Rigid Body- Derivation of Momentum Equations

In summary, the conversation discusses deriving the momentum equation for rigid body motion in a non-inertial frame. Two common methods are discussed, one in a stationary inertial frame and one in a moving inertial frame. There is some confusion about the interpretation of certain equations, but overall the conversation provides a thorough explanation of the topic.
  • #1
bradm707
5
0

Homework Statement


Hello,
I'm having some issues deriving the momentum equation for rigid body motion in a non-inertial frame. Consider a rigid body moving in a two dimensional (inertial XY) plane. A body fixed frame (origin denoted as B) is located off the center of gravity. The body has arbitrary motion in the XY plane and arbitrary rotation about the Bz axis (denoted as ω[itex]\hat{k}[/itex])

There are two common methods I've seen to derive solutions to this problems. The first should always be true and I've included it for completeness (i.e. just so people can double check I haven't gone completely insane), but the second analysis is what is giving me trouble.

Homework Equations


None

The Attempt at a Solution


3i. Analysis in a Stationary Inertial Frame (V[itex]_{x}[/itex][itex]\hat{I}[/itex], V[itex]_{y}[/itex][itex]\hat{J}[/itex]=0)
First consider an arbitrary point on the rigid body (denoted as A). There are three significant position vectors to consider: 1.) Vector from the inertial frame to the body fixed frame (r[itex]_{OB}[/itex]), 2.) Vector from body fixed frame to the arbitrary point (r[itex]_{BA}[/itex]), and 3.) Vector from the inertial frame to the arbitrary point (r[itex]_{OA}[/itex]).

The rate of change of r[itex]_{OA}[/itex] is simply
[itex]\frac{d}{dt}r_{OA}= \frac{d}{dt}r_{OB}+\frac{D}{Dt}r_{BA}:\frac{D}{Dt}[/itex] represents the rate of change of the the vector BA w.r.t. the inertial frame.

[itex]\frac{D}{Dt}r_{BA}= \frac{d}{dt}r_{BA}+ω×r_{BA}[/itex]

[itex]\Rightarrow\frac{d}{dt} r_{OA}= \frac{d}{dt}r_{OB}+\frac{d}{dt}r_{BA}+ω×r_{BA}[/itex]

and

[itex]\frac{d^{2}}{dt^{2}}r_{OA}= \frac{d^{2}}{dt^{2}}r_{OB}+\frac{d^{2}}{dt^{2}}r_{BA}+2ω×\frac{d}{dt}r_{BA}+\dot{ω}×r_{BA}+ω×(ω×r_{BA})[/itex]

With the assumption that the body is rigid, [itex]\frac{d}{dt}r_{BA}[/itex] and [itex]\frac{d^{2}}{dt^{2}}r_{BA}[/itex] equal 0. If point A represents an arbitrary point mass, the momentum of the point (viewed from the stationary inertial frame) is [itex]p=m_A a_A: a_A=\frac{d^{2}}{dt^{2}}r_{OA}[/itex] is the acceleration of point A viewed in the inertial frame. Summing over all points and assuming that all points share the same velocity and acceleration (rigid body assumption) gives

[itex]\dot{p}=m\frac{d^{2}}{dt^{2}}r_{OA}= m(\frac{d^{2}}{dt^{2}}r_{OB}+\dot{ω}×r_{B-cg}+ω×(ω×r_{B-cg}))[/itex]

B-cg is the position vector from the the body frame to the body center of gravity. Note that if B is located at the object cg [itex]r_{B-cg}=0[/itex] and the rate of change of momentum equals the acceleration of the body center of gravity.

Aside Question 1:
We can also note that we have described the rate of change (viewed in an inertial frame) of an arbitrary vector defined in a non-inertial reference frame. We can note that momentum is actually a vector defined in any frame (although the rate of change of momentum only equals the sum of the external forces in an inertial frame). I would have thought the rate of change of momentum (defined in the body fixed frame) w.r.t. a zero velocity inertial frame would be given by

[itex]\frac{d}{dt} p_{inertial}= \frac{d}{dt}p_{frame-motion}+\frac{d}{dt}p_{body}+ω×p_{body}[/itex]

but most of the references I've seen just have the last two terms on the RHS (a possible reason is given in 3ii). Additionally, I would think the momentum defined in the body fixed frame would be equal to zero. I think this because the body fixed frame should be translating with the same velocity of the body, and therefore the velocity vector in the body fixed frame is a zero magnitude vector. If someone could shed some light on this I'd appreciate it.

