Dynamics: Kinematics, Newton's Laws, and Relative Motion

Kinematics: describing motion

Kinematics relates position s, velocity v, and acceleration a without asking what causes the motion. The general differential relations from the Handbook are v = ds/dt, a = dv/dt, and the chain-rule form a ds = v dv, which is invaluable when acceleration depends on position rather than time.

When acceleration is constant, integrate once to get the three standard equations:

$v = v_0 + at$ $s = s_0 + v_0 t + \tfrac{1}{2}at^2$ $v^2 = v_0^2 + 2a(s - s_0)$

The third equation is time-free and is the fastest route when time is neither given nor asked. Worked example: a car brakes from 30 m/s to rest over 90 m. Using v² = v₀² + 2as: 0 = 30² + 2a(90), a = −5 m/s². Time to stop: 0 = 30 + (−5)t → t = 6 s.

Projectile motion

A projectile under gravity alone has a_x = 0 and a_y = −g, so horizontal velocity is constant and vertical motion is constant-acceleration. The two directions decouple and share only the elapsed time. A ball thrown at v₀ = 20 m/s at 30° has v₀x = 20cos30° = 17.3 m/s and v₀y = 20sin30° = 10 m/s. Time to apex: v_y = 0 = 10 − 9.81t → t = 1.02 s; total flight 2.04 s; range = 17.3(2.04) = 35.3 m. Maximum height = v₀y²/(2g) = 10²/19.62 = 5.1 m.

Curvilinear motion: normal-tangential coordinates

For motion along a curved path, normal-tangential (n-t) coordinates split acceleration into two physically distinct parts:

$a_t = \frac{dv}{dt} \quad (\text{speeds up/slows down}), \qquad a_n = \frac{v^2}{\rho} \quad (\text{turns, toward center})$

where ρ is the radius of curvature. Even at constant speed (a_t = 0) a turning body still accelerates because a_n ≠ 0 — this is why a car at steady speed on a curve needs friction. The total magnitude is a = √(a_t² + a_n²). A car at 25 m/s on a 200 m radius curve has a_n = 25²/200 = 3.13 m/s² directed toward the curve center.

Newton's second law and the kinetic diagram

Kinetics links force to motion through Newton's second law, ΣF = ma. The reliable FE method is to draw a free-body diagram (all real forces) beside a kinetic diagram (the ma vector), then equate components. In SI, m is in kg and the equation is direct. In USCS, weight and mass differ: ΣF = (W/g)a with g = 32.2 ft/s², so a 64.4 lb weight has mass W/g = 2 slugs.

Classic FE problem: a 1500 kg car rounds a flat 120 m radius curve at 20 m/s. Required centripetal force = m v²/ρ = 1500(400)/120 = 5000 N, supplied by friction f = μ_s N = μ_s mg. Minimum μ_s = (v²/ρ)/g = (400/120)/9.81 = 0.34.

Relative motion

When two bodies move, their velocities relate by vector addition:

$\vec{v}_B = \vec{v}_A + \vec{v}_{B/A}$

and the identical rule holds for accelerations. v_{B/A} is the velocity of B as seen from A. Treat each term as a vector: resolve into components, add, and recombine. Two cars, A heading east at 60 km/h and B north at 80 km/h, give v_{B/A} = v_B − v_A with magnitude √(60² + 80²) = 100 km/h, directed northwest relative to A. Distractor answers usually add the speeds arithmetically (140) — always go through components.

Position-dependent acceleration and dependent motion

When acceleration varies, do not reach for the constant-a equations. If a = f(t), integrate a dt to get velocity and again for position. If a = f(s) (acceleration depends on position, as in a spring or magnetic field), use the chain-rule form a ds = v dv, then integrate: ∫a ds = ∫v dv = ½v² + C. This separates the variables cleanly and is the only correct path when, for example, a = −kx (which integrates to the simple-harmonic v–x relation). Recognizing which integration applies is itself a tested skill.

Dependent (constrained) motion appears in pulley problems: the total length of an inextensible cable is constant, so the displacements of the connected blocks are linked. Differentiating the length constraint relates their velocities and accelerations. In a single fixed-and-movable pulley system where one block descends, the supported block may move at half the speed and half the acceleration — write the cable-length equation, differentiate twice, and the ratios fall out automatically rather than from guessing.

Rotating frames and the polar description

For a particle on a rotating arm, polar (r-θ) coordinates capture both radial and transverse effects. The Handbook gives a_r = r̈ − rθ̇² (the second term is the centripetal contribution) and a_θ = rθ̈ + 2ṙθ̇ (the 2ṙθ̇ term is the Coriolis acceleration that appears whenever a particle moves radially in a rotating frame). A bead sliding outward on a spinning rod feels a Coriolis component even though it moves in a straight line along the rod. Recognizing that a uniformly rotating arm (θ̇ constant, r constant) reduces to pure centripetal acceleration rθ̇² toward the center keeps these problems quick.

Always state your coordinate frame first — mixing rectangular and polar terms is a leading cause of wrong dynamics answers.