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

# Normal-tangential acceleration; nozzle stream curvature

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

- Asked in: 2074 Chaitra, 4+6 marks
- Video: [Curvilinear Motion: Normal and Tangential components (Learn to solve any problem)](https://www.youtube.com/watch?v=hwHy4Dewc7g) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/normal-tangential-acceleration-nozzle
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

Derive the relationship for normal and tangential components of acceleration for a particle moving in a curvilinear path. A nozzle discharges a stream of water in the direction as shown in the figure below with an initial velocity of 25 m/s. Determine the radius of curvature of the stream (i) as it leaves the nozzle (ii) at the maximum height of the stream. [Figure: nozzle at A; stream leaves at 25 m/s along a slope of 3 vertical to 4 horizontal.]

## Short answer

For curvilinear motion, tangential acceleration a_t = dv/dt and normal acceleration a_n = v²/ρ, where ρ is the radius of curvature. At the nozzle, a_n = g cosθ = 7.848 m/s², so ρ = v²/a_n = 79.6 m. At maximum height, v=20 m/s, a_n=g=9.81 m/s², ρ=40.8 m.

## Answer

#### Normal and tangential components of acceleration
Let a particle move on a curved path with speed v. Its velocity is always tangent to the path: **v = v e_t**, where e_t is the unit tangent vector.

Differentiating, a = dv/dt = (dv/dt) e_t + v (de_t/dt).

Over a small time dt the tangent turns through dθ, and de_t = dθ e_n (e_n is the unit normal pointing to the centre of curvature). If ρ is the radius of curvature, ds = ρ dθ = v dt, so dθ/dt = v/ρ. Therefore de_t/dt = (v/ρ) e_n and

**a = a_t e_t + a_n e_n, with a_t = dv/dt and a_n = v²/ρ**

- a_t changes the magnitude of the velocity (tangential component).
- a_n changes its direction and always points to the centre of curvature (normal component).
- Magnitude: a = √(a_t² + a_n²).

#### Nozzle
The stream leaves A at v = 25 m/s along a slope 3 : 4, so θ = tan⁻¹(3/4) = 36.87° above the horizontal (cos θ = 0.8, sin θ = 0.6). The only acceleration is g = 9.81 m/s² downward.

**(i) At the nozzle:** the normal component of g is a_n = g cos θ = 9.81 × 0.8 = 7.848 m/s².
ρ = v²/a_n = 25²/7.848 = **79.6 m**

**(ii) At the maximum height:** the velocity is horizontal, v = v_x = 25 cos θ = 20 m/s, and g is entirely normal, a_n = 9.81 m/s².
ρ = v²/a_n = 20²/9.81 = **40.8 m**

## Check yourself

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

1. In the derivation, why is a_n = v^2/ρ?
   A. Because de_t = dθ e_n and dθ/dt = v/ρ (from ds = ρ dθ = v dt)
   B. Because a_n = dv/dt
   C. Because ρ = v t
   D. Because e_n is along the tangent
   Answer: A. v times de_t/dt = v (dθ/dt) e_n = (v^2/ρ) e_n. (Key point: Origin of normal component)

2. What do a_t and a_n change?
   A. a_t changes direction, a_n changes speed
   B. Both change speed only
   C. a_t changes speed, a_n changes direction and points to the centre of curvature
   D. Both change direction only
   Answer: C. a_t = dv/dt changes the magnitude of velocity; a_n changes its direction. (Key point: Role of components)

3. At the nozzle (θ = 36.87 deg, v = 25 m/s), what is ρ?
   A. 25^2/9.81 = 63.7 m
   B. 25^2/7.848 = 79.6 m
   C. 25^2/5.886 = 106 m
   D. 20^2/9.81 = 40.8 m
   Answer: B. Only the component of g normal to the path counts: a_n = g cos θ = 7.848, so ρ = 625/7.848 = 79.6 m. (Key point: Radius at the nozzle)

4. At maximum height, why is ρ = 40.8 m?
   A. Velocity is horizontal (20 m/s) and all of g is normal, so ρ = 20^2/9.81
   B. Velocity is zero
   C. Speed is 25 m/s and a_n = g cos θ
   D. a_t = g at the top
   Answer: A. At the top v = v_x = 25 cos θ = 20 m/s and g is perpendicular to it, so a_n = 9.81. (Key point: Radius at top)

## Related in Applied Mechanics

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