---
title: "Angular momentum rate; 5 kg particle under F"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:angular-momentum-moment-force-5kg-particle"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/angular-momentum-moment-force-5kg-particle
site: BCT Engineering
---

# Angular momentum rate; 5 kg particle under F

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2072 Kartik, 4+6 marks
- Video: [Sect 15.6: Relationship between Moment of Force and Angular Momentum - Engineering dynamics](https://www.youtube.com/watch?v=TZs9JbEvwRQ) — Matthew Pusko
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/angular-momentum-moment-force-5kg-particle
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

Show that, "rate of change of angular momentum about a point is equal to moment of the force about the same point." The resultant external force acting on a 5 kg particle in space is F = (12t i − 24t² j + 40t³ k) N, where t is in seconds. The particle is initially at rest at origin. Determine the x component of acceleration, velocity and position at the instant of 5 sec.

## Short answer

Differentiating angular momentum H_O = r×m v gives dH_O/dt = v×m v + r×m a = r×F, so rate of change equals moment of force. For x-components: a_x = F_x/m = 2.4t, at t=5 s a_x=12 m/s²; velocity v_x = 1.2t² = 30 m/s; position x = 0.4t³ = 50 m.

## Answer

#### Proof: rate of change of angular momentum = moment of the force
Let a particle of mass m have position vector r from a fixed point O and velocity v = dr/dt. Its angular momentum about O is

H_O = r × m v

Differentiate with respect to time (product rule):

dH_O/dt = (dr/dt) × m v + r × m (dv/dt) = (v × m v) + r × m a

- v × m v = 0, because the cross product of a vector with itself is zero.
- m a = F (Newton's second law), where F is the resultant force.

Hence **dH_O/dt = r × F = M_O**: the rate of change of angular momentum about O equals the moment of the resultant force about the same point O.

#### Numerical part (m = 5 kg, F = 12t i − 24t² j + 40t³ k N, at rest at the origin)
- x-component of acceleration: aₓ = Fₓ/m = 12t/5 = 2.4t, so at t = 5 s: **aₓ = 12 m/s²**
- Velocity: vₓ = ∫2.4t dt = 1.2t², so at t = 5 s: **vₓ = 1.2(25) = 30 m/s**
- Position: x = ∫1.2t² dt = 0.4t³, so at t = 5 s: **x = 0.4(125) = 50 m**

**Answer: aₓ = 12 m/s², vₓ = 30 m/s, x = 50 m at t = 5 s.**

## Check yourself

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

1. In the proof, differentiating H_O = r x m v gives (v x m v) + (r x m a). Why is the first term zero?
   A. Because v is constant
   B. Because the cross product of a vector with itself is zero
   C. Because r is zero
   D. Because m is small
   Answer: B. v x m v = m(v x v) = 0, since parallel vectors have zero cross product. (Key point: Differentiating H_O)

2. What does the remaining term r x m a become, and what is the conclusion?
   A. r x F = M_O, so dH_O/dt equals the moment of the force about O
   B. r . F, a work term
   C. F / r, a torque per length
   D. m r, the first moment of mass
   Answer: A. Since m a = F, dH_O/dt = r x F = M_O. (Key point: Result of the proof)

3. For a_x = F_x/m = 2.4t from rest at the origin, which sequence gives x at t = 5 s?
   A. Differentiate twice to get x = 2.4
   B. Integrate once for v_x = 1.2t^2 (30 m/s), again for x = 0.4t^3 (50 m)
   C. x = a_x t^2/2 = 12 x 25/2 = 150 m
   D. x = v_x t = 30 x 5 = 150 m
   Answer: B. Acceleration varies with time, so integrate: v_x = 1.2t^2 and x = 0.4t^3. The constant-acceleration formulas do not apply. (Key point: Integrating for x-motion)

## Related in Applied Mechanics

- [Particle under F = 12t i − 24t² j − 40t³ k](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aforce-particle-ax-vy-impulse-momentum) — Applied Mechanics, chapter 9
- [Angular momentum; rate of change equals moment](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aangular-momentum-rate-equals-moment-proof) — Applied Mechanics, chapter 9
- [Linear and angular momentum of particle system](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Asystem-of-particles-linear-angular-momentum) — Applied Mechanics, chapter 9