Angular Acceleration & Integration: Rotating Platform | Engineering Dynamics Part 13

Welcome back to Engineering Dynamics with Prof. Gearford Orange from the F.I.T. Department of Mechanical Engineering! How do we calculate angular velocity when angular acceleration changes with time? In Engineering Dynamics Part 13, we’ll solve a rotational kinematics problem involving an industrial testing platform with two different acceleration phases. We’ll use integration, piecewise angular acceleration, and acceleration-time graphs to determine angular velocity and tangential speed. Step 1: Understanding the Problem A testing platform rotates counterclockwise around a fixed vertical axis. Given: * Sensor radius: r = 1.50 m * Initial angular velocity: ω₀ = 0.80 rad/s * Phase 1: Quadratically increasing acceleration, 0–6 s * Phase 2: Linearly decreasing acceleration, 6–14 s * Maximum angular acceleration: α = 0.90 rad/s² Find the sensor’s tangential speed at t = 6 s and t = 14 s. Step 2: Angular Acceleration and Integration Angular acceleration is the rate of change of angular velocity: α = dω/dt Therefore: ω(t) = ω₀ + ∫α(t) dt For a point at fixed radius, tangential speed is: v = r|ω| Step 3: Phase 1 — Quadratic Acceleration For 0 ≤ t ≤ 6 s: α(t) = 0.90(t/6)² α(t) = 0.025t² Integrating from t = 0: ω(t) = 0.80 + ∫₀ᵗ 0.025τ² dτ ω(t) = 0.80 + t³/120 At t = 6 s: ω(6) = 0.80 + 216/120 ω(6) = 2.60 rad/s Tangential speed: v(6) = 1.50 × 2.60 v(6) = 3.90 m/s Step 4: Phase 2 — Linear Acceleration From 6 to 14 seconds, acceleration decreases from 0.90 to zero. Slope: m = (0 − 0.90)/(14 − 6) m = −0.1125 Therefore: α(t) = 0.90 − 0.1125(t − 6) For 6 ≤ t ≤ 14 s, integrate using ω(6) = 2.60 rad/s: ω(t) = 2.60 + 0.90(t − 6) − 0.05625(t − 6)² At t = 14 s: ω(14) = 2.60 + 0.90(8) − 0.05625(8²) ω(14) = 2.60 + 7.20 − 3.60 ω(14) = 6.20 rad/s Tangential speed: v(14) = 1.50 × 6.20 v(14) = 9.30 m/s Step 5: Graphical Interpretation The area beneath an angular acceleration-time graph represents the change in angular velocity: Δω = ∫α(t) dt Phase 1 — Parabolic area: Δω₁ = 1.80 rad/s Phase 2 — Triangular area: Δω₂ = ½ × 8 × 0.90 Δω₂ = 3.60 rad/s Total: ω(14) = 0.80 + 1.80 + 3.60 ω(14) = 6.20 rad/s The graphical and analytical methods give the same result! Final Results Quantity t = 6 s t = 14 s Angular velocity 2.60 rad/s 6.20 rad/s Tangential speed 3.90 m/s 9.30 m/s Angular acceleration 0.90 rad/s² 0 rad/s² Key Takeaways * Integrating angular acceleration gives the change in angular velocity. * For multiple phases, the final velocity of one phase becomes the initial velocity of the next. * The area beneath an acceleration-time graph represents the accumulated velocity change. * Zero angular acceleration does not mean zero angular velocity! Thanks for watching, and see you in the next lesson! #EngineeringDynamics #AngularAcceleration #AngularVelocity #RotationalKinematics #Integration #RotationalMotion #RigidBodyDynamics #MechanicalEngineering #EngineeringMechanics #Kinematics #Dynamics