Polar Coordinates: Radar Tracking & Zero Acceleration | Engineering Dynamics Part 11

Welcome back to Engineering Dynamics with Prof. Gearford Orange from the F.I.T. Department of Mechanical Engineering! Can radial velocity and angular velocity change even when an object moves in a straight line at constant speed? In Engineering Dynamics Part 11, we’ll solve a radar-tracking problem using polar-coordinate kinematics. An aircraft flies horizontally at constant velocity while a stationary radar measures its changing distance and line-of-sight angle. We’ll calculate radial velocity, angular velocity, radial acceleration, and transverse acceleration—and discover why the total acceleration is exactly zero! Step 1: Understanding the Problem An aircraft flies horizontally at constant altitude. Given: * Horizontal position: x = 1,200 m * Altitude: h = 900 m * Horizontal velocity: ẋ = 120 m/s * Horizontal acceleration: ẍ = 0 Our objectives are to determine the aircraft’s polar position, velocity components, and acceleration. Step 2: Polar Position Using the Pythagorean theorem: r = √(x² + h²) r = √(1200² + 900²) r = 1,500 m Line-of-sight angle: θ = arctan(h/x) θ ≈ 36.87° Step 3: Radial Velocity From geometry: r² = x² + h² Differentiating with respect to time: 2rṙ = 2xẋ Therefore: ṙ = xẋ/r Substituting: ṙ = (1200 × 120)/1500 ṙ = +96 m/s The aircraft is moving farther from the radar station. Step 4: Angular Velocity Starting with: tan(θ) = h/x Differentiating and simplifying: θ̇ = −hẋ/r² Substituting: θ̇ = −(900 × 120)/1500² θ̇ = −0.048 rad/s The negative sign indicates that the line-of-sight angle is decreasing. Step 5: Checking the Total Speed Polar velocity is: v⃗ = ṙeᵣ + rθ̇eθ Radial component: vᵣ = +96 m/s Transverse component: vθ = 1500(−0.048) vθ = −72 m/s Total speed: |v⃗| = √(96² + 72²) |v⃗| = 120 m/s This matches the aircraft’s actual speed! Step 6: Radial Acceleration The polar radial acceleration equation is: aᵣ = r̈ − rθ̇² Differentiate rṙ = xẋ: ṙ² + rr̈ = ẋ² Therefore: r̈ = (ẋ² − ṙ²)/r r̈ = (120² − 96²)/1500 r̈ = 3.456 m/s² Now calculate radial acceleration: aᵣ = 3.456 − 1500(0.048²) aᵣ = 3.456 − 3.456 = 0 The radial acceleration is zero! Step 7: Transverse Acceleration The transverse acceleration equation is: aθ = rθ̈ + 2ṙθ̇ Differentiating the angular velocity relationship: θ̈ = 2hẋṙ/r³ Substituting: θ̈ = 2(900)(120)(96)/1500³ θ̈ = +0.006144 rad/s² Now calculate transverse acceleration: aθ = 1500(0.006144) + 2(96)(−0.048) aθ = 9.216 − 9.216 aθ = 0 Both acceleration components cancel exactly. Final Results * Radial distance: r = 1,500 m * Line-of-sight angle: θ = 36.87° * Radial velocity: ṙ = +96 m/s * Angular velocity: θ̇ = −0.048 rad/s * Transverse velocity: vθ = −72 m/s * Radial acceleration: aᵣ = 0 m/s² * Transverse acceleration: aθ = 0 m/s² Key Takeaway Changing radial distance and angular position do not necessarily mean an object is accelerating. Polar unit vectors rotate with the line of sight, so the complete acceleration equations must account for changes in coordinate directions. In particular, r̈ is not the same as radial acceleration, and θ̈ is not the transverse component of linear acceleration. Coming Up Next: Rigid Body Motion We’ve completed our introduction to particle motion in polar coordinates! Thanks for watching, and see you in the next lesson! #EngineeringDynamics #PolarCoordinates #RadarTracking #PolarKinematics #RadialVelocity #AngularVelocity #RadialAcceleration #CurvilinearMotion #EngineeringMechanics #MechanicalEngineering #Kinematics