omega}\times(\mathbf I \boldsymbol{\omega})[/tex]
     
  17. May 28, 2009 #16
    Hey, that's the first explanation of Euler's equations I've understood. Thanks, DH!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: About rotation and precession
Loading...