Theory of Machines — Dynamics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Flywheel: Stores energy as kinetic energy. Energy stored E = ½I(ω₂² – ω₁²). For flywheel design, find maximum fluctuation of speed (ω_max – ω_min). Coefficient of fluctuation C_s = (ω_max – ω_min)/ω_mean.

Governor: Maintains mean speed. Watt governor (centrifugal, radial) vs Porter governor (adds weight on sleeve). Hartnell governor (spring-controlled). Sensitivity = dN/dM where N = speed, M = mass of sleeve.

Balancing: Rotating mass imbalance → equal masses at 180° apart. Balancing of reciprocating masses (single-cylinder engine): partial balance using balance weight (~0.5 × reciprocating mass).

Vibration: Undamped free: ω_n = √(k/m) = √(g/δ_static). Damped: c_c = 2√(km) = 2mω_n. Logarithmic decrement δ = ln(x₁/x₂) = 2πξ/√(1–ξ²).

Critical speed: N_c = 60 × √(k/m) / (2π) = (60/2π)ω_n for single mass system.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Flywheel Analysis

Kinetic Energy and Speed Fluctuation

Energy stored in flywheel: E = ½Iω² where I = mass moment of inertia about axis

For a rim-type flywheel (concentrated rim): I = m × r² where m = rim mass, r = mean radius

For a solid disc flywheel: I = ½ m × r²

Coefficient of fluctuation of speed: C_s = (ω_max – ω_min) / ω_mean

Where ω_mean = (ω_max + ω_min)/2

Mean speed: ω_mean = 2ω_mean (wrong!) → correct: ω_mean = (ω_max + ω_min)/2

⚠️ GATE trap: Confusion between mean speed and average speed. Mean speed = (max + min)/2. Average speed over a cycle = (area under ω-t curve)/period.

Flywheel Design Problem

Given: Coefficient of fluctuation C_s, mean speed ω_mean, maximum fluctuation ΔE

Energy equation: ΔE = ½I(ω_max² – ω_min²) = ½I(ω_max – ω_min)(ω_max + ω_min) = ½I × (C_s × ω_mean) × (2ω_mean) ΔE = I × C_s × ω_mean²

Therefore: I = ΔE / (C_s × ω_mean²)

Example: ΔE = 5 kJ, C_s = 0.02, ω_mean = 100 rad/s I = 5000 / (0.02 × 10000) = 5000 / 200 = 25 kg·m²

Turning Moment Diagram

The turning moment diagram shows torque variation over one cycle:

1. Mean torque line drawn through the diagram area
2. Maximum positive fluctuation: Area above mean line = maximum energy storage needed
3. Maximum negative fluctuation: Area below mean line = energy released

For a single-cylinder engine (4-stroke):

• High torque during power stroke (expansion)
• Near-zero or negative torque during other strokes
• Large flywheel needed to smooth out these variations

Governors

Types of Governors

Watt Governor (Centrifugal, 1780): Simple two-link mechanism. Balls rise as speed increases. Height h = g/ω² (decreases with increasing ω — counter-intuitive from “rise” perspective). At ground pivot, h = (g/ω²) is the perpendicular distance from pivot to ball line.

Porter Governor: Watt governor with additional mass added to sleeve. Same height formula but heavier sleeve means higher sensitivity.

Hartnell Governor: Spring-controlled governor. Spring force balances centrifugal force. Natural frequency: ω_n = √(k/m_sleeve) where k = spring stiffness, m = equivalent sleeve mass

Hartog Governor: Springs control each ball separately — more sensitive.

Governor Terms

Sensitiveness: S = (N₁ – N₂)/(N_mean) where N₁ and N₂ are speeds at upper and lower extreme positions.

For Watt governor at height h: S = (2h/L) × (Δh/ω) … complex.

More practically for Porter governor: N₁ = √(g/W × (m+M) × L/m) … (too many variations — understand the principle)

💡 GATE Pattern: Governor questions often ask which governor is most sensitive for the same speed change, or to calculate required spring stiffness for Hartnell governor.

