BITS F441 Robotics — Comprehensive Solutions

Comprehensive Examination

BITS F441 — Robotics

26 December 2022 Closed Book 80 Marks 3 Hours

Question 1a

Lagrangian Mechanics — Single Link

10 Marks (2+4+1+1+2)

Given

Link length l = 0.35 m, Mass m = 0.25 kg, Revolute joint at one end, link rotating in a vertical plane.

Concept

Lagrange: The Lagrangian L = T − V, where T = kinetic energy, V = potential energy.
The Lagrangian Equation for generalized coordinate q_i:

d/dt(\partialL/\partialq̇ᵢ) - \partialL/\partialqᵢ = τᵢ

For a uniform rod pivoted at one end: Center of mass is at l/2 from pivot. Moment of inertia about pivot = ml²/3.

Formula Used

T = ½ I θ̇², I = ml²/3 (uniform rod about pivot)
V = mg(l/2)sinθ (height of CoM = (l/2)sinθ)
L = T − V
τ = d/dt(∂L/∂θ̇) − ∂L/∂θ

Step-by-Step Solution

Moment of Inertia:

I = ml²/3 = (0.25)(0.35)²/3 = (0.25)(0.1225)/3 = 0.010208 kg\cdotm²

Kinetic Energy:

T = ½Iθ̇² = ½(0.010208)θ̇² = 0.005104 θ̇² Joules

Potential Energy (taking pivot as datum, +ve upward):

V = mg(l/2)sinθ = (0.25)(9.81)(0.175)sinθ = 0.4291 sinθ Joules

Lagrangian:

L = T - V = 0.005104 θ̇² - 0.4291 sinθ

Apply Lagrangian Equation:

\partialL/\partialθ̇ = 0.010208 θ̇ d/dt(\partialL/\partialθ̇) = 0.010208 θ̈ \partialL/\partialθ = -0.4291 cosθ τ = 0.010208 θ̈ - (-0.4291 cosθ) τ = 0.010208 θ̈ + 0.4291 cosθ [N\cdotm]

Final Answer

Kinetic Energy

0.005104 θ̇² J

Potential Energy

0.4291 sinθ J

Lagrangian L

0.005104θ̇² − 0.4291sinθ

Torque τ

0.01021θ̈ + 0.4291cosθ N·m

Question 1b

Principle of Virtual Work — Reaction at C

5 Marks

Given

Beam AD with pin support at A, roller support at C.
Geometry: A — 2m — B — 4m — C — 2m — D
100 N load at B (2m from A), 200 N load at D (8m from A)

Concept

Principle of Virtual Work: For a system in equilibrium, the total virtual work done by all forces during any virtual displacement is zero.
Remove support C, impose virtual upward displacement δ at C. Beam rotates rigidly about A.

Step-by-Step Solution

Identify Virtual Displacement: Give C a virtual displacement δ upward. Beam rotates about A (pin).

At A (x=0): δ_A = 0 (fixed) At B (x=2): δ_B = (2/6)δ = δ/3 ↑ At C (x=6): δ_C = δ ↑ At D (x=8): δ_D = (8/6)δ = 4δ/3 ↑

Apply Virtual Work Principle (ΣW = 0):
Forces doing work: Rc (upward at C), 100N (downward at B), 200N (downward at D).

Rc\cdotδ_C - 100\cdotδ_B - 200\cdotδ_D = 0 Rc\cdotδ - 100\cdot(δ/3) - 200\cdot(4δ/3) = 0 Divide by δ: Rc - 100/3 - 800/3 = 0 Rc = 900/3 = 300 N

Final Answer

Reaction at C

Rc = 300 N (upward)

Question 2

3-DOF Manipulator — Transformation Matrices & Jacobian

20 Marks

Given

[0T1] = [C1 0 S1 0.2C1] [1T2] = [-S2 -C2 0 0] [S1 0 -C1 0.2S1] [C2 -S2 0 0] [0 1 0 0.4 ] [0 0 1 0] [0 0 0 1 ] [0 0 0 1] [2T3] = [1 0 0 0 ] (d3 variable \to Joint 3 is PRISMATIC) [0 1 0 0 ] [0 0 1 d3] [0 0 0 1 ]

Manipulator type: Revolute (J1) Revolute (J2) Prismatic (J3) = R-R-P

[0T2] Computation

Multiply [0T1] × [1T2] row by column:

