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Rigid Body Derivation of Momentum Equations 
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#1
Jun2412, 06:18 PM

P: 5

1. The problem statement, all variables and given/known data
Hello, I'm having some issues deriving the momentum equation for rigid body motion in a noninertial 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. 2. Relevant equations None 3. 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_{Bcg}+ω×(ω×r_{Bcg}))[/itex] Bcg 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_{Bcg}=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 noninertial 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_{framemotion}+\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] 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 


#2
Jun2512, 01:29 AM

HW Helper
Thanks
P: 4,842

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
Jun2512, 01:47 PM

P: 5

Ah that makes sense. Let me work with this for a little while, and if I have any issues I'll repost. Thanks!



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