---
title: "Pendulum bob with given tension"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:pendulum-bob-tension-velocity"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/pendulum-bob-tension-velocity
site: BCT Engineering
---

# Pendulum bob with given tension

Applied Mechanics (Dynamics) · Chapter 1: Curvilinear Motion of Particles

- Asked in: 2074 Asoj, 4 marks
- Video: [Principle of Work and Energy Pendulum Example - Engineering Dynamics](https://www.youtube.com/watch?v=3TpKlv-8p_0) — structurefree
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/pendulum-bob-tension-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

The bob of a 2 m pendulum describes an arc of circle in a vertical plane. If the tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity and acceleration of the bob in the given position. [Figure: pendulum of length 2 m, cord at $40^\circ$ to the vertical.]

## Short answer

From radial equilibrium T - W cosθ = (W/g) v²/L, with T = 2.5W, find v = 5.83 m/s. Normal acceleration is 17.01 m/s², tangential is 6.31 m/s², total acceleration 18.14 m/s².

## Answer

**Assumption (figure not shown):** the cord makes $\theta=40^\circ$ with the vertical, and the tension is $T=2.5W$. The bob moves on a circle of radius $L=2$ m, so use normal and tangential components.

```text
          O
          |\
          | \  L = 2 m
          |40\
          |   \
                O  bob (W)
```

#### Normal direction (towards O)
$$
T-W\cos\theta=\frac{W}{g}\frac{v^{2}}{L}
$$

$$
v^{2}=gL\,(2.5-\cos40^\circ)=9.81(2)(2.5-0.7660)=34.02\ \text{m}^2/\text{s}^2
$$

$$
v=5.833\ \text{m/s}
$$

#### Accelerations
$$
a_n=\frac{v^{2}}{L}=\frac{34.02}{2}=17.01\ \text{m/s}^2
$$

Tangential direction (only the weight component acts):

$$
W\sin\theta=\frac{W}{g}a_t\ \Rightarrow\ a_t=g\sin40^\circ=6.306\ \text{m/s}^2
$$

$$
a=\sqrt{a_n^{2}+a_t^{2}}=\sqrt{17.01^{2}+6.306^{2}}=18.14\ \text{m/s}^2
$$

The acceleration makes $\tan^{-1}(6.306/17.01)=20.3^\circ$ with the cord.

**Answer: $v=5.83$ m/s; $a=18.14$ m/s$^2$ ($a_n=17.01$, $a_t=6.31$ m/s$^2$).**

## Check yourself

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

1. Which equation gives the speed of the bob?
   A. T - W cos(theta) = (W/g)(v^2/L)
   B. T + W cos(theta) = (W/g)(v^2/L)
   C. T - W sin(theta) = (W/g)(v^2/L)
   D. T = W v^2 / L
   Answer: A. Along the cord towards O, net force T - W cos(theta) supplies the centripetal force. (Key point: Normal equation)

2. With T = 2.5W, theta = 40 degrees and L = 2 m, what is v?
   A. 5.83 m/s
   B. 4.43 m/s
   C. 17.0 m/s
   D. 6.31 m/s
   Answer: A. v^2 = gL(2.5 - cos40) = 9.81(2)(1.734) = 34.0, so v = 5.83 m/s. (Key point: Speed value)

3. What is the tangential acceleration?
   A. g cos(40) = 7.51 m/s^2
   B. g sin(40) = 6.31 m/s^2
   C. v^2/L = 17.0 m/s^2
   D. Zero, since speed is constant
   Answer: B. Only the weight component W sin(theta) acts along the tangent. (Key point: Tangential acceleration)

4. How is the total acceleration found?
   A. a = a_n + a_t = 23.3
   B. a = sqrt(a_n^2 + a_t^2) = 18.14 m/s^2
   C. a = a_n = 17.01 m/s^2
   D. a = a_t = 6.31 m/s^2
   Answer: B. The two components are perpendicular, so they combine by Pythagoras. (Key point: Total acceleration)

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