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!

Rigid Body Motion: Determing Angular Momentum Equations

  1. Jul 30, 2012 #1
    Hello,
    I'm trying to determine whether my work for the following rigid body motion is correct, and if so why I'm not able to verify equivalent formulations.

    For the sake of simplicity small angle approximations will be used, and cross terms were neglected.

    1. The problem statement, all variables and given/known data
    Consider the rigid body motion (shown in the attached schematic). The body rotates, represented in the schematic by its center of gravity(CGm), about the origin of X'Y'Z'. X''Y''Z'' represents a body fixed reference frame, which has an angular velocity of [itex]\dot{\Phi}\hat{i}''[/itex] with respect to the X'Y'Z' frame, and XYZ represents the inertial reference frame. Note that the X'Y'Z' frame also rotates with respect to the inertial frame with an angular velocity of [itex]\dot{\Psi}\hat{k}'[/itex]. Additionally, the body has a velocity V defined in the XY plane of the inertial frame.

    2. Relevant equations
    To derive the angular momentum first note that while the inertial frame is shown detached from the X'Y'Z' frame we can always find an inertial frame coincident and with the same instantaneous velocity as the X'Y'Z' frame. For clarity I left them separated in the schematic.

    My Issue:
    There should be two equivalent ways to define the angular momentum of the body in the X'Y'Z'.

    1.) Moment of momentum. Defining the momentum of the body in the X'Y'Z' frame and then taking the moment arm (also defined in the X'Y'Z' frame) should be the body's angular momentum.

    2.) Using the body fixed frame X''Y''Z'' define the angular momentum in the inertial frame due to changes in [itex]\Phi[/itex]. Because the inertial frame and X'Y'Z' frame are assumed coincident at the moment considered the basis vectors are interchangeable. Then add the angular momentum due to the forward velocity of the body.

    3. The attempt at a solution
    Consider the moment of momentum method. The momentum of the body is due to the changes in [itex]\Phi[/itex] (denoted [itex]P|_{roll}[/itex]) and the forward velocity of the body (denoted [itex]P|_{bulk}[/itex]).

    [itex]P|_{roll}=mV_{X'Y'Z'}=m\frac{\text{d}}{\text{dt}}r_{CGm}:r_{CGm}=h\sin{\Phi}\hat{j}'-h\cos{\Phi}\hat{k}' \approx h\Phi\hat{j}'-h\hat{k}'[/itex]

    [itex] P|_{roll}=m\frac{\text{d}}{\text{dt}}(h\Phi\hat{j}'-h\hat{k}')=mh\dot{\Phi}\hat{j}'[/itex]

    [itex]P|_{bulk}=m(V\cos{\beta}\hat{i}'+V\sin{\beta}\hat{j}') \approx mV\hat{i}'+mV\beta\hat{j}'[/itex]

    [itex]\implies P_{X'Y'Z'}=P|_{roll}+P|_{bulk}=mV\hat{i}'+(mV\beta+mh\dot{\Phi})\hat{j}'[/itex]

    The angular momentum should then be
    [itex]H_{X'Y'Z'}=r_{CGm}\times P_{X'Y'Z'}=(h\Phi\hat{j}'-h\hat{k}')\times (mV\hat{i}'+(mV\beta+mh\dot{\Phi})\hat{j}')=(mVh \beta+mh^2\dot{\Phi})\hat{i}'-mhV\hat{j}'-mhV\Phi\hat{k}'[/itex]

    Consider the body fixed frame method. The angular momentum of a body, about an arbitrary point in a body fixed frame, defined in the inertial frame is given by
    [itex]H_{XYZ}=H|_{roll}=\[
    \begin{matrix}
    Ixx & -Ixy & -Ixz \\
    -Iyx & Iyy & -Iyz \\
    -Izx & -Izy & Izz
    \end{matrix}\] \[
    \begin{matrix}
    \omega_x \\
    \omega_y \\
    \omega_z
    \end{matrix}
    \] [/itex]

    (sorry the matrix command isn't working for me)

    where [itex] \omega_x,\omega_y,\omega_z[/itex] is the rotation of the body with respect to the inertial frame. For the given example [itex] \omega_x=\dot{\Phi},\omega_y=0,\omega_z=\dot{\Psi}[/itex]

    [itex]\implies H|_{roll}= (Ixx \dot{\Phi}-Ixz \dot{\Psi})\hat{i}'+(-Izx \dot{\Phi}+ Izz \dot{\Psi})\hat{k}'[/itex]

    and

    [itex] H|_{bulk}=r_{CGm}\times P|_{bulk}=(h\Phi\hat{j}'-h\hat{k}') \times (mV\hat{i}'+mV\beta\hat{j}')=hmV \beta \hat{i}'-hmV\hat{j}'-mVh\Phi\hat{k}'[/itex]

    [itex]\implies H_{X'Y'Z'}=H|_{roll}+H_{bulk}=(Ixx \dot{\Phi}-Ixz \dot{\Psi}+hmV \beta) \hat{i}'-hmV\hat{j}'+(-Izx \dot{\Phi}+ Izz \dot{\Psi}-mVh\Phi)\hat{k}'[/itex]

    Summary
    Comparing [itex]H_{X'Y'Z'}[/itex] from the two different methods we can see that the formulations are similar, but they must be equivalent.

