Rigid-Body Rotation and Mechanical Vibrations

Mass Moment of Inertia

Where linear motion resists acceleration through mass m, rotation resists angular acceleration through the mass moment of inertia I, measured in kg·m² (SI) or slug·ft² (USCS). It is defined as I = ∫r² dm, so mass located far from the axis counts much more heavily — the r² weighting is why a flywheel concentrates material at its rim.

The FE Reference Handbook tabulates I for common bodies. Memorize the pattern, not every value:

Parallel-axis theorem (mass form): I = I_G + md², where I_G is taken about the axis through the mass center G and d is the distance to the new parallel axis. This is the exact analog of the area version used in Statics, and it is the single most common rotational-inertia trap on the FE — the tabulated value is almost always the centroidal one, so you must add md² to use any off-center pivot.

Rotational Kinetics, Energy, and Momentum

The rotational form of Newton's second law about the mass center is ΣM_G = I_G α, where α is angular acceleration (rad/s²). For a body pinned at a fixed point O, you may instead write ΣM_O = I_O α using the inertia about O directly.

The rotational analogs of the energy and momentum quantities are exact mirrors of the linear ones:

A body that both translates and rotates carries both energy terms: KE_total = ½mv_G² + ½I_G ω². For rolling without slipping, the contact point is the instantaneous center, v_G = rω links the two, and friction does no work because the contact point has zero velocity.

Worked Example — Rolling Cylinder Energy

A solid cylinder (m = 10 kg, r = 0.2 m) rolls without slipping at v_G = 3 m/s. Its total kinetic energy is ½mv² + ½Iω² with I = ½mr² and ω = v/r. So KE = ½(10)(3²) + ½(½·10·0.2²)(3/0.2)² = 45 + ½(0.2)(225) = 45 + 22.5 = 67.5 J. The rotational share (one-third here) is exactly the kind of detail FE distractors omit.

Free Vibration and Natural Frequency

A mass m on a spring of stiffness k, displaced and released, obeys mẍ + kx = 0. Its motion is sinusoidal at the undamped natural frequency:

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

Stiffer springs vibrate faster; heavier masses vibrate slower. Trap: ωₙ uses √(k/m) directly, but the period T uses √(m/k) — invert carefully. For a small-angle simple pendulum of length L, ωₙ = √(g/L), independent of mass.

Worked Example — Natural Frequency

A 4 kg mass sits on a spring with k = 1600 N/m. Then ωₙ = √(1600/4) = √400 = 20 rad/s, and fₙ = 20/(2π) ≈ 3.18 Hz. Quadrupling k would only double ωₙ, because frequency scales with the square root of stiffness.

Damping and Resonance

Real systems lose energy. Adding a damper c gives mẍ + cẋ + kx = 0, and behavior is set by the damping ratio ζ = c/(2√(km)) = c/c_c, where c_c = 2√(km) is the critical damping coefficient.

• ζ < 1 (underdamped): decaying oscillation; the damped frequency is ω_d = ωₙ√(1−ζ²).
• ζ = 1 (critically damped): returns to equilibrium fastest with no overshoot — the target for door closers and instruments.
• ζ > 1 (overdamped): slow non-oscillating return.

When a system is driven by a periodic force whose frequency approaches ωₙ, the response amplitude peaks sharply — resonance. Engineers either add damping or shift ωₙ (by changing m or k) so operating speeds avoid it. On the FE, expect to compute ωₙ or fₙ from k and m, classify damping from ζ, or identify resonance as the cause of a runaway amplitude.

Angular Kinematics and the Rotation–Vibration Link

Before applying ΣM = Iα you often need angular kinematics, the rotational mirror of the particle-kinematics relations. For constant angular acceleration α: ω = ω₀ + αt, θ = ω₀t + ½αt², and ω² = ω₀² + 2αθ, with ω in rad/s and θ in radians. A point at radius r on a spinning body has tangential speed v = rω, tangential acceleration aₜ = rα, and normal (centripetal) acceleration aₙ = rω² = v²/r directed toward the axis. Trap: even at constant ω (α = 0) a rotating point still has nonzero centripetal acceleration, so a net inward force is required — forgetting aₙ is a frequent error.

Worked Example — Spin-Up

A wheel accelerates from rest at α = 4 rad/s² for 3 s. Then ω = 0 + 4(3) = 12 rad/s and θ = ½(4)(3²) = 18 rad. A point 0.5 m out now moves at v = 0.5(12) = 6 m/s with aₙ = 0.5(12²) = 72 m/s².

Forced Vibration and Transmissibility

When a harmonic force F₀sin(Ωt) drives the spring–mass–damper, the steady-state amplitude depends on the frequency ratio r = Ω/ωₙ. The dynamic amplification factor (for light damping) is roughly 1/|1 − r²|: far below resonance (r ≪ 1) the response tracks the static deflection F₀/k; near r = 1 it spikes; far above (r ≫ 1) it falls off.

Vibration isolation exploits this — mounting equipment on soft springs so ωₙ sits well below the disturbing frequency (r > √2) transmits less force to the foundation. This is why isolating a machine means lowering its natural frequency, not raising it. On the FE, recognize that operating at r ≈ 1 is dangerous and that isolation requires Ω > √2·ωₙ.