---
title: "Particle motion in polar coordinates"
subject: "Engineering Mechanics"
chapter: 2
key: "semester_3/engineering_mechanics:polar-motion-numerical-particle"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/polar-motion-numerical-particle
site: BCT Engineering
---

# Particle motion in polar coordinates

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

- Asked in: year not given, 6 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/engineering_mechanics/polar-motion-numerical-particle
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-2

## Question

A particle moves in a plane such that its polar coordinates vary with time t (in seconds) as r = 2 + 0.5t² (m) and θ = 0.4t² (rad). At t = 2 s, determine the magnitude and direction (relative to the radial line) of the velocity and the acceleration of the particle.

## Short answer

At t=2 s: r=4 m, ṙ=2 m/s, r̈=1 m/s², θ̇=1.6 rad/s, θ̈=0.8 rad/s². v_r=2 m/s, v_θ=6.4 m/s, v=6.71 m/s at 72.6° to radial. a_r= –9.24 m/s², a_θ=9.60 m/s², a=13.32 m/s² at 133.9° to radial.

## Answer

**Given:** $r = 2 + 0.5t^2$, $\theta = 0.4t^2$, at $t = 2$ s.

#### Derivatives at t = 2 s

| Quantity | Expression | Value |
|---|---|---|
| $r$ | $2 + 0.5t^2$ | 4.00 m |
| $\dot r$ | $t$ | 2.00 m/s |
| $\ddot r$ | $1$ | 1.00 m/s² |
| $\dot\theta$ | $0.8t$ | 1.60 rad/s |
| $\ddot\theta$ | $0.8$ | 0.80 rad/s² |

#### Velocity

$$
\begin{aligned}
v_r &= \dot r = 2.00\ \text{m/s} \\
v_\theta &= r\dot\theta = 4 \times 1.6 = 6.40\ \text{m/s} \\
v &= \sqrt{2^2 + 6.4^2} = 6.71\ \text{m/s}
\end{aligned}
$$

Angle from the radial line: $\tan\phi = v_\theta / v_r = 6.4/2$, so $\phi = 72.6^\circ$.

#### Acceleration

$$
\begin{aligned}
a_r &= \ddot r - r\dot\theta^2 = 1 - 4(1.6)^2 = -9.24\ \text{m/s}^2 \\
a_\theta &= r\ddot\theta + 2\dot r\dot\theta = 4(0.8) + 2(2)(1.6) = 9.60\ \text{m/s}^2 \\
a &= \sqrt{(-9.24)^2 + 9.6^2} = 13.32\ \text{m/s}^2
\end{aligned}
$$

Since $a_r$ is negative and $a_\theta$ positive, the acceleration lies in the second quadrant of the $(e_r, e_\theta)$ axes, at $133.9^\circ$ measured from the $+e_r$ direction toward $+e_\theta$ (that is, $46.1^\circ$ from the $-e_r$ direction).

**Answer: v = 6.71 m/s at 72.6° to the radial line; a = 13.32 m/s² at 133.9° to the radial line.**

## Check yourself

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

1. At t = 2 s, with r = 2 + 0.5t^2 and theta = 0.4t^2, what are r and theta' ?
   A. r = 4 m, theta' = 1.6 rad/s
   B. r = 3 m, theta' = 0.8 rad/s
   C. r = 4 m, theta' = 0.8 rad/s
   D. r = 2 m, theta' = 1.6 rad/s
   Answer: A. r = 2 + 0.5(4) = 4 m and theta' = 0.8t = 1.6 rad/s. (Key point: Derivatives)

2. What is the speed at t = 2 s?
   A. 2.0 m/s
   B. 6.40 m/s
   C. 6.71 m/s
   D. 8.40 m/s
   Answer: C. v_r = 2 and v_theta = r theta' = 6.4, so v = sqrt(4 + 40.96) = 6.71 m/s. (Key point: Velocity)

3. The radial acceleration component is found from:
   A. r'' + r theta'^2 = 11 m/s^2
   B. r'' - r theta'^2 = -9.24 m/s^2
   C. r theta'' = 3.2 m/s^2
   D. r'' - theta'^2 = -1.56 m/s^2
   Answer: B. a_r = r'' - r theta'^2 = 1 - 4(1.6)^2 = -9.24 m/s^2. (Key point: Radial acceleration)

4. With a_r = -9.24 and a_theta = 9.60, where does the acceleration point?
   A. In the first quadrant of the (e_r, e_theta) axes
   B. In the second quadrant, 133.9 deg from +e_r toward +e_theta
   C. Along -e_theta
   D. Along +e_r
   Answer: B. Negative radial and positive transverse components put it in the second quadrant, at about 133.9 deg from +e_r. (Key point: Acceleration direction)

## Related in Engineering Mechanics

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