---
title: "Dynamic equilibrium; velocity and acceleration in polar"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:dynamic-equilibrium-polar-velocity-acceleration"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/dynamic-equilibrium-polar-velocity-acceleration
site: BCT Engineering
---

# Dynamic equilibrium; velocity and acceleration in polar

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2071 Shrawan, 2+8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/dynamic-equilibrium-polar-velocity-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

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.

## 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.

## Answer

#### 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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is added to the real forces to put a moving particle in dynamic equilibrium?
   A. An inertia force -m a, opposite to acceleration
   B. A force equal to the velocity
   C. A force equal to m g
   D. A friction force
   Answer: A. F - m a = 0 treats the body as in equilibrium with an imaginary inertia force opposite to the acceleration (D'Alembert). (Key point: Dynamic equilibrium)

2. How are theta_dot and theta_ddot found for theta = t^2/3?
   A. Differentiate once and twice: theta_dot = 2t/3, theta_ddot = 2/3
   B. Integrate: t^3/9
   C. theta_dot = theta/t
   D. Both equal pi/2
   Answer: A. Differentiating gives 2t/3 and 2/3. With t = 2.171 s at theta = pi/2, theta_dot = 1.447 rad/s. (Key point: Finding time and rates)

3. Which gives the speed in polar coordinates?
   A. v = r_dot + r theta_dot
   B. v = sqrt(r_dot^2 + (r theta_dot)^2)
   C. v = r theta_dot only
   D. v = sqrt(r_dot^2 - r^2 theta_dot^2)
   Answer: B. v_r = r_dot and v_theta = r theta_dot are perpendicular, so v = sqrt(7.236^2 + 11.366^2) = 13.47 m/s. (Key point: Polar velocity components)

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