[0T2] = [0T1][1T2] = [-C1S2 -C1C2 S1 0.2C1] [-S1S2 -S1C2 -C1 0.2S1] [ C2 -S2 0 0.4 ] [ 0 0 0 1 ]

💡Key check: 3rd row of [0T1] is [0,1,0,0.4]. This picks out row 2 of [1T2] for 3,1 entry → C2. ✓

[0T3] Computation

[0T3] = [0T2][2T3]. Since [2T3] only adds d3 along z₂:

Position of EE = R₀₂\cdot[0,0,d3]ᵀ + t₀₂ R₀₂\cdot[0,0,d3]ᵀ = [S1\cdotd3, -C1\cdotd3, 0]ᵀ [0T3] = [-C1S2 -C1C2 S1 0.2C1 + S1\cdotd3] [-S1S2 -S1C2 -C1 0.2S1 - C1\cdotd3] [ C2 -S2 0 0.4 ] [ 0 0 0 1 ]

Jacobian Derivation

Expressions:

Revolute joint i: Jᵢ = [⁰zᵢ₋₁ \times (⁰Pₙ - ⁰Pᵢ₋₁)] ⁰zᵢ₋₁ is z-axis of frame {i-1} in {0} [ ⁰zᵢ₋₁ ] Prismatic joint i: Jᵢ = [⁰zᵢ₋₁] [ 0 ] ⁰Pᵢ₋₁ = translation column of [⁰Tᵢ₋₁]

Joint 1 (Revolute):

⁰z₀ = [0, 0, 1]ᵀ ⁰P₀ = [0, 0, 0]ᵀ ⁰Pₙ = [0.2C1+S1d3, 0.2S1-C1d3, 0.4]ᵀ J₁_lin = z₀ \times Pₙ = [0,0,1] \times [0.2C1+S1d3, 0.2S1-C1d3, 0.4] = [-(0.2S1-C1d3), (0.2C1+S1d3), 0] = [C1d3 - 0.2S1, 0.2C1 + S1d3, 0]ᵀ J₁ = [C1d3 - 0.2S1, 0.2C1 + S1d3, 0, 0, 0, 1]ᵀ

Joint 2 (Revolute):

⁰z₁ = 3rd col of [0T1] rotation = [S1, -C1, 0]ᵀ ⁰P₁ = [0.2C1, 0.2S1, 0.4]ᵀ ⁰Pₙ - ⁰P₁ = [S1d3, -C1d3, 0]ᵀ J₂_lin = [S1,-C1,0] \times [S1d3,-C1d3,0] = [-C1\cdot0 - 0\cdot(-C1d3), 0\cdotS1d3 - S1\cdot0, S1(-C1d3) - (-C1)(S1d3)] = [0, 0, 0]ᵀ (z₁ ∥ displacement vector \to no linear contribution) J₂ = [0, 0, 0, S1, -C1, 0]ᵀ

Joint 3 (Prismatic):

⁰z₂ = 3rd col of [0T2] rotation = [S1, -C1, 0]ᵀ J₃ = [S1, -C1, 0, 0, 0, 0]ᵀ

Full Manipulator Jacobian:

J1 J2 J3 [C1d3-0.2S1 0 S1 ] [0.2C1+S1d3 0 -C1 ] [ 0 0 0 ] [ 0 S1 0 ] [ 0 -C1 0 ] [ 1 0 0 ]

Numerical Substitution (θ₁=30°, θ₂=45°, d₃=0.4 m)

C1 = cos30° = 0.866, S1 = sin30° = 0.500 C2 = cos45° = 0.707, S2 = sin45° = 0.707 J₁: [C1d3-0.2S1] = [0.866\times0.4 - 0.2\times0.5] = [0.3464-0.1] = 0.2464 [0.2C1+S1d3] = [0.2\times0.866 + 0.5\times0.4] = [0.1732+0.2] = 0.3732 J₂: [0, 0, 0, 0.5, -0.866, 0]ᵀ J₃: [0.5, -0.866, 0, 0, 0, 0]ᵀ J = [0.2464 0 0.500] [0.3732 0 -0.866] [0 0 0 ] [0 0.500 0 ] [0 -0.866 0 ] [1 0 0 ]

