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Dynamic equilibrium; velocity and acceleration in polar

Short answer

Dynamic equilibrium is when an inertia force -m a is added to real forces, so ΣF - m a = 0, allowing statics equations. For r=5θ, θ=t²/3 at θ=π/2, radial and transverse components give velocity magnitude 13.47 m/s and acceleration magnitude 29.28 m/s².

Define the dynamic equilibrium. Determine the velocity and acceleration of the particle, if it moves along a curved path defined by r = 5θ and θ = t²/3, where r is in meters and t is in seconds. Given that the instant angle is θ = π/2.

Dynamic equilibrium

For a particle of mass m acted on by a resultant force F, Newton's law gives F = m a. Writing this as F − m a = 0 shows that the particle is in equilibrium if an imaginary inertia force (−m a), opposite to the acceleration, is added to the real forces (D'Alembert's principle). The equations of statics ΣFx = 0, ΣFy = 0 and ΣM = 0 can then be used for a body in motion; this state is called dynamic equilibrium.

Velocity and acceleration in polar coordinates

r = 5θ and θ = t²/3. When θ = π/2 = 1.5708 rad:

  • t² = 3θ = 4.712, so t = 2.171 s
  • θ̇ = 2t/3 = 1.4472 rad/s, θ̈ = 2/3 = 0.6667 rad/s²
  • r = 5θ = 7.854 m, ṙ = 5θ̇ = 7.236 m/s, r̈ = 5θ̈ = 3.333 m/s²

Velocity

  • v_r = ṙ = 7.236 m/s
  • v_θ = rθ̇ = 7.854 × 1.4472 = 11.366 m/s
  • v = √(7.236² + 11.366²) = 13.47 m/s, at tan⁻¹(11.366/7.236) = 57.5° from the radial direction

Acceleration

  • a_r = r̈ − rθ̇² = 3.333 − 7.854(2.094) = −13.12 m/s²
  • a_θ = rθ̈ + 2ṙθ̇ = 7.854(0.6667) + 2(7.236)(1.4472) = 5.236 + 20.944 = 26.18 m/s²
  • a = √(13.12² + 26.18²) = 29.28 m/s²

Answer: v = 13.47 m/s, a = 29.28 m/s².

✓ Answer checked by Claude Opus 5.5 (high effort)

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