---
title: "Pendulum bob velocity and acceleration; momentum"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:pendulum-bob-linear-angular-momentum-tension"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/pendulum-bob-linear-angular-momentum-tension
site: BCT Engineering
---

# Pendulum bob velocity and acceleration; momentum

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

- Asked in: 2072 Chaitra, 2+8 marks; 2069 Chaitra, 2+8 marks (asked 2 times)
- Video: [Velocity and Acceleration of Pendulum](https://www.youtube.com/watch?v=NN_2MckMSm4) — Indra Lal Subedi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/pendulum-bob-linear-angular-momentum-tension
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

Define the linear momentum and angular momentum. Find the velocity and the acceleration of the bob in the given position. The bob of a 2 m pendulum describes an arc of a circle in a vertical plane, which is shown in figure below. If the tension in the cord is 2.5 times the weight of the bob for the position shown. [Figure: pendulum with pivot O, cord length 2 m, cord at 30° to the vertical, bob B.]

## Short answer

Linear momentum = m v; angular momentum about O = r × m v. For the bob: T − W cos30° = m v²/L gives v = 5.66 m/s. Tangential acceleration = g sin30° = 4.91 m/s², normal = v²/L = 16.03 m/s²; total acceleration = 16.76 m/s².

## Answer

#### Linear and angular momentum
- **Linear momentum** of a particle of mass m and velocity v: L = m v (kg·m/s), a vector along the velocity.
- **Angular momentum** about a point O: H_O = r × m v, with magnitude m v r sin φ (kg·m²/s); for motion in a circle of radius r, H_O = m v r.

#### Pendulum bob
The bob moves on a circle of radius L = 2 m. At the position shown the cord makes θ = 30° with the vertical and the tension is T = 2.5W. Let W = mg.

```text
        O
        |\
        | \ 2 m
        |30\
        |   \ T
        |    B   (mass m, weight W)
```

**Normal (toward O) direction:** T − W cos 30° = m v²/L
2.5W − 0.866W = (W/g) v²/2
1.634 g × 2 = v², so v² = 1.634 × 9.81 × 2 = 32.06
**v = 5.66 m/s** (along the tangent to the circle, perpendicular to the cord)

**Tangential direction:** W sin 30° = m a_t, so a_t = g sin 30° = 4.905 m/s²

**Normal acceleration:** a_n = v²/L = 32.06/2 = 16.03 m/s² toward O

**Total acceleration:** a = √(16.03² + 4.905²) = **16.76 m/s²**, at tan⁻¹(4.905/16.03) = 17.0° to the cord.

**Answer: v = 5.66 m/s; a = 16.76 m/s².**

## Check yourself

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

1. Which correctly defines the angular momentum of a particle about point O?
   A. m v, the mass times velocity
   B. r x m v, the moment of linear momentum about O
   C. m a r, mass times acceleration times distance
   D. F x r, force times distance
   Answer: B. H_O = r x m v, the moment of the linear momentum about O (m v r for circular motion). (Key point: Angular momentum definition)

2. At the 30 degree position, which equation along the cord (toward O) gives the speed?
   A. T + W cos 30 = m v^2/L
   B. T - W sin 30 = m v^2/L
   C. T - W cos 30 = m v^2/L
   D. T = m v^2/L only
   Answer: C. Toward O the net force is T minus the weight component along the cord, W cos 30, and it equals m v^2/L. (Key point: Normal-direction equation)

3. What produces the tangential acceleration of the bob, and what is its value?
   A. Tension T, so a_t = 2.5g
   B. The weight component W sin 30, so a_t = g sin 30 = 4.905 m/s^2
   C. The weight component W cos 30, so a_t = g cos 30
   D. It is zero because the cord is taut
   Answer: B. Tension acts along the cord, so only the weight component W sin 30 acts along the tangent: a_t = g sin 30 = 4.905 m/s^2. (Key point: Tangential acceleration)

4. How is the total acceleration of the bob found once a_n = 16.03 and a_t = 4.905 m/s^2 are known?
   A. a = a_n + a_t = 20.9 m/s^2
   B. a = a_n - a_t
   C. a = sqrt(a_n^2 + a_t^2) = 16.76 m/s^2
   D. a = a_n only, since the bob is on a circle
   Answer: C. The two components are perpendicular, so they combine by Pythagoras: a = 16.76 m/s^2. (Key point: Total acceleration)

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