Joint Torques [τ] = [J]ᵀ [ℱ]

[ℱ] = [10, 25, 0, 10, 20, 0]ᵀ τ₁ = (0.2464)(10) + (0.3732)(25) + 0 + 0 + 0 + 1(0) = 2.464 + 9.33 = 11.79 N\cdotm τ₂ = 0 + 0 + 0 + (0.5)(10) + (-0.866)(20) + 0 = 5 - 17.32 = -12.32 N\cdotm τ₃ = (0.5)(10) + (-0.866)(25) + 0 + 0 + 0 + 0 = 5 - 21.65 = -16.65 N

Final Answer

Joint 1 Torque τ₁

11.79 N·m

Joint 2 Torque τ₂

−12.32 N·m

Joint 3 Force τ₃

−16.65 N

Question 3

C² Trajectory Planning — Via Point

10 Marks

Given

Joint 1 of R-R-P manipulator. Start at rest (θ=0°, θ̇=0).
Via Point: t=4s, θ=24°, θ̇=32°/s
Goal Point: t=8s (4s after via), θ=60°, θ̇=60°/s

Concept

A C²-smooth trajectory uses cubic polynomials for each segment, ensuring continuity of position, velocity AND acceleration at the via point. Each segment: θ(t) = a₀ + a₁t + a₂t² + a₃t³ (4 unknowns, 4 boundary conditions per segment).

Segment 1 (0 ≤ t ≤ 4 s)

BCs: θ(0)=0°, θ̇(0)=0, θ(4)=24°, θ̇(4)=32°/s

a₀ = 0 (from θ(0)=0) a₁ = 0 (from θ̇(0)=0) θ(4): 16a₂ + 64a₃ = 24 \to a₂ + 4a₃ = 1.5 ... (i) θ̇(4): 8a₂ + 48a₃ = 32 \to a₂ + 6a₃ = 4.0 ... (ii) (ii)-(i): 2a₃ = 2.5 \to a₃ = 1.25 (i): a₂ = 1.5 - 4(1.25) = 1.5 - 5 = a₂ = -3.5 θ₁(t) = -3.5t² + 1.25t³ [degrees] θ̇₁(t) = -7t + 3.75t² [degrees/s] θ̈₁(t) = -7 + 7.5t [degrees/s²]

✅Verify: θ₁(4) = −3.5(16)+1.25(64) = −56+80 = 24° ✓ | θ̇₁(4) = −28+60 = 32°/s ✓

Segment 2 (4 ≤ t ≤ 8 s, let τ = t−4)

BCs: θ(0)=24°, θ̇(0)=32°/s, θ(4)=60°, θ̇(4)=60°/s

b₀ = 24, b₁ = 32 θ(4): 24 + 32(4) + 16b₂ + 64b₃ = 60 16b₂ + 64b₃ = 60 - 24 - 128 = -92 b₂ + 4b₃ = -5.75 ... (iii) θ̇(4): 32 + 8b₂ + 48b₃ = 60 b₂ + 6b₃ = 3.5 ... (iv) (iv)-(iii): 2b₃ = 9.25 \to b₃ = 4.625 (iii): b₂ = -5.75 - 4(4.625) = b₂ = -24.25 θ₂(τ) = 24 + 32τ - 24.25τ² + 4.625τ³ [°], τ = t-4 θ̇₂(τ) = 32 - 48.5τ + 13.875τ² [°/s] θ̈₂(τ) = -48.5 + 27.75τ [°/s²]

✅Verify: θ₂(4) = 24+128−388+296 = 60° ✓ | θ̇₂(4) = 32−194+222 = 60°/s ✓

Final Answer

Segment 1 (0 ≤ t ≤ 4 s):

θ(t) = -3.5t² + 1.25t³ θ̇(t) = -7t + 3.75t² θ̈(t) = -7 + 7.5t

Segment 2 (4 ≤ t ≤ 8 s, τ = t−4):

θ(τ) = 24 + 32τ - 24.25τ² + 4.625τ³ θ̇(τ) = 32 - 48.5τ + 13.875τ² θ̈(τ) = -48.5 + 27.75τ

Question 4

2-DOF RR Manipulator — DH Parameters, Transforms & Velocities

20 Marks

Given

Two revolute joints (planar), a₁ = a₂ = 0.25 m, both links in horizontal plane rotating about vertical axes.

