Mechatronics (ME-503 (A)) - Unit 4 Short Notes

UNIT 4: DYNAMICS OF MACHINERY - GOVERNORS, FLYWHEELS, BALANCING & FRICTION DRIVES

1.0 GOVERNORS (HIGH PRIORITY)

1.1 Function & Classification

• Function: Automatically controls the speed of an engine/machine by regulating the fuel/steam supply in response to load changes.

• Classification:

  • Centrifugal Governors: Use centrifugal force of rotating masses (balls).

    • Examples: Watt, Porter, Proell, Hartnell.

  • Inertia Governors: Use inertia forces of a rotating mass (e.g., swinging pendulum).

1.2 Watt Governor

• Construction: Two balls attached to arms, pivoted on a rotating spindle. Arms connected to a sleeve via links.

• Working: As speed ↑, balls fly outwards due to centrifugal force → sleeve rises → reduces fuel supply.

• Derivation (Height h vs. Speed N):

  Let:

  m = mass of each ball, r = radius of rotation of ball, h = height of governor (vertical distance from pivot to ball center), ω = angular velocity, T = tension in arm.

  At equilibrium (ball considered as particle):

  Centrifugal force = mω²r

  Resolving vertically: T cosθ = mg

  Resolving horizontally: T sinθ = mω²r

  ⇒ tanθ = ω²r / g

  From geometry: tanθ = r / h

  ∴ r/h = ω²r / g ⇒ h = g / ω²

  Since ω = 2πN/60:

$$ \boxed{h = \frac{g}{\left( \frac{2\pi N}{60} \right)^2} = \frac{3660}{N^2} \text{ (if } h \text{ in mm, } N \text{ in rpm)} } $$

**Proof:** `h ∝ 1/N²` ✓

1.3 Porter Governor

• Key Difference: An additional central load W is placed on the sleeve. Lower arms are connected to the sleeve.

• Derivation (Equilibrium Speed):

  Consider forces on one ball and its lower arm.

  Let m = ball mass, M = central load (sleeve + additional load), L = length of upper arm, l = length of lower arm, r = radius of rotation, θ = inclination of upper arm.

  From geometry: r = (L sinθ) + (l sinφ) but typically l sinφ is small.

  Vertical force balance on sleeve:

  W + 2T cosθ = Mg ...(1)

  For ball: T sinθ = mω²r ...(2)

  Solving (1) & (2) and using r ≈ L sinθ:

$$ \boxed{ N^2 = \frac{(M+2m)g}{4\pi^2 L \cos\theta} \cdot \frac{1}{\cos\theta} \approx \frac{(M+2m)g}{4\pi^2 L} \sec^3\theta } $$

(More precise: `r = L sinθ + l sinφ`, `φ` from lower arm geometry).

1.4 Proell Governor

• Key Difference: The pivot of the lower arms is offset from the axis of rotation by a distance d. Lower arms are extended to connect to balls.

• Derivation (Min/Max Speed):

  At minimum radius r_min, lower arm extensions are parallel to the axis.

  Let L = length of upper arm, l = length of lower arm extension, d = offset of lower arm pivot.

  Geometry: r_min = L sinθ_min + l (since extension is horizontal)

  Force equations similar to Porter, but r expression differs.

  Final expression:

$$ \boxed{ N^2 = \frac{(M+2m)g}{4\pi^2} \cdot \frac{(L + l \cos\theta)}{L r \cos\theta} } $$

**Result:** For same `θ` range, Proell has **higher sensitiveness** than Porter.

1.5 Governor Terminology & Characteristics [HIGH]

[!TIP] Exam Focus: Derive expressions for h (Watt), N (Porter/Proell). Be prepared to compare sensitiveness of Porter vs. Proell mathematically. Hunting and stability are very frequent 2-mark questions.

