Kinematics of particles and rigid bodies, mass moments of inertia, force-acceleration, and work, energy, and power.
2 concepts
A. Kinematics
Kinematics, Newton's Second Law, and Work-Energy
Describe motion with the constant-acceleration and projectile equations, relate force to acceleration through F=ma, then short-circuit the algebra with work-energy and power.
Dynamics splits cleanly into two questions: how a body moves (kinematics) and why it moves that way (kinetics). On the FE Civil exam this topic is worth 4-6 questions, and the fastest way to lose those points is to reach for a force balance when a single energy equation would have finished the problem in one line. The FE Reference Handbook hands you every equation in its Dynamics chapter, so the exam is not testing recall — it is testing whether you can pick the right tool and keep your signs, units, and reference frames straight. This concept builds that judgment: rectilinear and curvilinear kinematics, projectile and relative motion, ΣF=ma, and the work-energy-power chain that ties forces to speeds without ever solving for time.
Position, velocity, acceleration
Kinematics starts with a position vector r(t) and two derivatives: velocity v=dr/dt and acceleration a=dv/dt. In one dimension these are just v=s˙ and a=s¨. The single most useful relation drops the time variable entirely — multiply a=dv/dt by ds/ds to get ads=vdv, which lets you connect speed and position directly. Whenever a problem gives you a distance and asks for a speed (or vice versa) without mentioning time, this is the equation to reach for.
v=dtdr,a=dtdv,ads=vdv
Constant-acceleration equations
When acceleration is constant — gravity, steady braking, a uniform thrust — the calculus collapses into three algebraic workhorses. Pick the one that avoids the variable you do not have: use v=v0+at when you know time, and the time-free form v2=v02+2a(s−s0) when you do not. A free-falling body is simply this set with a=g pointed down. Choose a positive direction first and hold it for the whole problem; a sign flip mid-solution is the most common kinematics error.
v=v0+at,s=s0+v0t+21at2,v2=v02+2a(s−s0)
Projectile motion
Projectile motion is two independent constant-acceleration problems sharing one clock. Horizontally there is no acceleration, so x=(v0cosθ)t; vertically gravity acts, so vy=v0sinθ−gt and y=(v0sinθ)t−21gt2. The link between the two axes is time. Apex occurs when vy=0, and for launch and landing at the same elevation the total flight time is tf=2v0sinθ/g. Never apply a -style range formula blindly when the launch and landing heights differ — go back to the component equations.
x=(v0cosθ)t,y=(v0sinθ)t−21gt2
Curvilinear paths and relative motion
On a curved path it is cleaner to resolve acceleration into normal and tangential components: the tangential part at=v˙ changes speed, while the normal part an=v2/ρ (always pointing toward the center of curvature) changes direction. For circular motion of radius r these become v=rω, at=rα, and an=rω2. When two bodies move at once, work in a translating frame: vA=vB+vA/B. Read 'the velocity of A relative to B' as exactly that subscript order, and add the vectors head-to-tail rather than guessing signs.
at=dtdv,an=ρv2=rω2,vA=vB+vA/B
Newton's second law
Kinetics enters through ΣF=ma — the resultant of all applied forces equals mass times acceleration, written as one scalar equation per direction. Draw a free-body diagram first, every time: on an incline, resolve weight into components along and perpendicular to the surface so that ΣF=mgsinθ−μkN along the slope and N=mgcosθ across it. In USCS remember that weight in pounds-force is not mass; convert with m=W/g so the units close (lbf=slug⋅ft/s2).
ΣFx=max,ΣFy=may
Work, energy, and the work-energy theorem
The work-energy theorem is the great shortcut: the net work done on a particle equals its change in kinetic energy, U1→2=ΔT=21mv22−21mv12. Because work depends on distance, not time, you skip the time variable entirely — ideal for 'speed after sliding d' problems. Bookkeep work by source: a constant force does U=Fscosθ, gravity does U=−WΔy (negative going up), and a spring does U=−21k(s22−s12). Friction always removes energy, so its work is negative.
U1→2=ΔT=21mv22−21mv12
Conservation of energy and power
When only conservative forces act, mechanical energy is conserved: T1+V1=T2+V2, with gravitational potential Vg=mgh and elastic potential Ve=21ks2. If nonconservative forces (friction, an applied push) are present, carry them as a work term: T1+V1+U1→2=T2+V2. Power is the time rate of doing work, P=dU/dt=F⋅v, and mechanical efficiency is ε=Pout/Pin. Watch the units: 1hp=550ft-lbf/s=746W.
