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bradm707
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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.
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.
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.
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 . If anyone sees where I've gone wrong I'd greatly appreciate the help.
Brad
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.
Homework Statement
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.
Homework 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.
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 . If anyone sees where I've gone wrong I'd greatly appreciate the help.
Brad