1.6 Comparative Analysis: Porter vs. Proell

Proof: Sensitiveness(Proell) > Sensitiveness(Porter)

For same θ₁, θ₂, L, l, d:

In Proell, r is larger for same θ due to offset d → denominator in N² expression larger → ΔN smaller? Actually, sensitiveness (N₁-N₂)/N depends on d(CF)/dr. Proell's CF curve is flatter near equilibrium → less ΔN for same Δr? Wait, higher sensitiveness means larger (N₁-N₂) for same N. Proell has larger range of r for same θ range, so for same Δθ, Δr is larger, leading to larger ΔN. Hence Proell is more sensitive.

2.0 FLYWHEELS & ENERGY FLUCTUATION (HIGH PRIORITY)

2.1 Function & Difference from Governor

2.2 Turning Moment Diagram

• For 4-stroke IC engine: One working stroke (power) every 2 revolutions (720°).

• Diagram: Plot of crank torque vs. crank angle.

• Interpretation:

  • Area above mean torque line = Energy supplied during power stroke.

  • Area below mean torque line = Energy consumed during other strokes (compression, suction, exhaust).

  • Net area over one cycle (720°) = 0 (mean torque line is such that work done = work against resistance).

  • Fluctuation: Difference between max and min energy levels from a reference.

2.3 Fluctuation of Energy & Speed

• Coefficient of Fluctuation of Energy (β_E):

$$ \boxed{ \beta_E = \frac{\Delta E}{E_{\text{mean}}} } $$

`ΔE` = max energy fluctuation (from diagram), `E_mean` = mean energy (area under mean torque line per cycle).

• Coefficient of Fluctuation of Speed (β_N):

$$ \boxed{ \beta_N = \frac{N_{\text{max}} - N_{\text{min}}}{N_{\text{mean}}} } $$

• Relationship:

  From energy conservation: ΔE = I ω Δω (approx for small Δω).

  Since ω = 2πN/60:

$$ \Delta E = I \cdot \frac{2\pi}{60} \cdot \Delta N \quad \Rightarrow \quad \Delta N = \frac{60 \Delta E}{2\pi I \omega} = \frac{60 \Delta E}{2\pi I (2\pi N/60)} = \frac{900 \Delta E}{\pi^2 I N} $$

Also, `β_N = ΔN / N`:

$$ \boxed{ \beta_N = \frac{900 \Delta E}{\pi^2 I N^2} } \quad \text{or} \quad \boxed{ \Delta E = \frac{\pi^2}{900} \beta_N I N^2 } $$

2.4 Flywheel Design & Analysis

• Mass Moment of Inertia (I):

$$ I = m k^2 $$

`m` = mass of flywheel, `k` = radius of gyration.

• Design Formula:

$$ \boxed{ I = \frac{\Delta E}{\omega^2 \beta_N} = \frac{\Delta E}{(2\pi N/60)^2 \beta_N} = \frac{3660 \Delta E}{N^2 \beta_N} } $$

(If `ΔE` in N-m, `N` in rpm, `I` in N-m-s²).

• Numerical Approach:

  1. Find ΔE from turning moment diagram (scale given: area in mm² → convert to N-m).

  2. Given N, β_N (or N_max, N_min), find I or k.

  3. If m given, find k = √(I/m).

[!TIP] Common Pitfall: Ensure consistent units. Convert diagram scale correctly: Area (mm²) × (Scale_Torque) × (Scale_Angle in rad)? Actually: ΔE (N-m) = Area (mm²) × (Torque_Scale N-m/mm) × (Angle_Scale rad/mm)? No! If vertical scale: 1 mm = S_T N-m, horizontal: 1 mm = S_θ rad, then Area (mm²) = (S_T × S_θ) × (Actual Area in N-m-rad). But ΔE is area between curves in Torque-Angle diagram, so:

`ΔE (N-m) = [Area (mm²)] × [Torque_Scale (N-m/mm)] × [Angle_Scale (rad/mm)]`.

Often `Angle_Scale` given in `°/mm` → convert to rad: `rad/mm = (π/180) × (°/mm)`.

