Bedford & Fowler — Chapter 18, Planar Dynamics of Rigid Bodies

Chapter 18 — Planar Dynamics of Rigid Bodies: Decision Map

Forces and moments on a rigid body — how does it move?

General selection Moment of inertia Translation Fixed-axis rotation General plane motion
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Planar Equations of Motion — §18.1–18.2
ΣFx = m(aG)x,  ΣFy = m(aG)y
ΣMG = IGα
Free-body diagram = kinetic diagram (maG at G, plus IGα)
Details
Or, about any point P: ΣMP = m(aG)xdy − m(aG)ydx + IGα (kinetic moment).
Mass Moment of Inertia — Ch. 18 Appendix
I = ∫ r² dm
I = IG + m d²
k = √(I/m)
Details
Rod (about G): 1/12 ml²;  rod (about an end): 1/3 ml².
Thin disk (about G, ⊥): ½ mR²;  ring: mR²;  sphere: 2/5 mR².
Composite body: locate G, then add IG,i + midi² for each part.
The parallel-axis theorem works only from an axis through G.
How Does the Body Move?
Classify the motion (Chapter 17), then specialize the equations
Translation — §18.2
α = 0
ΣF = maG,  ΣMG = 0
Details
All forces must balance their moments about G: this sets the wheel loads of an accelerating cart.

Slide or tip?
Assume it slides (F = μkN), solve for where N acts, check it lies inside the base.
General Plane Motion — §18.2
Translation and rotation together
Rotation About a Fixed Axis — §18.2
ΣFn = m rGω²
ΣFt = m rGα
ΣMO = IOα,  IO = IG + m rG²
Details
Use n (toward O) and t at G. The pin reactions and m rGω² pass through O, so ΣMO gives α directly.
α depends on position? Integrate ω dω = α dθ to get ω.
ΣMO = IOα only when O is the fixed axis.
Rolling Without Slipping
aG = rα
Ff ≤ μsN  (static, unknown)
ΣMIC = IICα
Check Ff ≤ μsN at the end.
Details
ΣMIC = IICα holds for a round wheel with G at its center, rolling without slipping — not in general.
Slipping
Ff = μkN  (kinetic, known)
aG ≠ rα
Details
aG and α are independent: use ΣFx for aG and ΣMG = IGα for α.
Reached when the no-slip check fails (required friction > μsN).
Other Bodies and Linked Points
aA = aG + α × rA/G − ω²rA/G
Details
Three equations of motion plus Chapter 17 kinematics when more unknowns remain.
Released from rest: ω = 0 at that instant, so the −ω²r term vanishes.