---
title: "Dependent motion; radius of curvature of particle path"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:dependent-motion-particle-ax-ay"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/dependent-motion-particle-ax-ay
site: BCT Engineering
---

# Dependent motion; radius of curvature of particle path

Applied Mechanics · Chapter 8: Kinematics of Particles and Rigid Body

- Asked in: 2072 Kartik, 3+7 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/dependent-motion-particle-ax-ay
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

What do you mean by dependent motion of particles? Illustrate it with suitable example. A particle starting from origin is subjected to acceleration such that ax = −2 m/sec² and ay = −5 m/sec². The initial velocity is 60 m/sec directed at a slope of 30° w.r.t. horizontal. Compute the radius of curvature at the end of 3 sec. Also determine its position at the end of 3 sec.

## Short answer

Dependent motion: e.g., blocks connected by a cord with constraint x_A + 2y_B = L, so v_A = -2v_B. For the particle, at t=3 s: v_x=45.96 m/s, v_y=15 m/s, a=5.385 m/s², a_n=4.133 m/s², ρ=v²/a_n≈566 m; position (146.9 m, 67.5 m).

## Answer

#### Dependent motion of particles
The motion of one particle is dependent on that of another when the two are connected by an inextensible cord (or rigid link) so that their positions are related by a constraint equation.

Example: block A on a horizontal table is tied by a cord that passes over a fixed pulley to a block B hanging below a movable pulley. The cord length is constant: x_A + 2y_B = L (x_A measured from the fixed pulley, y_B downward). Differentiating: v_A = −2v_B and a_A = −2a_B, so if B goes down at 1 m/s, A moves at 2 m/s toward the pulley. One coordinate fixes the other.

#### Curved motion with constant acceleration
Data: a_x = −2 m/s², a_y = −5 m/s², u = 60 m/s at 30°: u_x = 60 cos 30° = 51.96 m/s, u_y = 60 sin 30° = 30 m/s. At t = 3 s:

- v_x = u_x + a_x t = 51.96 − 2(3) = **45.96 m/s**
- v_y = u_y + a_y t = 30 − 5(3) = **15.00 m/s**
- Speed v = √(45.96² + 15²) = 48.35 m/s, direction 18.1° above the x-axis.
- Total acceleration a = √(2² + 5²) = 5.385 m/s².
- Tangential component: a_t = (a_x v_x + a_y v_y)/v = (−2 × 45.96 − 5 × 15)/48.35 = −3.453 m/s²
- Normal component: a_n = √(a² − a_t²) = √(29 − 11.92) = 4.133 m/s²
- Radius of curvature: ρ = v²/a_n = 48.35²/4.133 = **565.6 m ≈ 566 m**

#### Position at t = 3 s
- x = u_x t + ½a_x t² = 51.96(3) − ½(2)(9) = 155.88 − 9 = **146.88 m**
- y = u_y t + ½a_y t² = 30(3) − ½(5)(9) = 90 − 22.5 = **67.50 m**

**Answer: ρ ≈ 566 m; position (146.9 m, 67.5 m) from the origin.**

## Check yourself

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

1. For block A on a table linked over a fixed pulley to B under a movable pulley, if B goes down 1 m/s, what does A do?
   A. Moves 0.5 m/s toward the pulley
   B. Moves 1 m/s toward the pulley
   C. Moves 2 m/s toward the pulley
   D. Stays still
   Answer: C. x_A + 2y_B = L gives v_A = -2v_B. (Key point: Constraint equation)

2. With u_x = 51.96, u_y = 30, a_x = -2, a_y = -5, what are v_x and v_y at t = 3 s?
   A. 45.96 and 15.00 m/s
   B. 57.96 and 45 m/s
   C. 51.96 and 30 m/s
   D. 45.96 and 25 m/s
   Answer: A. v = u + at in each direction: 51.96 - 6 = 45.96 and 30 - 15 = 15. (Key point: Velocity components at 3 s)

3. How is ρ obtained?
   A. ρ = v/a
   B. ρ = v^2/a_n where a_n = sqrt(a^2 - a_t^2)
   C. ρ = v^2/a_t
   D. ρ = a^2/v
   Answer: B. Curvature depends only on the normal acceleration, found by removing the tangential part a_t = (a·v)/v from the total acceleration. (Key point: Finding radius of curvature)

4. What is the x coordinate at t = 3 s?
   A. 155.88 m
   B. 146.88 m
   C. 164.88 m
   D. 137.88 m
   Answer: B. x = u_x t + 1/2 a_x t^2 = 155.88 - 9 = 146.88 m. (Key point: Position at 3 s)

## Related in Applied Mechanics

- [Nozzle stream curvature; dependent motion](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Anozzle-curvature-dependent-motion) — Applied Mechanics, chapter 8
- [Vibrating particle velocity, acceleration and path](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Avibrating-particle-x-100-sin-pi-t) — Applied Mechanics, chapter 8
- [Normal, tangential acceleration; radius of curvature](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aposition-vector-normal-tangential-radius-curvature) — Applied Mechanics, chapter 8
- [Normal-tangential acceleration; nozzle stream curvature](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Anormal-tangential-acceleration-nozzle) — Applied Mechanics, chapter 8
- [Ball tossed upward from window; dependent motion](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aball-tossed-vertical-window-dependent-motion) — Applied Mechanics, chapter 8
- [Particle with a = v³; projectile motion equations](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aacceleration-a-v3-displacement-projectile-equations) — Applied Mechanics, chapter 8

## Related in other subjects

- [Radius of curvature of x³+y³=3axy at origin](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Acurvature-origin-folium) — Mathematics I, chapter 1, asked 3×
- [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×
- [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
- [Velocity and acceleration of a particle](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Avelocity-and-acceleration-of-a-particle) — Mathematics III, chapter 2