3.0 BALANCING OF RECIPROCATING & ROTATING MASSES (VERY HIGH PRIORITY)

3.1 Fundamental Concepts

• Primary vs. Secondary:

  • Primary: Due to unbalanced mass at crank radius r. Frequency = crank frequency (ω). Force = m ω² r cosθ.

  • Secondary: Due to obliquity of connecting rod (λ = r/L). Frequency = 2ω. Force = m ω² r (r/L) cos2θ.

• Dynamically Equivalent System: A system of two masses (one at crank radius, one at λr) that produces same primary & secondary forces as the original reciprocating mass.

3.2 Balancing of Single-Cylinder Engines

• Rotating mass: Fully balanced by counterweight opposite to crank.

• Reciprocating mass: Only primary can be balanced fully (by same counterweight). Secondary remains unbalanced.

  • Usually balance fraction (1 - λ) or (1 - λ²)? Actually, to balance primary only, add mass m_r at radius r opposite to crank. This balances primary reciprocating force but introduces secondary force from this balancing mass? No, the balancing mass is rotating, so it only produces primary force. The unbalanced reciprocating mass m_r (1 - cosθ?) Wait:

  Let m_R = reciprocating mass.

  Primary force: F_p = m_R ω² r cosθ.

  To balance, add rotating mass m_b at radius r such that m_b ω² r = m_R ω² r ⇒ m_b = m_R. But this balances only primary. The secondary force from m_R remains: F_s = m_R ω² r λ cos2θ.

  Often, balance only a fraction (e.g., 2/3) to reduce hammer blow in locomotives.

• Residual Unbalanced Force:

  If balance fraction β (e.g., β=1 for full primary balance, β=0 for no balance):

$$ F_{\text{res}} = m_R ω² r [ (1-β) \cosθ + λ \cos2θ ] $$

3.3 Balancing of Multi-Cylinder In-Line Engines

• Crank Arrangement: 180° (2-cyl), 90° (4-cyl), 120° (3-cyl radial), etc.

• Primary Unbalanced Force:

  For n cylinders, spacing l between adjacent cylinders.

$$ F_{p,x} = \sum_{i=1}^n m_r ω² r \cos(θ + α_i) \quad \text{(along crank direction)} $$

$$ F_{p,y} = \sum_{i=1}^n m_r ω² r \sin(θ + α_i) \quad \text{(perpendicular)} $$

`α_i` = phase angle of i-th crank.

*   **Condition for complete primary balance:** `∑ cosα_i = 0` and `∑ sinα_i = 0`.

• Secondary Unbalanced Force:

$$ F_{s,x} = \sum m_r ω² r λ \cos(2θ + 2α_i) \quad F_{s,y} = \sum m_r ω² r λ \sin(2θ + 2α_i) $$

• Swaying Couple:

  • Primary Swaying Couple (T_p): Couple about vertical axis due to unbalanced primary forces.

$$ T_p = m_r ω² r \cdot \text{Algebraic sum of } (x_i \cosα_i - y_i \sinα_i) \cdot \sinθ \quad ? $$

    Actually for in-line engines (cylinders along x-axis, `y_i=0`):

$$ T_p = m_r ω² r \left[ \sum_{i=1}^n x_i \sinα_i \right] \sinθ $$

    where `x_i` = distance of i-th cylinder from reference.

*   **Secondary Swaying Couple (`T_s`):**

$$ T_s = m_r ω² r λ \left[ \sum_{i=1}^n x_i \sin2α_i \right] \sin2θ $$

• Hammer Blow: Vertical unbalanced force at wheel-rail contact (for locomotives). Due to unbalanced mass m_u rotating at radius r:

$$ \text{Hammer Blow} = m_u ω² r \quad \text{(at speed `ω`)} $$

Must be < allowable limit to prevent rail damage.

• Variation in Tractive Effort: Horizontal unbalanced force at wheel-rail contact, affects adhesion.