    Moment of Momentum:
    [itex]H_{X'Y'Z'}=(mVh \beta+mh^2\dot{\Phi})\hat{i}'-mhV\hat{j}'-mhV\Phi\hat{k}'[/itex]

    Body Fixed Frame:
    [itex]H_{X'Y'Z'}(Ixx \dot{\Phi}-Ixz \dot{\Psi}+mVh \beta) \hat{i}'-mhV\hat{j}'+(-Izx \dot{\Phi}+ Izz \dot{\Psi}-mhV\Phi)\hat{k}'[/itex]

    The only way for the previous formulation to always be true is for the following to hold
    1.) [itex]mh^2\dot{\Phi}=Ixx \dot{\Phi}-Ixz \dot{\Psi}[/itex]
    2.) [itex]-Izx \dot{\Phi}+ Izz \dot{\Psi}=0[/itex]

    While there are instances where these two conditions could be met, it's much more likely I've made a mistake :smile:. If anyone sees where I've gone wrong I'd greatly appreciate the help.
    Brad
     

    Attached Files:

  2. jcsd
  3. Jul 31, 2012 #2
    Assume for the sake of the argument that V is identically zero. Then the angular momentum in X'Y'Z', derived per the second method, would be identical with that in XYZ. That cannot be correct, because they rotate relative to each other.
     
  4. Jul 31, 2012 #3
    Hello,

    I appreciate you taking the time to look at my problem.

    Considering the second method I presented, if you assume that there is no forward velocity then

    [itex]H|_{bulk}=0 \implies H_{X'Y'Z'}=H|_{roll}: H|_{roll}=(Ixx \omega_x + -Ixz\omega_z)\hat{i}+ (-Izx \omega_x + Izz \omega_z)\hat{k}[/itex]

    If I am understanding you correctly the issue you've brought up is changing from the inertial frame, basis vectors [itex]\hat{i},\hat{j},\hat{k}[/itex], to the X'Y'Z' basis vectors, [itex]\hat{i}',\hat{j}',\hat{k}'[/itex].

    Now the trick, as I've seen it, is to realize that an inertial frame does not imply a zero velocity (in a non-rigorous explanation: inertial frames are equivalent as long as the motion of the frame has only constant velocity). Therefore we can select the inertial frame of interest to be one that is coincident with the X'Y'Z' frame and has the same instantaneous velocity. Therefore when we consider the angular momentum vector (technically pseudo-vector) viewed from the inertial XYZ, the projection of the vector along the X'Y'Z' frame is simply the original angular momentum vector with changed basis vectors.

    Does that make sense?

    Brad
     
  5. Jul 31, 2012 #4
    Imagine you are standing on a pole of the Earth (= origin of X'Y'Z'). The angular momentum of the Earth, as measured in your co-rotating frame of reference, is zero. The angular momentum of the Earth, as measured in the inertial frame of reference (neglecting the Earth orbital motion), say in the Earth's center of mass, or even the one that is momentarily coincident with X'Y'Z', is not zero.

    You, however, equate the momenta in the two reference frames.

    Have a look here: http://en.wikipedia.org/wiki/Rotating_reference_frame#Time_derivatives_in_the_two_frames
     
  6. Jul 31, 2012 #5
    Ultimately, I want to relate the angular momentum to the sum of the torques. Towards that end we have to discuss semantics a little bit. We need to be careful a a vector viewed from and defined in. I drew up a simple schematic for discussion purposes. Consider the point A, mass m, translates due to external forces with V=V(t) and a=a(t). X'Y' remains coincident with point A throughout.

    If you considered the basic kinematic equations
    [itex] \frac{\text{d}}{\text{dt}} P|_{XY}=\sum{Forces}=\frac{\text{d}}{\text{dt}}P|_{X'Y'}+\omega \times P|_{X'Y'}[/itex]

    where [itex] \omega [/itex] would be the rotation of the X'Y' with respect to the inertial frame.

    If I understand your previous post, to define [itex] P_{X'Y'} [/itex] we'd consider the momentum of the particle, defined in the X'Y' frame. Since the frame is coincident with the particle at all times the velocity of the particle, defined in the X'Y' frame, is zero and therefore [itex] P_{X'Y'}=0 \implies \frac{\text{d}}{\text{dt}} P|_{XY}=0 \neq ma[/itex].

    However, what I'm saying is that [itex] P_{X'Y'} [/itex] is the momentum vector, defined in the XY frame, as viewed from the X'Y' frame. This would give, in this case, [itex] P_{X'Y'}=V\cos{\theta}\hat{i}'+V\sin{\theta}\hat{j}' [/itex].

    Applying the same to the example you gave, I agree that the angular momentum defined in the co-rotating frame would be zero. However, if we're going to consider the sum of the torques on a body in the co-rotating frame, then the angular momentum, viewed in the co-rotating frame, would not be zero.

    I provided a bit more rigor in this post (I initially had the same thoughts you presented):
    https://www.physicsforums.com/showthread.php?p=3971049#post3971049

    Newton's Laws seem to cause me trouble to no end!
     

    Attached Files:

  7. Aug 1, 2012 #6
    Just because the momentum happens to be zero at any given instant in any given frame does not mean its derivative is zero in that same frame. A momentarily co-incident inertial frame might cancel out some velocities, but it cannot cancel out their derivatives.

    Coming back, when you have a body rotating about an axis which is itself rotating in an inertial frame, then the frame in which the mentioned axis is fixed will only register the rotation of the body about the axis - just because the motion is defined as such!

    Forces (and torques) are a different matter. To deal with forces in a rotating frame, you would need to add inertial forces to the real forces acting in the frame.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rigid Body Motion: Determing Angular Momentum Equations
  1. Rigid body motion (Replies: 1)

  2. Rigid body motion (Replies: 14)

Loading...