3ii. Analysis in Moving Inertial Frame (V[itex]_{x}[/itex][itex]\hat{I}[/itex], V[itex]_{y}[/itex][itex]\hat{J}[/itex]=constant (≠0)
We can note that all inertial frames differ at most by a constant velocity. If, at each instant in time, an inertial frame with the same velocity as the rigid body is considered, the rate of change of an arbitrary vector (defined in the body fixed frame) w.r.t. the inertial frame is given by

[itex]\frac{d}{dt} r_{OA}= [/itex][STRIKE]\frac{d}{dt}r_{OB}[/STRIKE][itex]+\frac{d}{dt}r_{BA}+ω×r_{BA}[/itex]

where [itex] \frac{d}{dt}r_{OB} [/itex] is zero because the inertial frame has the same velocity at the instant considered. This would give that the rate of change of momentum is

[itex]\frac{d}{dt} p_{inertial}=\frac{d}{dt}p_{body}+ω×p_{body}[/itex]

but I still don't know why the momentum in the body fixed frame isn't equal to zero. This is the formulation that also leads to the Euler Equations (what I'm also trying to derive).

I'm not really sure how this formulation is helpful, but it is very prevalent in the mechanical systems modeling references I've found. If anyone has thoughts on why this formulation is preferred, or good references they'd be appreciated.

Thanks,
Brad
 
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  • #2
I think part of the difficulty is giving the correct interpretation to the equation ([itex]\frac{dA}{dt}[/itex])[itex]_{I}[/itex]=([itex]\frac{dA}{dt}[/itex])[itex]_{B}[/itex] + ωxA, which says that the rate of change of a vector A as viewed in the inertial frame (I) equals the rate of change of the same vector as viewed in the body frame (B) plus ωxA.

Now let the vector A be the momentum of the body relative to the inertial frame, [itex]P_{I}[/itex]. Then we would have ([itex]\frac{dP_{I}}{dt}[/itex])[itex]_{I}[/itex]=([itex]\frac{dP_I}{dt}[/itex])[itex]_{B}[/itex] + ωx[itex]P_{I}[/itex].

Note that the same vector [itex]P_{I}[/itex] appears throughout. In particular, the first term on the right side of the equation is not ([itex]\frac{dP _ B}{dt}[/itex])[itex]_{B}[/itex] where [itex]P_{B}[/itex] is the momentum of the body with respect to the body (which would be zero).

It can be very confusing.
 
  • #3
Ah that makes sense. Let me work with this for a little while, and if I have any issues I'll re-post. Thanks!
 

1. What is the derivation of momentum equations for rigid bodies?

The derivation of momentum equations for rigid bodies is a fundamental concept in physics and engineering. It involves the application of Newton's laws of motion to a rigid body, which is a physical object that maintains its shape and size even when subjected to external forces.

2. What are the equations used in the derivation of momentum equations for rigid bodies?

The equations used in the derivation of momentum equations for rigid bodies are the linear momentum equation, the angular momentum equation, and the moment of inertia equation. These equations describe the relationships between the forces, mass, and motion of a rigid body.

3. How is the linear momentum equation derived for rigid bodies?

The linear momentum equation for rigid bodies is derived by applying Newton's second law of motion, which states that the net force acting on a body is equal to the rate of change of its momentum. This equation takes into account both the translational motion and the rotational motion of the rigid body.

4. What is the significance of the angular momentum equation in the derivation of momentum equations for rigid bodies?

The angular momentum equation is important in the derivation of momentum equations for rigid bodies because it takes into account the rotational motion of the body. It relates the angular momentum of the body to the external torques acting on it, and provides a way to analyze the rotational motion of a rigid body.

5. How is the moment of inertia equation used in the derivation of momentum equations for rigid bodies?

The moment of inertia equation is used in the derivation of momentum equations for rigid bodies to describe the resistance of a body to rotational motion. It is a measure of how much force is required to rotate a body around a given axis, and is crucial in determining the rotational behavior of a rigid body.

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