3.4 Balancing of Locomotives (Two-Cylinder, Uncoupled)

• Crank Arrangement: Usually 90°.

• Derivations (for 90° crank, cylinder spacing l, crank radius r):

  • Primary Swaying Couple:

    α₁=0°, α₂=90°, x₁ = -l/2, x₂ = +l/2

    ∑ x_i sinα_i = (-l/2) sin0 + (l/2) sin90 = l/2

$$ \boxed{ T_p = \frac{1}{2} m_r ω² r l \sinθ } \quad \text{(Max at θ=90°)} $$

*   **Hammer Blow:** Unbalanced vertical force from **both cylinders**.

    Vertical component of primary force: `F_{p,y} = m_r ω² r (sinθ + cosθ)`

$$ \boxed{ \text{Hammer Blow} = m_r ω² r \sqrt{2} \sin(θ+45°) } \quad \text{(Max = } \sqrt{2} m_r ω² r \text{)} $$

*   **Variation in Tractive Effort:** Horizontal unbalanced force:

    `F_{p,x} = m_r ω² r (cosθ - sinθ)`

$$ \boxed{ \text{Tractive Effort Variation} = m_r ω² r \sqrt{2} \cos(θ+45°) } \quad \text{(Max = } \sqrt{2} m_r ω² r \text{)} $$

[!TIP] Exam Focus: Derive expressions for swaying couple & hammer blow for 90° crank. Numerical: Given m_r, r, l, ω, find max values. Also: Fraction of reciprocating mass to balance given hammer blow limit:

$$ \text{If balance fraction } \beta \text{ (mass balanced = } \beta m_r \text{), then unbalanced mass } m_u = (1-\beta)m_r. $$

Max hammer blow = √2 (1-β) m_r ω² r ≤ allowable.

4.0 FRICTION DRIVES: CLUTCHES & BRAKES (HIGH PRIORITY)

4.1 Friction Clutches

• Single Plate Clutch:

  • Construction: Two discs (one on driving shaft, one on driven) with friction lining. Pressure plate & springs apply axial force F.

  • Assumptions:

    1. Uniform Pressure (p constant): F = π (R² - r²) p.

    2. Uniform Wear (p r = constant): F = 2π p_m R r (R - r) / \ln(R/r), where p_m = mean pressure.

  • Torque (T) Derivation:

    • Uniform Pressure:

      dT = μ p · 2πr dr · r = 2π μ p r² dr

      T = ∫_r^R 2π μ p r² dr = (2/3) π μ p (R³ - r³)

      Using F = π p (R² - r²):

$$ \boxed{ T = \frac{2}{3} μ F \frac{R^3 - r^3}{R^2 - r^2} } \quad \text{or} \quad T = μ F R_m \text{ with } R_m = \frac{2}{3} \frac{R^3 - r^3}{R^2 - r^2} $$

    *   **Uniform Wear:**

        `p = c / r`, `F = ∫_r^R 2π c dr = 2π c (R - r)` ⇒ `c = F / [2π(R-r)]`

        `T = ∫_r^R 2π μ (c/r) r² dr = 2π μ c ∫_r^R r dr = π μ c (R² - r²)`

$$ \boxed{ T = \frac{1}{2} μ F \frac{R^2 - r^2}{R \ln(R/r)} } \quad \text{or} \quad T = μ F R_m \text{ with } R_m = \frac{R^2 - r^2}{2 R \ln(R/r)} $$

*   **Numerical:** Given `P`, `N`, `F_max`, `μ`, find `R`, `r`, `b` (face width = `R-r`). Often assume `R = 2r` or `R/r = 1.5` to simplify.

• Conical Clutch:

  • Construction: Friction surfaces on cone. Axial force F → normal force N = F / (2 sin(α/2)) (for single cone, symmetric).

  • Torque Derivation:

    Consider mean radius R_m (or integrate).

    dT = μ N · 2π r dr, but N varies with r? Actually, pressure may be uniform or wear uniform. Usually assume uniform pressure.