Force Analysis — Porter Governor

At equilibrium angle θ: Centrifugal force on each ball = m × ω² × r = m × ω² × L × sin θ Component along arm = m × ω² × L × sin θ × cos θ

Nomenclature: L = length of link from pivot to center of gravity m = mass of each rotating ball M = mass of sleeve θ = inclination of arm from vertical r = L sin θ = horizontal radius

At speed N (RPM), equilibrium when: m ω² L sin θ cos θ = m g sin θ + M g (L sin θ)/P … (P = link ratio)

Simplified: Equilibrium when ω² × constant = g × constant This leads to: N ∝ √(1/h) where h = height from pivot to plane of rotation

Balancing of Rotating Masses

Single-Plane Balancing (In-Line)

For masses m₁ at radius r₁ and m₂ at radius r₂ rotating at same angular position (same phase): Static balance: m₁r₁ = m₂r₂ (vector, in same direction)

For N masses at angles θ₁, θ₂…: ∑mᵢrᵢ cos θᵢ = 0 (x-direction) ∑mᵢrᵢ sin θᵢ = 0 (y-direction)

Two equations → can balance with two known mass locations. Minimum two masses needed.

⚠️ GATE Trap: Students forget to use the vector sum (both x and y components) when balancing more than two masses.

Balancing Procedure

1. Choose scale for mass-radius product (m×r) vectors
2. Draw vectors for each rotating mass in order of angular position
3. Complete the polygon — last vector closes the polygon
4. The closing vector (reversing direction) gives the required counterbalance

Resultant unbalance = vector sum of all m×r vectors.

Example: Two Masses at 90°

Masses: m₁=5 kg at r₁=0.2m at θ=0°, m₂=3 kg at r₂=0.15m at θ=90°

m₁r₁ vector = 1.0 × 0° = (1.0, 0) m₂r₂ vector = 0.45 × 90° = (0, 0.45)

Resultant = (1.0, 0.45) Required balance mass m₃ at r₃: m₃r₃ = 1.0i + 0.45j |m₃r₃| = √(1² + 0.45²) = √(1 + 0.2025) = √1.2025 = 1.097 kg·m Angle = tan⁻¹(0.45/1.0) = 24.2° from m₁ direction

Balancing of Reciprocating Masses

Primary and Secondary Unbalance

Primary unbalance: 1st harmonic of reciprocating force = m × ω² × r × cos θ This is the dominant component (largest magnitude).

Secondary unbalance: 2nd harmonic = m × ω² × r × (r/(2l)) × cos 2θ Small for small r/l ratio (connecting rod).

Balancing a Single-Cylinder Engine

Fully balanced would require m_p = m (all reciprocating mass). Practically impossible.

Partial balance using balance weight on crank: Balance weight added to crank = (m × r) × (2/π) × ω² … roughly 0.5 × m_p reciprocating mass is balanced.

Standard practice: 50% of reciprocating mass is balanced by a counterweight on the crank (for inline engines).

Residual unbalanced force = 0.5 × m_p × ω² × cos θ (at primary frequency)

Multi-Cylinder Engine Balancing

Inline engine (i cylinders, equally spaced):

• Primary force: Complete balance if cylinder spacing = 720°/i × k. For i=4, 90° crank spacing → primary balanced ✓
• Secondary force: May still be unbalanced — depends on cylinder arrangement

V-engine (i cylinders each side):

• If V-angle = 90° and cylinders opposite cancel secondary partially
• Primary balance: depends on crank arrangement

⚠️ GATE exam note: Two-cylinder inline engines cannot be fully balanced. V-twin engines can be partially balanced depending on firing order and crank phase.

Vibration Analysis

Undamped Free Vibration

Equation of motion: mẍ + kx = 0 Natural frequency (ω_n): ω_n = √(k/m) rad/s

Period: T = 2π/ω_n = 2π√(m/k)

Static deflection method: If a mass m deflects δ under its own weight: δ = mg/k Therefore: ω_n = √(g/δ) — very useful for spring-supported systems.

Frequency in Hz: f_n = ω_n/2π = (1/2π)√(k/m)

Damped Free Vibration

Equation: mẍ + cẋ + kx = 0 Damping ratio: ξ = c/c_c where c_c = 2√(km) = 2mω_n

Damping	Condition	Behavior
Underdamped	ξ < 1	Oscillatory decay
Critically damped	ξ = 1	Fastest non-oscillatory return
Overdamped	ξ > 1	Slow non-oscillatory decay

Underdamped case (most common): ω_d = ω_n√(1 – ξ²) — damped natural frequency

Logarithmic decrement: δ = ln(x₁/x₂) = 2πξ/√(1–ξ²) If ξ is small (δ < 0.3): ξ ≈ δ/(2π)

From test data: ξ = δ/(√(δ² + 4π²))

⚠️ GATE trick: Using the wrong formula. Students often solve for ξ from δ incorrectly. Remember: δ = 2πξ/√(1–ξ²). So ξ²(δ² + 4π²) = 4π² ξ² → No: rearrange gives ξ = δ/√(δ² + 4π²).

