---
title: "Car on circular curve, angular rate"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:car-circular-curve-constant-speed"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/car-circular-curve-constant-speed
site: BCT Engineering
---

# Car on circular curve, angular rate

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

- Asked in: 2081 Baishakh, 4 marks
- Video: [12-167) The radial & transverse components of velocity & acceleration of a car](https://www.youtube.com/watch?v=PKVsLMwjtQw) — RoadToEngineering 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/car-circular-curve-constant-speed
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

The car travels along the circular curve of radius $r = 400$ m with a constant speed of $v = 30$ m/s. Determine the angular rate of rotation $\dot\theta$ of the radial line $r$ and the magnitude of the car's acceleration.

## Short answer

Angular rate θ̇ = v/r = 30/400 = 0.075 rad/s. With constant speed, acceleration is purely centripetal: a = v²/r = 30²/400 = 2.25 m/s².

## Answer

The car moves on a circle with constant speed, so the speed is entirely transverse: $v=r\dot\theta$.

#### Angular rate
$$
\dot\theta=\frac{v}{r}=\frac{30}{400}=0.075\ \text{rad/s}
$$

#### Acceleration
With $r$ constant and $v$ constant: $\dot r=\ddot r=0$ and $\ddot\theta=0$.

$$
\begin{aligned}
a_r&=\ddot r-r\dot\theta^{2}=-400(0.075)^{2}=-2.25\ \text{m/s}^2\\
a_\theta&=r\ddot\theta+2\dot r\dot\theta=0
\end{aligned}
$$

So the acceleration is purely centripetal, $a=\dfrac{v^2}{r}=\dfrac{30^2}{400}=2.25\ \text{m/s}^2$, directed towards the centre.

**Answer: $\dot\theta=0.075$ rad/s; $a=2.25$ m/s$^2$.**

## Check yourself

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

1. How is the angular rate of the radial line found for a car at constant speed on a circle?
   A. theta_dot = v r
   B. theta_dot = v / r
   C. theta_dot = v^2 / r
   D. theta_dot = r / v
   Answer: B. On a circle the whole speed is transverse, v = r theta_dot, so theta_dot = 30/400 = 0.075 rad/s. (Key point: Angular rate)

2. For constant r and constant speed, which quantities are zero?
   A. r_dot, r_double_dot and theta_double_dot
   B. Only r_dot
   C. Only theta_dot
   D. theta_dot and theta_double_dot
   Answer: A. r is constant and theta_dot is constant, so r_dot, r_double_dot and theta_double_dot vanish. (Key point: Zero terms)

3. What is the transverse acceleration a_theta of the car?
   A. 2.25 m/s^2
   B. 0.075 m/s^2
   C. 0
   D. -2.25 m/s^2
   Answer: C. a_theta = r theta_double_dot + 2 r_dot theta_dot = 0 since both terms contain a zero factor. (Key point: Transverse acceleration)

4. What is the car's acceleration magnitude and direction?
   A. 2.25 m/s^2 towards the centre
   B. 2.25 m/s^2 along the velocity
   C. 0.075 m/s^2 towards the centre
   D. 30 m/s^2 towards the centre
   Answer: A. a = v^2/r = 900/400 = 2.25 m/s^2, purely centripetal. (Key point: Magnitude of acceleration)

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