    For uniform pressure p:

    F = 2π p R_m (R - r) / sinα (approx, from area of frustum).

    T = (2/3) μ p π (R³ - r³) (same as plate).

    Combining:

$$ \boxed{ T = \frac{2}{3} μ F \frac{R^3 - r^3}{R^2 - r^2} \cdot \frac{1}{\sin\alpha} } \quad \text{(if uniform pressure)} $$

    Often use mean radius `R_m` and effective `μ / sinα`.

*   **Power lost in friction:** `P_loss = μ N ω R_m`? Actually, friction heat = `T ω`.

4.2 Friction Brakes

• Band Brake (Simple & Differential):

  • Construction: Flexible band around drum. One end fixed, other end pulled by lever.

  • Theory: T₁ / T₂ = e^(μθ), where θ = angle of embrace (radians), T₁ = tight side tension, T₂ = slack side.

  • Braking Torque (T_b): T_b = (T₁ - T₂) R_drum.

  • Simple Band Brake: T₂ fixed, T₁ from lever: T₁ = P × (l / b) (moment about fulcrum).

  • Differential Band Brake: Band passes over two pulleys on drum. Lever pulls at point between. Condition for self-holding: (T₁/T₂) > (b/a) where a,b = lever arms.

• Internal Expanding Shoe Brake (Double Shoe):

  • Construction: Two shoes inside drum, pivoted at one end, forced outward by spring or hydraulic cylinder.

  • Analysis:

    For each shoe, normal force N from actuation. Friction force f = μ N at mean radius R_m.

    Braking torque T = 2 × (μ N R_m).

    Pressure distribution: Assume uniform pressure p or uniform wear (p ∝ 1/cosφ).

    Given max allowable pressure p_max, find shoe width b:

    N = ∫ p b R dφ over contact angle θ.

    For uniform pressure: N = p b R θ.

    Then T = 2 μ p b R² θ.

    Also, spring force F_spring relates to N via geometry.

4.3 Friction Concepts

• Friction Circle:

  • Definition: Circle of radius r_f in journal bearing where resultant reaction acts, assuming Coulomb friction.

  • Significance: Used to find frictional torque on journal.

  • Derivation:

    Journal radius r, load W, friction coefficient μ, angle of friction φ (tanφ = μ).

    Normal pressure distributed → resultant R passes through center? No, due to friction, resultant is shifted.

    For uniform pressure, resultant R is at center. Friction force F_f = μ W acts tangentially. The line of action of R is shifted by e = (μ r)? Actually, the frictional torque T = μ W r.

    The friction circle has radius r_f such that T = W r_f. So:

$$ \boxed{ r_f = \frac{T}{W} = \mu r } $$

    But more precisely, `r_f = r sinφ`? Since `μ = tanφ`, `μ r = r tanφ`. For small `φ`, `sinφ ≈ tanφ ≈ φ`. So `r_f ≈ μ r`.

    **Exact:** For **uniform wear**, pressure `p ∝ 1/cosθ`, resultant shifts. Friction circle radius `r_f = r \sin\phi` where `φ` is **angle of friction** (`tanφ = μ`). This is the **maximum possible** shift.

[!TIP] Clutch/Brake Numerical Steps:

1. Identify assumption (uniform pressure/wear). Usually uniform wear for clutches (new lining), uniform pressure for brakes (rigid shoes).

2. Write F (axial force) in terms of p.

3. Write T in terms of p (integrate).

4. Eliminate p to get T in terms of F.

5. Use P = Tω to find unknowns.

5.0 BEARINGS (MEDIUM PRIORITY)

5.1 Conical Pivot Bearings

• Construction: Conical surface on shaft or housing.

• Assumptions: Uniform pressure p or uniform wear (p r = constant).