DH Parameters

Link | θᵢ | dᵢ | aᵢ | αᵢ 1 | θ₁* | 0 | 0.25m | 0° 2 | θ₂* | 0 | 0.25m | 0° (* = variable joint angle)

Transformation Matrices

[0T1] = [C1 -S1 0 0.25C1] [S1 C1 0 0.25S1] [0 0 1 0 ] [0 0 0 1 ] [1T2] = [C2 -S2 0 0.25C2] [S2 C2 0 0.25S2] [0 0 1 0 ] [0 0 0 1 ] [0T2] = [0T1][1T2] = [C12 -S12 0 0.25C1+0.25C12] [S12 C12 0 0.25S1+0.25S12] [0 0 1 0 ] [0 0 0 1 ] where C12 = cos(θ₁+θ₂), S12 = sin(θ₁+θ₂)

Velocity Expressions & Computation

Angular velocity of link i: ⁰ωᵢ = ⁰ωᵢ₋₁ + θ̇ᵢ \cdot ⁰zᵢ₋₁ Linear velocity of link i: ⁰vᵢ = ⁰vᵢ₋₁ + ⁰ωᵢ \times ⁰rᵢ₋₁,ᵢ

Link 1:

⁰ω₁ = [0, 0, θ̇₁]ᵀ ⁰v₁ = ⁰ω₁ \times r₁ = [0,0,θ̇₁] \times [0.25C1, 0.25S1, 0] = [0\cdot0 - θ̇₁\cdot0.25S1, θ̇₁\cdot0.25C1 - 0\cdot0, 0] = [-0.25S1\cdotθ̇₁, 0.25C1\cdotθ̇₁, 0]ᵀ

Link 2 (End Effector):

⁰ω₂ = [0, 0, θ̇₁+θ̇₂]ᵀ ⁰v₂ = ⁰ω₂ \times ⁰p₂ ⁰p₂ = [0.25C1+0.25C12, 0.25S1+0.25S12, 0]ᵀ ⁰v₂ = [0,0,(θ̇₁+θ̇₂)] \times [0.25C1+0.25C12, 0.25S1+0.25S12, 0] = [-(0.25S1\cdotθ̇₁ + 0.25S12(θ̇₁+θ̇₂))] [ (0.25C1\cdotθ̇₁ + 0.25C12(θ̇₁+θ̇₂))] [ 0 ]

Manipulator Velocity Matrix (Jacobian form):

[ẋ ] [-0.25S1-0.25S12 -0.25S12] [θ̇₁] [ẏ ] = [ 0.25C1+0.25C12 0.25C12] [θ̇₂] [ω_z] [ 1 1 ]

Question 5a

Gear Train System — Transfer Function

12 Marks

Given

1V → Motor shaft = 120 rpm
Pinion speed = 2.5 × Gear speed (motor drives pinion → gear reduces speed)
Link at 0.5 rpm → Load moves 8 cm/min

Step-by-Step Solution

Motor Gain:

Km = 120 rpm / 1 V = 120 rpm/V

Gear Train Gain:
Pinion is driven by motor. ω_pinion = 2.5 × ω_gear. Motor → Pinion → Gear → Link.

K_gt = ω_gear / ω_pinion = 1/2.5 = 0.4

Link-Load Gain:

K_ll = (8 cm/min) / (0.5 rpm) = 16 cm per revolution

Overall Transfer Function:

K_total = Km \times K_gt \times K_ll = 120 \times (1/2.5) \times 16 = 120 \times 0.4 \times 16 = 768 cm/min per Volt

Supply Voltage for load at 76.8 m/min:

76.8 m/min = 7680 cm/min V = 7680 / 768 = 10 V

Final Answer

Motor Gain Km

120 rpm/V

Gear Ratio Kgt

0.4 (= 1/2.5)

Link-Load Kll

16 cm/rpm

Overall Gain

768 cm/min/V

Voltage for 76.8 m/min

V = 10 V

Question 5b

Strain Gauge — Load on Link

3 Marks

Given

Strain gauge resistance R = 100 Ω, Change ΔR = 0.2 Ω
Modulus of Elasticity E = 200 GPa = 200 × 10⁹ Pa
Cross-sectional area A = 7.5 × 10⁻⁶ m²

