Work, Energy, Impulse, Momentum, Rotation, and Vibration

Work-energy method

The work-energy theorem equates the net work done on a body to its change in kinetic energy:

$U_{1\to2} = \Delta KE = \tfrac{1}{2}mv_2^2 - \tfrac{1}{2}mv_1^2$

This is the fastest tool whenever force acts over a distance and you do not need time. Work of a constant force is F·d (along the displacement); spring work is ½k(x₁² − x₂²); gravity work is −WΔh (or +Wh on the way down). When only conservative forces (gravity, springs) act, mechanical energy is conserved: KE₁ + PE₁ = KE₂ + PE₂.

Worked example: a 2 kg block slides from rest down a frictionless 5 m incline at 30°. Drop height h = 5 sin30° = 2.5 m. Energy conservation: ½mv² = mgh → v = √(2gh) = √(2·9.81·2.5) = 7.0 m/s. Add kinetic friction μ_k = 0.20: friction force = μ_k mg cos30° = 0.20(2)(9.81)(0.866) = 3.40 N; friction work = −3.40(5) = −17.0 J. Then ½(2)v² = mgh − 17.0 = 49.05 − 17.0 = 32.0 J → v = 5.66 m/s.

Impulse-momentum and impact

The impulse-momentum principle integrates Newton's law over time:

$\int F\,dt = m v_2 - m v_1$

When no external impulse acts on a system, linear momentum is conserved — the basis of collision problems. For two colliding bodies, momentum conservation gives one equation; the coefficient of restitution supplies the second:

$e = \frac{v'_B - v'_A}{v_A - v_B}$

with e = 1 for a perfectly elastic impact (kinetic energy conserved) and e = 0 for a perfectly plastic impact (bodies move together). A 3 kg cart at 4 m/s strikes a stationary 1 kg cart and they couple (e = 0): common velocity = (3·4)/(3+1) = 3 m/s. The kinetic energy lost = ½(3)(16) − ½(4)(9) = 24 − 18 = 6 J, dissipated in the coupling.

Rigid-body rotation

For rotation about a fixed axis, the angular analog of Newton's law is ΣM = Iα, where I is the mass moment of inertia and α the angular acceleration. Rotational kinetic energy is ½Iω², and angular work-energy and angular impulse-momentum (∫M dt = I Δω) mirror their linear forms. For a body rolling without slipping, the contact-point constraint gives v = rω and a = rα, linking translation and rotation.

General plane motion combines both: total KE = ½mv_G² + ½I_G ω². A solid cylinder (I = ½mr²) rolling down an incline has acceleration a = g sinθ / (1 + I/mr²) = g sinθ/(1.5) = ⅔ g sinθ — less than a frictionless sliding block because energy also spins it up.

Single-degree-of-freedom vibration

Free undamped vibration of a mass-spring system obeys mẍ + kx = 0, giving simple harmonic motion at the natural frequency:

$\omega_n = \sqrt{\frac{k}{m}} \;(\text{rad/s}), \qquad f_n = \frac{\omega_n}{2\pi}, \qquad T = \frac{1}{f_n}$

A simple pendulum has ω_n = √(g/L); note mass cancels. Worked example: m = 4 kg on k = 1600 N/m gives ω_n = √(1600/4) = 20 rad/s, f_n = 20/2π = 3.18 Hz, period T = 0.314 s. Adding viscous damping introduces ζ = c/(2√(km)); ζ < 1 is underdamped (oscillates and decays), ζ = 1 critically damped (fastest no-overshoot return), ζ > 1 overdamped. The damped frequency ω_d = ω_n√(1 − ζ²) is slightly below ω_n.

Power, efficiency, and conservative-force traps

Power is the time rate of doing work, P = dU/dt = F·v for a force moving at velocity v, or P = Tω for a torque on a rotating shaft. A 1200 N tractive force at 25 m/s delivers P = 30 000 W = 30 kW. Mechanical efficiency η = P_out/P_in is always less than one; friction and other losses make up the difference, and the work-energy bookkeeping must include that lost work as a negative term. A common FE trap is to apply energy conservation when friction is present — friction is non-conservative, so use the general form U_external + U_friction = ΔKE, never the conservative KE + PE balance, unless the surface is explicitly frictionless.

Angular momentum and gyroscopic intuition

The angular-momentum principle, ∫M dt = ΔH where H = Iω, governs spinning bodies and is conserved when net external moment is zero. A skater pulling in her arms reduces I, so ω rises to keep Iω constant — a direct conservation-of-angular-momentum result that occasionally appears in FE qualitative items. For a rigid body, H about its mass center is I_G ω.

Mass moment of inertia reference values

Rotational problems require the correct I; the Handbook tabulates the standard ones (axis through the mass center unless noted):

Use the parallel-axis theorem I = I_G + m d² to shift to an off-center axis — a rod swinging about its end has (1/12)mL² + m(L/2)² = (1/3)mL², which is also the pendulum result. Picking the wrong axis (center vs end) is the most frequent rotation error, so confirm where the rotation axis actually is before selecting a formula.