• Derivations:

  Let R₁ = inner radius, R₂ = outer radius, α = cone angle (half-angle? Usually total cone angle = 2α? Clarify: If cone angle is 120°, then half-angle α = 60°? Actually, cone angle is angle between sides. For calculations, use α as half-angle? In textbooks: α is angle between bearing surface and axis. So if cone angle is 120°, then α = 60°? No: cone angle = 2α? Check: In conical pivot, α is angle with vertical axis. If cone angle = 120°, then α = 60°? Actually, if cone is wide, α is small? Let's define: α = angle between bearing surface and horizontal? Better: α = included angle? Standard: α = half-angle of cone? In many texts: α = angle between the bearing surface and the axis of the shaft. So if cone angle (included) = 120°, then α = 60°. But then sinα and cosα appear.

  • Uniform Pressure:

    Area A = π (R₂² - R₁²) / sinα (frustum area).

    F = p A = p π (R₂² - R₁²) / sinα.

    Torque T = ∫ μ p · 2π r dr · (r / sinα)? Actually, friction force dF_f = μ p dA, lever arm = r / sinα? Wait: The radius of rotation for friction is r / sinα? No, the moment arm for friction about axis is r (tangential force at radius r). But the normal force acts perpendicular to surface. The friction force is μ dN, and its moment about axis is μ dN × r. So:

    dT = μ p · (2π r dr / sinα) · r = (2π μ p / sinα) r² dr

    T = (2π μ p / sinα) ∫_{R₁}^{R₂} r² dr = (2π μ p / (3 sinα)) (R₂³ - R₁³)

    Substitute p from F:

$$ \boxed{ T = \frac{2}{3} μ F \frac{R_2^3 - R_1^3}{R_2^2 - R_1^2} \cdot \frac{1}{\sin\alpha} } \quad \text{(Uniform Pressure)} $$

*   **Uniform Wear:** `p r = constant = c`.

    `F = ∫ p dA = ∫ (c/r) (2π r dr / sinα) = (2π c / sinα) (R₂ - R₁)`

    `T = ∫ μ p r dA = ∫ μ c (2π dr / sinα) = (2π μ c / sinα) (R₂ - R₁)`

    So `T = μ F`? Wait:

    `F = (2π c / sinα) (R₂ - R₁)`

    `T = (2π μ c / sinα) (R₂ - R₁) = μ F`

    That gives `T = μ F`? That can't be right because dimensionally `T` should have length. I missed the **radius** in torque! Torque element: `dT = (friction force) × (radius) = (μ dN) × r`. But `dN = p dA = (c/r) dA`. So `dT = μ (c/r) dA × r = μ c dA`. And `dA = 2π r dr / sinα`. So `dT = μ c (2π r dr / sinα)`. Then `T = (2π μ c / sinα) ∫ r dr = (π μ c / sinα) (R₂² - R₁²)`.

    Now `F = (2π c / sinα) (R₂ - R₁)`.

    So:

$$ \boxed{ T = \frac{1}{2} μ F \frac{R_2^2 - R_1^2}{R_2 - R_1} \cdot \frac{1}{\sin\alpha} = \frac{1}{2} μ F \frac{R_2 + R_1}{\sin\alpha} } \quad \text{(Uniform Wear)} $$

*   **Power Loss:** `P_loss = T ω`.

5.2 Collar Bearings (Collar Friction)

• Construction: Shaft with one or more collars (discs) bearing on a flat surface.

• Assumptions: Uniform pressure p or uniform wear (p r = constant).

• Derivations:

  Let R₁ = inner radius (shaft), R₂ = outer radius (collar), n = number of collars.

