---
title: "Radial and transverse velocity and acceleration"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:radial-transverse-velocity-acceleration"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/radial-transverse-velocity-acceleration
site: BCT Engineering
---

# Radial and transverse velocity and acceleration

Applied Mechanics (Dynamics) · Chapter 1: Curvilinear Motion of Particles

- Asked in: 2080 Bhadra, 4 marks
- Video: [Curvilinear Motion Polar Coordinates (Learn to solve any question)](https://www.youtube.com/watch?v=MiFYsG7kRVQ) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/radial-transverse-velocity-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

Derive the radial and transverse components of velocity and acceleration of a body moving in a curvilinear path.

## Short answer

Velocity components: v_r = ṙ, v_θ = rθ̇. Acceleration components: a_r = r̈ − rθ̇², a_θ = rθ̈ + 2ṙθ̇, obtained from r = r e_r with e_ṙ = θ̇ e_θ, e_θ̇ = −θ̇ e_r.

## Answer

A particle moving on a plane curve is located by polar coordinates $(r,\theta)$. Using the unit vectors $\hat e_r$ (along the radius vector) and $\hat e_\theta$ (perpendicular to it, in the direction of increasing $\theta$), velocity and acceleration are found by differentiating $\vec r = r\hat e_r$.

```text
        y
        |        P
        |       /|
        |    r / | e_theta
        |     /  |
        |    /   -> e_r along OP
        |   /\ theta
        +--------------- x
        O
```

#### Unit-vector derivatives
$$
\hat e_r=\cos\theta\,\hat i+\sin\theta\,\hat j,\qquad \hat e_\theta=-\sin\theta\,\hat i+\cos\theta\,\hat j
$$

Differentiating with respect to time:

$$
\dot{\hat e}_r=\dot\theta\,\hat e_\theta,\qquad \dot{\hat e}_\theta=-\dot\theta\,\hat e_r
$$

#### Velocity
$$
\vec v=\frac{d}{dt}(r\hat e_r)=\dot r\,\hat e_r+r\dot\theta\,\hat e_\theta
$$

So $v_r=\dot r$ and $v_\theta=r\dot\theta$.

#### Acceleration
$$
\begin{aligned}
\vec a=\frac{d\vec v}{dt}&=\ddot r\,\hat e_r+\dot r\,\dot{\hat e}_r+(\dot r\dot\theta+r\ddot\theta)\hat e_\theta+r\dot\theta\,\dot{\hat e}_\theta\\
&=\ddot r\,\hat e_r+\dot r\dot\theta\,\hat e_\theta+(\dot r\dot\theta+r\ddot\theta)\hat e_\theta-r\dot\theta^{2}\hat e_r
\end{aligned}
$$

Collecting terms:

$$
\vec a=(\ddot r-r\dot\theta^{2})\,\hat e_r+(r\ddot\theta+2\dot r\dot\theta)\,\hat e_\theta
$$

- Radial component: $a_r=\ddot r-r\dot\theta^{2}$
- Transverse component: $a_\theta=r\ddot\theta+2\dot r\dot\theta$
- Magnitude: $a=\sqrt{a_r^{2}+a_\theta^{2}}$

Here $-r\dot\theta^2$ is the centripetal term and $2\dot r\dot\theta$ is the Coriolis term. For motion on a circle ($r$ constant) $a_r=-r\dot\theta^2$ and $a_\theta=r\ddot\theta$.

## Check yourself

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

1. What are the radial and transverse components of velocity?
   A. v_r = r theta_dot, v_theta = r_dot
   B. v_r = r_dot, v_theta = r theta_dot
   C. v_r = r_dot, v_theta = r theta_double_dot
   D. v_r = r_double_dot, v_theta = r theta_dot
   Answer: B. Differentiating r e_r gives r_dot e_r + r theta_dot e_theta. (Key point: Velocity components)

2. In polar coordinates, what are the time derivatives of the unit vectors?
   A. de_r/dt = theta_dot e_theta and de_theta/dt = -theta_dot e_r
   B. de_r/dt = theta_dot e_r and de_theta/dt = theta_dot e_theta
   C. de_r/dt = -theta_dot e_theta and de_theta/dt = theta_dot e_r
   D. de_r/dt = 0 and de_theta/dt = 0 because they have unit length
   Answer: A. Rotating e_r by dtheta tilts it towards e_theta, while e_theta tilts towards -e_r. (Key point: Unit-vector derivatives)

3. Which expression is the radial component of acceleration?
   A. r_double_dot + r theta_dot^2
   B. r_double_dot - r theta_dot^2
   C. r theta_double_dot + 2 r_dot theta_dot
   D. r_double_dot - 2 r_dot theta_dot
   Answer: B. Collecting e_r terms gives a_r = r_double_dot - r theta_dot^2, with -r theta_dot^2 being the centripetal term. (Key point: Radial acceleration)

4. For motion on a circle (r constant), what are a_r and a_theta?
   A. a_r = -r theta_dot^2 and a_theta = r theta_double_dot
   B. a_r = r theta_dot^2 and a_theta = 2 r theta_dot
   C. a_r = 0 and a_theta = r theta_double_dot
   D. a_r = -r theta_dot^2 and a_theta = 0 always
   Answer: A. With r_dot = r_double_dot = 0 the Coriolis term vanishes and the radial term is purely centripetal. (Key point: Circular motion special case)

## Related in Applied Mechanics (Dynamics)

- [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×
- [Coriolis acceleration; rod AB and BD velocity](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acoriolis-define-rod-ab-bd-velocity) — Applied Mechanics (Dynamics), chapter 4
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