Step-by-Step Solution

Gauge Factor (GF = 2 for metal wire strain gauge):

GF = (ΔR/R) / ε \to ε = (ΔR/R) / GF = (0.2/100) / 2 = 0.001 (strain)

Stress from Hooke's Law:

σ = E \times ε = 200\times10⁹ \times 0.001 = 2\times10⁸ Pa = 200 MPa

Force:

F = σ \times A = 2\times10⁸ \times 7.5\times10⁻⁶ = 1500 N

Final Answer

Load on the link

F = 1500 N

Comprehensive Examination

BITS F441 — Robotics

10 May 2023 Closed Book 80 Marks 3 Hours

Question 1

Lagrangian Mechanics — Single Link (Different Values)

10 Marks

Given

Link length l = 0.4 m, Mass m = 0.5 kg, Revolute joint at base.

Solution (same method as Paper 1 Q1a)

Moment of Inertia:

I = ml²/3 = (0.5)(0.4)²/3 = (0.5)(0.16)/3 = 0.02667 kg\cdotm²

Kinetic Energy:

T = ½Iθ̇² = ½(0.02667)θ̇² = 0.01333 θ̇² J

Potential Energy:

V = mg(l/2)sinθ = (0.5)(9.81)(0.2)sinθ = 0.981 sinθ J

Lagrangian:

L = T - V = 0.01333 θ̇² - 0.981 sinθ

Torque:

d/dt(\partialL/\partialθ̇) = 0.02667 θ̈ \partialL/\partialθ = -0.981 cosθ τ = 0.02667 θ̈ - (-0.981 cosθ) = 0.02667 θ̈ + 0.981 cosθ N\cdotm

Final Answer

0.01333 θ̇² J

0.981 sinθ J

0.01333θ̇²−0.981sinθ

Torque τ

0.02667θ̈+0.981cosθ N·m

Question 2

C² Trajectory — RRP Joint 2, Via Point

15 Marks

Given

Joint 2 of 3-DOF RRP. Total time = 6s. θ: 15° → 45°. Final velocity = 18 deg/s (starts from rest).
Via point at t=3s: θ=25°, θ̇=5 deg/s

Segment 1 (0 ≤ t ≤ 3 s)

BCs: θ(0)=15°, θ̇(0)=0, θ(3)=25°, θ̇(3)=5 deg/s

a₀ = 15, a₁ = 0 θ(3): 15 + 9a₂ + 27a₃ = 25 \to 9a₂ + 27a₃ = 10 \to a₂ + 3a₃ = 10/9 ... (i) θ̇(3): 6a₂ + 27a₃ = 5 \to a₂ + 4.5a₃ = 5/6 ... (ii) (ii)-(i): 1.5a₃ = 5/6 - 10/9 = 15/18 - 20/18 = -5/18 a₃ = -5/(18\times1.5) = -5/27 \approx -0.1852 a₂ = 10/9 - 3(-5/27) = 10/9 + 15/27 = 30/27 + 15/27 = 45/27 = 5/3 \approx 1.667 θ₁(t) = 15 + (5/3)t² - (5/27)t³ [°] θ̇₁(t) = (10/3)t - (5/9)t² [°/s] θ̈₁(t) = (10/3) - (10/9)t [°/s²]

✅Check: θ₁(3)=15+(5/3)(9)−(5/27)(27) = 15+15−5 = 25° ✓ | θ̇₁(3)=(10/3)(3)−(5/9)(9)=10−5=5°/s ✓

Segment 2 (3 ≤ t ≤ 6 s, let τ = t−3)

BCs: θ(0)=25°, θ̇(0)=5 deg/s, θ(3)=45°, θ̇(3)=18 deg/s

b₀ = 25, b₁ = 5 θ(3): 25 + 15 + 9b₂ + 27b₃ = 45 \to 9b₂ + 27b₃ = 5 \to b₂+3b₃ = 5/9 ... (iii) θ̇(3): 5 + 6b₂ + 27b₃ = 18 \to 6b₂ + 27b₃ = 13 \to b₂+4.5b₃ = 13/6 ... (iv) (iv)-(iii): 1.5b₃ = 13/6 - 5/9 = 39/18 - 10/18 = 29/18 b₃ = 29/(18\times1.5) = 29/27 \approx 1.074 b₂ = 5/9 - 3(29/27) = 15/27 - 87/27 = -72/27 = -8/3 \approx -2.667 θ₂(τ) = 25 + 5τ - (8/3)τ² + (29/27)τ³ [°], where τ = t-3 θ̇₂(τ) = 5 - (16/3)τ + (29/9)τ² [°/s] θ̈₂(τ) = -(16/3) + (58/27)τ [°/s²]