T1+V1+U1→2=T2+V2,P=F⋅v
Exam strategy
Decide on the method before touching algebra. If the question asks for a speed given a distance (or a distance given a speed) and never mentions time, use work-energy — it is almost always two or three lines. If it asks for time, or for the force/acceleration itself, use ΣF=ma with a clean free-body diagram. Reserve the constant-acceleration kinematics set for when acceleration is genuinely constant and you need the explicit time history. For projectiles, split into x and y immediately and let time be the bridge. Lock in a positive direction at the start, label every g as 9.81m/s2 or 32.2ft/s2, and in USCS convert weight to mass before writing ma.
Key equations
Velocity and accelerationv=dtds,a=dtdv=vdsdv
Definitions for rectilinear motion; the last form (from ads=vdv) eliminates time. s in m or ft, v in m/s or ft/s, a in m/s² or ft/s².
Work of common forcesUF=Fscosθ,Ugrav=−WΔy,Uspring=−21k(s22−s12)
Potential energyVg=mgh,Ve=21ks2
Energy conservation with nonconservative workT1+V1+U1→2=T2+V2
Power and efficiencyP=dtdU=F⋅v,ε=PinPout
Worked examples
Projectile range and maximum height
Problem. A projectile leaves the ground at v0=25m/s at θ=40∘ above horizontal, landing back at ground level. Find the horizontal range and the maximum height. Use g=9.81m/s2.
Solution. Resolve the launch velocity: vx=25cos40∘=19.15m/s and vy0=25sin40∘=16.07m/s
R=vxtf=19.15×3.276=62.7m
Crate sliding down a rough incline (F = ma plus kinematics)
Problem. A 50 kg crate is released from rest on a 20∘ incline with kinetic friction coefficient μk=0.25. Find its acceleration and its speed after sliding 8.0 m down the slope.
Solution. Free-body diagram along the incline (positive down-slope): ΣF=mgsinθ−μkN
Speed at the bottom of a chute (work-energy with friction)
Problem. A 10 kg package starts from rest and slides 5.0 m down a straight chute, dropping 3.0 m in elevation, with μk=0.30. Find its speed at the bottom.
Solution. Geometry: the chute rises 3.0 m over 5.0 m, so sinθ=3/5=0.6
Common pitfalls
•Confusing weight and mass in USCS: a 'W = 100 lbf' block has mass m=W/g=100/32.2=3.11 slug. Plugging 100 directly into ΣF=ma overstates the force by a factor of g.
•Reaching for ΣF=ma when no force or time is asked. If the question relates a speed to a distance, work-energy is faster and avoids solving a differential equation.
•Sign errors on gravity's work: Ugrav=−WΔy is negative when the body rises and positive when it falls. Always pick a datum and a positive direction first.
•Using a horizontal-launch range formula when launch and landing elevations differ. Split into x and y components and let time be the link instead.
•Treating friction work as conservative or assigning it the wrong sign. Friction always opposes motion, so its work is negative and energy is lost — it cannot be stored as potential energy.
•Mixing degrees and radians in circular motion. v=rω, at=rα, and an=rω2
•Forgetting the normal acceleration on a curve: even at constant speed a body on a curved path accelerates, an=v2/ρ toward the center. Only at=0
References
NCEES FE Reference Handbook — Dynamics (Kinematics, Particle Kinetics, Work and Energy)
Hibbeler, Engineering Mechanics: Dynamics — particle kinematics and energy methods
Beer, Johnston, Vector Mechanics for Engineers: Dynamics
B. Mass moments of inertia
Mass Moments of Inertia and Rotation
Compute mass moment of inertia for standard shapes, shift axes with the parallel-axis theorem, and apply ΣM=Iα and rolling-without-slipping to rotating rigid bodies.
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Valid only when a is constant. Use the time-free third form when t is neither given nor asked.
Horizontal velocity constant (ax=0); vertical motion under ay=−g. Time links the two axes.
ρ = radius of curvature (m or ft); an points toward the center. For a circle ρ=r.
ω in rad/s, α in rad/s², r in m or ft. Angles must be in radians.
Translating-frame vector addition; vA/B is the velocity of A as seen from B.
. In USCS,
m=W/g
in slugs.
Net work of all forces equals change in kinetic energy. Distance-based, time-free.
θ = angle between force and displacement; gravity work negative when rising; spring work negative when stretched or compressed from its free length.
Gravitational (h above a datum) and elastic (s = deflection from free length, k in N/m or lbf/ft).
U1→2 is the work of nonconservative forces (negative for friction). Reduces to T1+V1=T2+V2 when only gravity/springs act.
P in watts (J/s) or ft-lbf/s; 1hp=746W=550ft-lbf/s.
.
Time of flight (same launch and landing elevation):