---
title: "Normal, tangential acceleration; radius of curvature"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:position-vector-normal-tangential-radius-curvature"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/position-vector-normal-tangential-radius-curvature
site: BCT Engineering
---

# Normal, tangential acceleration; radius of curvature

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

- Asked in: 2067 Asar, 10 marks
- Video: [Mathematical Tools | Lecture 9 | radius of curvature and normal acceleration](https://www.youtube.com/watch?v=zdPt7qlQn3Y) — Physics Educator
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/position-vector-normal-tangential-radius-curvature
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

The motion of a particle is defined by the position vector r = 3t²i + 4t³j + 5t⁴k where r is in meter and t is in second. Find the normal and tangential component of acceleration and the principal radius of curvature at the instant when t = 4 secs.

## Short answer

At t=4 s, v = 1294.54 m/s, a = 964.81 m/s². Tangential acceleration a_t = a·v/v = 963.6 m/s². Normal acceleration a_n = √(a²−a_t²) = 48.9 m/s². Radius of curvature ρ = v²/a_n = 3.43×10⁴ m.

## Answer

#### Velocity and acceleration vectors
r = 3t² i + 4t³ j + 5t⁴ k
- v = dr/dt = 6t i + 12t² j + 20t³ k
- a = dv/dt = 6 i + 24t j + 60t² k

At t = 4 s:
- v = 24 i + 192 j + 1280 k (m/s), so v = √(24² + 192² + 1280²) = 1294.54 m/s
- a = 6 i + 96 j + 960 k (m/s²), so a = √(6² + 96² + 960²) = 964.81 m/s²

#### Tangential component
a_t = a·v/v = (6 × 24 + 96 × 192 + 960 × 1280)/1294.54 = (144 + 18432 + 1228800)/1294.54 = **963.57 m/s²**

#### Normal component
a_n = √(a² − a_t²) = √(964.81² − 963.57²) = **48.93 m/s²**

#### Radius of curvature
ρ = v²/a_n = 1294.54²/48.93 = **34 250 m ≈ 34.25 km**

**Answer: a_t = 963.6 m/s², a_n = 48.9 m/s², ρ ≈ 3.43 × 10⁴ m.**

## Check yourself

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

1. How are velocity and acceleration obtained from r = 3t^2 i + 4t^3 j + 5t^4 k?
   A. Integrate r once for v and again for a
   B. Differentiate r once for v and differentiate v for a
   C. Differentiate r twice for v and three times for a
   D. Multiply r by t for v and by t^2 for a
   Answer: B. v = dr/dt and a = dv/dt, giving v = 6t i + 12t^2 j + 20t^3 k and a = 6 i + 24t j + 60t^2 k. (Key point: Differentiate position for v and a)

2. How is the tangential component of acceleration found from the vectors a and v?
   A. a_t = a x v / v
   B. a_t = a · v / v (projection of a on the velocity direction)
   C. a_t = |a| - |v|
   D. a_t = v^2 / a
   Answer: B. The tangential component is the projection of a along v, so a_t = (a·v)/v, using the dot product. (Key point: Tangential component formula)

3. Once a and a_t are known, how is the normal component a_n obtained?
   A. a_n = a + a_t
   B. a_n = a_t / a
   C. a_n = sqrt(a^2 - a_t^2)
   D. a_n = a x a_t
   Answer: C. a_t and a_n are perpendicular components of a, so a^2 = a_t^2 + a_n^2 and a_n = sqrt(a^2 - a_t^2). (Key point: Normal component from a and a_t)

4. Which relation gives the radius of curvature at t = 4 s (a_n = 48.93 m/s^2)?
   A. rho = v^2 / a_n
   B. rho = a_n / v^2
   C. rho = v / a_t
   D. rho = a_t^2 / a_n
   Answer: A. The normal acceleration is a_n = v^2/rho, so rho = v^2/a_n = 1294.54^2/48.93, about 34.25 km. (Key point: Radius of curvature)

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