✅Check: θ₂(3)=25+15−24+29=45° ✓ | θ̇₂(3)=5−16+29/3... let me verify: 5−(16/3)(3)+(29/9)(9)=5−16+29=18°/s ✓

Question 3a

Singularities from Jacobian

4 Marks

Given Jacobian

[J] = [(-0.25S₁₂-0.3S₁) -0.25S₁₂ 0] [ (0.25C₁₂+0.3C₁) 0.25C₁₂ 0] [ 0 0 1] [ 0 0 0] [ 0 0 0] [ 1 1 0]

Concept

Singularity occurs when det(J) = 0, i.e., the Jacobian loses rank. Extract the position Jacobian (upper 3×3):

Step-by-Step Solution

Extract position Jacobian (rows 1–3):

Jp = [(-0.25S₁₂-0.3S₁) -0.25S₁₂ 0] [ (0.25C₁₂+0.3C₁) 0.25C₁₂ 0] [ 0 0 1]

Expand along 3rd column (cofactor):

det(Jp) = 1 \times [(-0.25S₁₂-0.3S₁)(0.25C₁₂) - (-0.25S₁₂)(0.25C₁₂+0.3C₁)] = -0.0625 S₁₂C₁₂ - 0.075 S₁C₁₂ + 0.0625 S₁₂C₁₂ + 0.075 S₁₂C₁ = 0.075(S₁₂C₁ - S₁C₁₂) Using trig identity: sin(A-B) = sinA cosB - cosA sinB S₁₂C₁ - C₁₂S₁ = sin(θ₁+θ₂-θ₁) = sin(θ₂) ∴ det(Jp) = 0.075 sin(θ₂)

Singularity condition:

det(J) = 0 \to 0.075 sin(θ₂) = 0 \to sin(θ₂) = 0 \to θ₂ = 0° or θ₂ = \pm180°

Physical Meaning

θ₂ = 0°: Both links fully aligned (extended) — the manipulator is at its workspace boundary. One degree of freedom is lost.
θ₂ = ±180°: Link 2 folds exactly back on link 1 — another boundary singularity.

Final Answer

Singularity when θ₂ = 0° or ±180°

Question 3b

Gear Train System (Same as Paper 1 Q5a)

9 Marks

Solution

Identical problem to Paper 1 Q5a — refer to that solution above.

120 rpm/V

Kgt

0.4

Kll

16 cm/rpm

Voltage

10 V

Question 3c

Strain Gauge — Load on Link (Different Values)

4 Marks

Given

R = 120 Ω, ΔR = 0.125 Ω, E = 210 GPa, A = 7.255 mm² = 7.255×10⁻⁶ m²

Step-by-Step Solution

Strain (GF = 2):

ε = (ΔR/R) / GF = (0.125/120) / 2 = 0.001042/2 = 5.208\times10⁻⁴

Stress:

σ = E \times ε = 210\times10⁹ \times 5.208\times10⁻⁴ = 1.0938\times10⁸ Pa

Force:

F = σ \times A = 1.0938\times10⁸ \times 7.255\times10⁻⁶ = \approx 793.5 N \approx 794 N

Final Answer

Load on link

F ≈ 794 N

Question 4

2-DOF RR Manipulator (Same Structure as Paper 1 Q4)

20 Marks

Solution

Identical configuration to Paper 1 Q4 — refer to that solution for full derivation of DH parameters, [0T1], [1T2], [0T2], and velocity expressions.