  • Uniform Pressure:

    Area A = π (R₂² - R₁²)

    F = n p A = n p π (R₂² - R₁²)

    Torque T = ∫ μ p · 2π r dr · r = n (2π μ p) ∫_{R₁}^{R₂} r² dr = n (2π μ p / 3) (R₂³ - R₁³)

$$ \boxed{ T = \frac{2}{3} μ n F \frac{R_2^3 - R_1^3}{R_2^2 - R_1^2} } $$

*   **Uniform Wear:**

    `p = c / r`

    `F = n ∫ (c/r) 2π r dr = 2π n c (R₂ - R₁)`

    `T = n ∫ μ c 2π r dr = 2π n μ c \frac{R_2^2 - R_1^2}{2} = π n μ c (R₂² - R₁²)`

$$ \boxed{ T = \frac{1}{2} μ n F \frac{R_2^2 - R_1^2}{R_2 - R_1} = \frac{1}{2} μ n F (R_2 + R_1) } $$

*   **Power Absorbed:** `P = T ω`.

*   **Number of Collars:** Given `F`, `p_max`, find `n` from `F = n p_max A`.

6.0 DYNAMOMETERS (MEDIUM PRIORITY)

6.1 Function & Classification

• Function: Measure torque and power output of engines/motors.

• Types:

  • Absorption Dynamometers: Absorb engine power (as heat). E.g., Prony brake, Rope brake, Hydraulic (water).

  • Transmission Dynamometers: Transmit power while measuring. E.g., Epicyclic train dynamometer, Torsion dynamometer.

  • Torsion Dynamometer: Measures angle of twist in a shaft (using strain gauges, optical methods). Torque T = (G J / L) θ, where θ = twist angle, G = mod. rigidity, J = polar moment, L = length between gauges.

6.2 Torsion Dynamometer

• Construction: Shaft with strain gauges mounted at 45° to axis (where shear stress max). Or optical (mirror on shaft, light beam deflection).

• Working: Torque causes shear strain → change in gauge resistance (strain gauge) → measured as voltage. Or twist angle θ measured optically.

• Power Calculation:

  T = (G J / L) θ (from torsion of circular shaft).

  P = T ω.

7.0 ADVANCED TOPICS (LOWER PRIORITY)

7.1 Four-Bar Mechanism Analysis (Kinematics)

• Instantaneous Center (IC) Method:

  • Velocity: v = ω × r from IC. For any point on a link, velocity perpendicular to line joining point to IC. Magnitude v = ω_link × distance.

  • Acceleration: Use relative acceleration method:

    a_B = a_A + α × r_{B/A} - ω² r_{B/A} (for rotating link).

    Or from IC: a_t = α r, a_n = ω² r.

• Numerical: Given ω_A, find ω_B, ω_C, α_B, α_C. Use loop closure equations or IC method.

7.2 Cams & Followers (Application)

• Radial Cam with Offset Follower:

  • Lift (L): Vertical displacement of follower.

  • Pressure Angle (φ): Angle between follower motion and normal to cam profile. Must be < 30°.

  • Derivation of Follower Acceleration:

    For simple harmonic motion: s = (h/2)(1 - cos(πθ/β))

    v = ds/dt = (hπ ω / (2β)) sin(πθ/β)

    a = dv/dt = (hπ² ω² / (2β²)) cos(πθ/β)

• Critical Speed (Lift-off):

  When spring force = inertia force of follower + contact force? Actually, lift-off occurs when normal force between cam and follower becomes zero.

  Condition: F_spring - m a = 0? For vertical follower moving up:

  F_spring = m(g + a) if accelerating upward? Actually, free body diagram:

  Forces on follower: Spring force F_s (down if spring compressed), weight mg (down), contact force N (up from cam). Equation: N + F_s - mg = m a (up positive).

  Lift-off: N = 0 ⇒ F_s - mg = m a ⇒ a = (F_s - mg)/m.

  Given F_s varies with lift? Usually spring force = k (s + s₀) where s₀ is initial compression.

  Find ω such that max a exceeds (F_s - mg)/m at some θ.

FINAL REMINDER: Past papers show Governors, Flywheels, Balancing (especially locomotives), and Friction Drives are the core. Master derivations and numericals from these. Always state assumptions clearly (uniform pressure/wear). Convert units meticulously.