Critical Speed of Rotors

Single Mass-Shaft System

Rayleigh: Assumes deflection shape is close to static deflection. Approximation.

Whirling speed: Speed at which lateral vibration amplitude becomes large (resonance).

Critical speed formula (simple): N_c = (60/2π) × √(k/m) = (60/2π) × ω_n

For shaft with disk at center: I_disk causes gyroscopic effects at high speed (affects critical speed in 3D).

Bending critical speed: N_c = (60/2π) × √(g/δ_static) — same as natural frequency formula.

💡 GATE Pattern: Critical speed questions often involve finding the speed at which resonance occurs for a shaft with a disk. Convert deflection to natural frequency, then to RPM.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Flywheel — Torque-Time Diagram Integration

Work-energy method: Area under T-θ diagram = change in kinetic energy = ½I(ω_max² – ω_min²)

For a complete cycle: Net work = 0 (system returns to same state). Mean torque line ensures this.

Graphical method:

1. Draw T-θ curve (mean line horizontal)
2. Identify maximum positive area (A_max) above mean
3. Maximum energy fluctuation = A_max (in appropriate units)
4. Use I = ΔE/(C_s × ω_mean²)

Composite flywheel design: I_total = Σ I_i for each segment. For compound flywheels, calculate contribution from each part.

Governor — Stability and Hunting

Hunting: Oscillation about set speed if too sensitive. Occurs when governor is too sensitive (S too high) relative to system response.

Isochronous governor: Zero speed droop (N₁ = N₂) — theoretically perfect but practically unstable. Requires spring.

Watt governor: Not isochronous — speed increases continuously as balls rise. Speed droop = (N₁-N₂)/N_mean.

Porter governor with gravity: Speed droop ≈ M/(M + m) … heavier sleeve reduces droop (less hunting).

Hartnell governor spring calculation: k = (m × ω² × r) / x where m = ball mass, ω = speed, r = arm radius, x = displacement

Required spring force at operating point = centrifugal force of balls at that speed.

Multi-Cylinder Balancing — In-Line Engines

Four-cylinder inline (90° crank spacing):

• Primary forces: Balanced (centers of mass coincide)
• Primary couples: Zero (symmetric)
• Secondary forces: Balanced if firing order distributes
• Secondary couples: May exist (need calculation)

Six-cylinder inline (120° crank spacing):

• Primary and secondary both fully balanced

V-8 engine balancing: Depends on configuration (cross-plane, flat-plane). Most V8s have some residual unbalance requiring counterweights.

Vibration — Forced Vibration with Damping

Equation: mẍ + cẋ + kx = F₀ sin ωt

Steady-state amplitude: X = F₀/√((k – mω²)² + (cω)²)

Phase angle: φ = tan⁻¹(cω/(k – mω²))

Magnification factor (MF): MF = X/(F₀/k) = 1/√((1 – r²)² + (2ξr)²) where r = ω/ω_n

At resonance (r=1): X_max = F₀/(2ξk) = (static deflection)/(2ξ) Peak amplitude inversely proportional to damping!

⚠️ GATE trap: Forgetting that at resonance, the phase lag is 90° and amplitude is theoretically infinite for zero damping (which never happens in reality).

Gyroscopic Effects

Gyroscopic couple: C = I × ω × Ω

• I = mass moment of inertia of rotor
• ω = spin speed of rotor
• Ω = precession rate (angular velocity of axis rotation)

Direction (right-hand rule): Point thumb along spin axis, fingers curl in direction of rotation. Palm shows precession direction.

Practical application: In aircraft, engine torque causes pitching moment. In ships, roll coupling creates yaw instability at certain speeds.

For a disc-type rotor: C = (m × r_g²) × ω × Ω where r_g = radius of gyration.

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Example Problem

GATE 2018: A shaft carries a disc of mass 50 kg at midspan. Deflection under static load = 0.02 m. Find critical speed.

Solution:

Step 1: Find static deflection δ = 0.02 m (given under disc weight)

Step 2: Natural frequency ω_n = √(g/δ) = √(9.81/0.02) = √(490.5) ≈ 22.1 rad/s

Step 3: Critical speed in RPM N_c = (60/2π) × ω_n = (60/2π) × 22.1 = 211.2 RPM

💡 Note: This is the first critical speed (bending mode). Higher modes exist but are less common in GATE.

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