DH Parameters: a₁=a₂=0.25m, d₁=d₂=0, α₁=α₂=0° [0T2] = [C12 -S12 0 0.25C1+0.25C12] [S12 C12 0 0.25S1+0.25S12] [0 0 1 0 ] [0 0 0 1 ] ω₁ = [0, 0, θ̇₁]ᵀ ω₂ = [0, 0, θ̇₁+θ̇₂]ᵀ v₁ = [-0.25 S1 θ̇₁, 0.25 C1 θ̇₁, 0]ᵀ v₂ = [-(0.25S1θ̇₁+0.25S12(θ̇₁+θ̇₂)), (0.25C1θ̇₁+0.25C12(θ̇₁+θ̇₂)), 0]ᵀ

Question 5

3-DOF Manipulator — Full Jacobian Analysis

18 Marks

Given

[0T1]=[C1 0 S1 0.15C1] [1T2]=[C2 0 -S2 0] [2T3]=[1 0 0 0 ] [S1 0 -C1 0.15S1] [S2 0 C2 0] [0 1 0 0 ] [0 1 0 0.4 ] [0 -1 0 0] [0 0 1 d3] [0 0 0 1 ] [0 0 0 1] [0 0 0 1 ] Joint types: Revolute (J1) + Revolute (J2) + Prismatic (J3) = R-R-P

[0T2] = [0T1][1T2]

Element-wise multiplication: (1,1): C1C2+0+0 = C1C2 (1,2): 0+0+S1(-1) = -S1 (1,3): C1(-S2)+0+0 = -C1S2 (1,4): 0.15C1 (2,1): S1C2 = S1C2 (2,2): 0+0+(-C1)(-1) = C1 (2,3): S1(-S2)+0 = -S1S2 (2,4): 0.15S1 (3,1): 0+S2+0 = S2 (3,2): 0+0+0 = 0 (3,3): 0+C2+0 = C2 (3,4): 0.4 [0T2] = [C1C2 -S1 -C1S2 0.15C1] [S1C2 C1 -S1S2 0.15S1] [S2 0 C2 0.4 ] [0 0 0 1 ]

[0T3] = [0T2][2T3]

R_03 = R_02 (same rotation) p_03 = R_02 \cdot [0,0,d3]ᵀ + p_02 3rd column of R_02 = [-C1S2, -S1S2, C2]ᵀ R_02\cdot[0,0,d3]ᵀ = [-C1S2\cdotd3, -S1S2\cdotd3, C2\cdotd3]ᵀ [0T3] = [C1C2 -S1 -C1S2 0.15C1-C1S2\cdotd3] [S1C2 C1 -S1S2 0.15S1-S1S2\cdotd3] [S2 0 C2 0.4+C2\cdotd3 ] [0 0 0 1 ]

Jacobian Derivation

Joint 1 (Revolute):

z₀=[0,0,1]ᵀ, p_n=[0.15C1-C1S2d3, 0.15S1-S1S2d3, 0.4+C2d3]ᵀ J₁_lin = z₀\timesp_n = [-( 0.15S1-S1S2d3), (0.15C1-C1S2d3), 0] = [S1(S2d3-0.15), C1(0.15-S2d3), 0]ᵀ J₁ = [S1(S2d3-0.15), C1(0.15-S2d3), 0, 0, 0, 1]ᵀ

Joint 2 (Revolute):

z₁ = 3rd col of R_01 = [S1, -C1, 0]ᵀ p₁ = [0.15C1, 0.15S1, 0.4]ᵀ r = p_n - p₁ = [-C1S2d3, -S1S2d3, C2d3]ᵀ J₂_lin = z₁\timesr = [S1,-C1,0] \times [-C1S2d3,-S1S2d3,C2d3] = [(-C1)(C2d3)-0(-S1S2d3), 0(-C1S2d3)-S1(C2d3), S1(-S1S2d3)-(-C1)(-C1S2d3)] = [-C1C2d3, -S1C2d3, -(S1²+C1²)S2d3] = [-C1C2d3, -S1C2d3, -S2d3]ᵀ J₂ = [-C1C2d3, -S1C2d3, -S2d3, S1, -C1, 0]ᵀ

Joint 3 (Prismatic):

z₂ = 3rd col of R_02 = [-C1S2, -S1S2, C2]ᵀ J₃ = [-C1S2, -S1S2, C2, 0, 0, 0]ᵀ

Full Symbolic Jacobian:

J₁ J₂ J₃ [ S1(S2d3-0.15) -C1C2d3 -C1S2 ] [ C1(0.15-S2d3) -S1C2d3 -S1S2 ] [ 0 -S2d3 C2 ] [ 0 S1 0 ] [ 0 -C1 0 ] [ 1 0 0 ]

