---
title: "Radial and transverse components of acceleration"
subject: "Engineering Mechanics"
chapter: 2
key: "semester_3/engineering_mechanics:polar-components-velocity-acceleration"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/polar-components-velocity-acceleration
site: BCT Engineering
---

# Radial and transverse components of acceleration

Engineering Mechanics · Chapter 2: Kinetics of Particles: Force, Mass and Acceleration

- Asked in: year not given, 8 marks
- Video: [Introduction to Curvilinear Motion in Polar Coordinates (aka radial and transverse components)](https://www.youtube.com/watch?v=65jiyQ92R7I) — structurefree
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/polar-components-velocity-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-2

## Question

Derive expressions for the radial and transverse components of velocity and acceleration of a particle moving along a plane curve. Hence write the equations of motion of the particle in polar coordinates.

## Short answer

Velocity components: v_r = ṙ, v_θ = rθ̇. Acceleration: a_r = r̈ – rθ̇², a_θ = rθ̈ + 2ṙθ̇. Equations of motion: ΣF_r = m(r̈ – rθ̇²), ΣF_θ = m(rθ̈ + 2ṙθ̇). For central forces, angular momentum r²θ̇ is constant.

## Answer

Polar coordinates $(r,\theta)$ locate a particle by its distance $r$ from the pole O and the angle $\theta$ of the radius vector from a fixed line. Two unit vectors are used: $\mathbf{e}_r$ along the radius vector and $\mathbf{e}_\theta$ perpendicular to it (in the direction of increasing $\theta$).

```text
          y
          |      P
          |    / |
          |  /   | e_theta
          | / r  |-> e_r
          |/ th
     O----+------------ x
```

#### Rate of change of unit vectors
Because the directions of $\mathbf{e}_r$ and $\mathbf{e}_\theta$ rotate with $\theta$:

$$
\dot{\mathbf{e}}_r = \dot\theta\,\mathbf{e}_\theta \qquad \dot{\mathbf{e}}_\theta = -\dot\theta\,\mathbf{e}_r
$$

#### Velocity
The position vector is $\mathbf{r} = r\,\mathbf{e}_r$. Differentiating:

$$
\mathbf{v} = \dot r\,\mathbf{e}_r + r\dot{\mathbf{e}}_r = \dot r\,\mathbf{e}_r + r\dot\theta\,\mathbf{e}_\theta
$$

So $v_r = \dot r$ (radial) and $v_\theta = r\dot\theta$ (transverse), with $v = \sqrt{v_r^2 + v_\theta^2}$.

#### Acceleration
Differentiating again:

$$
\begin{aligned}
\mathbf{a} &= \ddot r\,\mathbf{e}_r + \dot r\dot\theta\,\mathbf{e}_\theta + (\dot r\dot\theta + r\ddot\theta)\,\mathbf{e}_\theta - r\dot\theta^2\,\mathbf{e}_r \\
&= (\ddot r - r\dot\theta^2)\,\mathbf{e}_r + (r\ddot\theta + 2\dot r\dot\theta)\,\mathbf{e}_\theta
\end{aligned}
$$

$$
a_r = \ddot r - r\dot\theta^2 \qquad a_\theta = r\ddot\theta + 2\dot r\dot\theta
$$

The term $2\dot r\dot\theta$ is the Coriolis component; $r\dot\theta^2$ is the centripetal term.

#### Equations of motion
Applying Newton's second law $\Sigma \mathbf{F} = m\mathbf{a}$ along the two directions:

$$
\Sigma F_r = m(\ddot r - r\dot\theta^2) \qquad \Sigma F_\theta = m(r\ddot\theta + 2\dot r\dot\theta)
$$

For a central force ($\Sigma F_\theta = 0$) the second equation gives $\dfrac{d}{dt}(r^2\dot\theta) = 0$, i.e. constant angular momentum $h = r^2\dot\theta$ (Kepler's law of areas).

## Check yourself

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

1. How do the polar unit vectors change with time?
   A. e_r' = theta' e_theta and e_theta' = -theta' e_r
   B. e_r' = -theta' e_theta and e_theta' = theta' e_r
   C. Both are constant
   D. e_r' = r' e_theta and e_theta' = r' e_r
   Answer: A. The unit vectors rotate with the radius vector, giving e_r' = theta' e_theta and e_theta' = -theta' e_r. (Key point: Unit vector derivatives)

2. The radial and transverse velocity components are:
   A. v_r = r theta', v_theta = r'
   B. v_r = r', v_theta = r theta'
   C. v_r = r'', v_theta = r theta''
   D. v_r = r' + r theta', v_theta = 0
   Answer: B. Differentiating r e_r gives r' e_r + r theta' e_theta. (Key point: Velocity components)

3. The transverse component of acceleration is:
   A. r theta'' - r' theta'
   B. r'' - r theta'^2
   C. r theta'' + 2 r' theta'
   D. r theta'^2 + 2 r'
   Answer: C. a_theta = r theta'' + 2 r' theta'. The radial one is r'' - r theta'^2. (Key point: Acceleration components)

4. For a central force, sum F_theta = 0. What follows?
   A. r' is constant
   B. r^2 theta' is constant (constant angular momentum)
   C. theta' is zero
   D. The speed is constant
   Answer: B. Setting r theta'' + 2 r' theta' = 0 gives d/dt (r^2 theta') = 0, so h = r^2 theta' is constant. (Key point: Central force)

## Related in Engineering Mechanics

- [Particle motion in polar coordinates](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Apolar-motion-numerical-particle) — Engineering Mechanics, chapter 2
- [Equations of motion for plane motion](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aequations-plane-motion-rigid-body) — Engineering Mechanics, chapter 6
- [Plane motion and relative velocity](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aplane-motion-types-relative-velocity) — Engineering Mechanics, chapter 5

## Related in other subjects

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- [Velocity and acceleration of moving particle at t=1](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Avelocity-acceleration-moving-particle-t) — Mathematics, chapter 5
- [Radial and transverse acceleration derivation](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aradial-transverse-acceleration) — Applied Mechanics (Dynamics), chapter 1, asked 2×
- [Velocity and acceleration of particle on curve](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Avelocity-acceleration-particle-curve) — Mathematics, chapter 5, asked 2×