Numerical: θ₁=30°, θ₂=45°, d₃=0.2 m

S1=0.5, C1=0.866, S2=0.707, C2=0.707, d3=0.2 J₁: S1(S2d3-0.15) = 0.5(0.707\times0.2-0.15) = 0.5(0.1414-0.15) = 0.5(-0.0086) = -0.0043 C1(0.15-S2d3) = 0.866(0.15-0.1414) = 0.866\times0.0086 = 0.00745 \to J₁ = [-0.0043, 0.00745, 0, 0, 0, 1]ᵀ J₂: -C1C2d3 = -0.866\times0.707\times0.2 = -0.1224 -S1C2d3 = -0.5\times0.707\times0.2 = -0.0707 -S2d3 = -0.707\times0.2 = -0.1414 \to J₂ = [-0.1224, -0.0707, -0.1414, 0.5, -0.866, 0]ᵀ J₃: -C1S2 = -0.866\times0.707 = -0.6124 -S1S2 = -0.5\times0.707 = -0.3536 C2 = 0.707 \to J₃ = [-0.6124, -0.3536, 0.707, 0, 0, 0]ᵀ Numerical J = [-0.0043 -0.1224 -0.6124] [ 0.00745 -0.0707 -0.3536] [ 0 -0.1414 0.707 ] [ 0 0.5 0 ] [ 0 -0.866 0 ] [ 1 0 0 ]

Joint Torques [τ] = [J]ᵀ [ℱ]

[ℱ] = [18, 22, 12, 0, 0, 0]ᵀ τ₁ = J₁ᵀ\cdotF = (-0.0043)(18) + (0.00745)(22) + 0(12) + 0+0+(1)(0) = -0.0774 + 0.1639 = 0.0865 N\cdotm τ₂ = J₂ᵀ\cdotF = (-0.1224)(18) + (-0.0707)(22) + (-0.1414)(12) + 0.5(0) + (-0.866)(0) + 0 = -2.203 - 1.555 - 1.697 = -5.455 N\cdotm τ₃ = J₃ᵀ\cdotF = (-0.6124)(18) + (-0.3536)(22) + (0.707)(12) = -11.023 - 7.779 + 8.484 = -10.318 N

Final Answer

τ₁

0.086 N·m

τ₂

−5.455 N·m

τ₃ (force)

−10.318 N

Study Plan

Chapters Appearing in Both Papers

6 Chapters Identified Attach slides → Get notes

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Once you attach the lecture slides for each chapter below, I'll generate comprehensive formula sheets, concept notes, and worked examples for each. All topics from both papers are covered here.

CH 01

Lagrangian Dynamics

Lagrangian L=T−V, Euler-Lagrange equation, torque derivation for single & multi-link systems, kinetic & potential energy of robot links

Dec'22 Q1aMay'23 Q1

CH 02

Forward Kinematics — DH Parameters & Transformation Matrices

Denavit-Hartenberg convention, link-joint transformation matrix (i−1)Tᵢ, frame assignment, chaining [0T2], [0T3] for multi-DOF manipulators

Dec'22 Q2, Q4May'23 Q4, Q5

CH 03

Velocity Kinematics — Jacobian Matrix

Jacobian for revolute & prismatic joints, z-axis cross product method, [0P_{i-1}] relation, singularity analysis (det J=0), joint torque from [τ]=[J]ᵀ[F]

Dec'22 Q2, Q4May'23 Q3a, Q4, Q5

CH 04

Trajectory Planning

C² smooth trajectories, cubic polynomial segments, via point conditions, position/velocity/acceleration time history for R-R-P manipulators

Dec'22 Q3May'23 Q2

CH 05

Robot Actuators & Control — Gear Train Transfer Function

Motor gain, gear ratio, link-to-load gain, block diagram, overall transfer function, supply voltage calculation

Dec'22 Q5aMay'23 Q3b

CH 06

Sensors — Strain Gauges

Gauge factor (GF), strain from ΔR/R, stress from Hooke's law (σ=Eε), axial load from F=σA, metal wire vs semiconductor gauges

Dec'22 Q5bMay'23 Q3c

CH 07

Statics — Principle of Virtual Work

Virtual displacement, virtual work equation ΣFᵢ·δᵢ=0, reaction at roller/pin supports, beam equilibrium

Dec'22 Q1b

Next Steps

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