---
title: "Angular velocity from rod data"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:velocity-acceleration-rod-angle"
source: https://www.bctengineer.com/q/semester_4/numerical_method/velocity-acceleration-rod-angle
site: BCT Engineering
---

# Angular velocity from rod data

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2068 Bhadra, 4 marks; 2068 Magh, 4 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/velocity-acceleration-rod-angle
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

The following table gives the angle in radians (θ) through which a rotating rod has turned for various values of time in seconds (t). Find the angular velocity and angular acceleration at t = 0.2.
| t | 0 | 0.2 | 0.4 | 0.6 | 0.8 |
|---|---|---|---|---|---|
| θ | 0 | 0.122 | 0.493 | 0.123 | 2.022 |

## Short answer

Assuming a corrected θ(0.6)=1.123, use Newton's forward formula with p=1. Differentiating yields angular velocity ω(0.2) ≈ 1.224 rad/s and angular acceleration α(0.2) ≈ 6.225 rad/s².

## Answer

**Data note.** The printed θ at t = 0.6 is 0.123, which breaks the smooth trend (0, 0.122, 0.493, …, 2.022; the second differences are almost constant only if θ(0.6) = 1.123). It is taken as a misprint of 1.123.

The time steps are equal (h = 0.2) and t = 0.2 is the second tabulated point, so Newton's forward formula with t₀ = 0 and p = 1 is used. Angular velocity ω = dθ/dt and angular acceleration α = d²θ/dt².

#### Difference table

| x | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 0 | 0.000 | 0.122 | 0.249 | 0.010 | 0.000 |
| 0.2 | 0.122 | 0.371 | 0.259 | 0.010 |  |
| 0.4 | 0.493 | 0.630 | 0.269 |  |  |
| 0.6 | 1.123 | 0.899 |  |  |  |
| 0.8 | 2.022 |  |  |  |  |

Take t₀ = 0, h = 0.2, so p = (0.2 − 0)/0.2 = 1.00. Differentiating Newton's forward formula with respect to t gives

```text
dθ/dt  = (1/h)[Δθ₀ + (2p−1)/2·Δ²θ₀ + (3p²−6p+2)/6·Δ³θ₀ + (4p³−18p²+22p−6)/24·Δ⁴θ₀]
d²θ/dt² = (1/h²)[Δ²θ₀ + (p−1)Δ³θ₀ + (6p²−18p+11)/12·Δ⁴θ₀]
```

#### First derivative

| Term | Coefficient | Value of difference | Product |
|---|---|---|---|
| Δθ₀ | 1.0000 | 0.122 | 0.1220 |
| Δ²θ₀ | 0.5000 | 0.249 | 0.1245 |
| Δ³θ₀ | -0.1667 | 0.010 | -0.0017 |
| Δ⁴θ₀ | 0.0833 | 0.000 | 0.0000 |

Sum = 0.2448, so dθ/dt = 0.2448/0.2 = 1.2242

#### Second derivative

| Term | Coefficient | Value of difference | Product |
|---|---|---|---|
| Δ²θ₀ | 1.0000 | 0.249 | 0.2490 |
| Δ³θ₀ | 0.0000 | 0.010 | 0.0000 |
| Δ⁴θ₀ | -0.0833 | 0.000 | 0.0000 |

Sum = 0.2490, so d²θ/dt² = 0.2490/0.2² = 6.2250

**Answer: ω(0.2) ≈ 1.224 rad/s, α(0.2) ≈ 6.225 rad/s²**

(With the printed 0.123 taken literally the same method gives ω ≈ 3.724 rad/s and α ≈ -2.108 rad/s², which is not physically consistent with the rest of the table.)

## Check yourself

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

1. What do ω and α represent?
   A. Both are θ
   B. ω = θ/t, α = ω/t
   C. ω = dθ/dt, α = d²θ/dt²
   D. ω = d²θ/dt², α = dθ/dt
   Answer: C. Velocity is first, acceleration second derivative. (Key point: Meaning)

2. Why use Newton's forward derivative formula at t = 0.2?
   A. t is the end
   B. Equal steps, with t = 0.2 the second point and p = 1
   C. Unequal steps
   D. Needs integration
   Answer: B. t0 = 0, h = 0.2, p = 1. (Key point: Method)

3. Why is θ(0.6) = 0.123 treated as a misprint?
   A. h varies
   B. t is wrong
   C. Second differences are nearly constant only if it is 1.123
   D. It's negative
   Answer: C. 1.123 gives smooth differences. (Key point: Data note)

## Related in Numerical Method

- [Velocity, acceleration at 0.2 s](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Avelocity-acceleration-distance-0-2) — Numerical Method, chapter 5
- [Velocity and acceleration at t=3](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Avelocity-acceleration-distance-time) — Numerical Method, chapter 5
- [Velocity, acceleration at t=0.1](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Avelocity-acceleration-position-time) — Numerical Method, chapter 5
- [Difference table (3.0 to 3.8)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Adifference-table-3-0) — Numerical Method, chapter 4
- [Newton interpolation: steam table](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Anewton-steam-table) — Numerical Method, chapter 4
- [Forward difference table](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aforward-difference-table-1-6) — Numerical Method, chapter 4

## Related in other subjects

- [Collar sliding on rotating arm, acceleration](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acollar-on-rotating-arm-acceleration) — Applied Mechanics (Dynamics), chapter 1, asked 2×
- [Collar on arm: velocity and relative acceleration](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acollar-on-rotating-arm-velocity-relative) — Applied Mechanics (Dynamics), chapter 1, asked 2×
- [Rod and spring, angular velocity at vertical](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Arod-spring-pivot-15kg-explain-impulse) — Applied Mechanics (Dynamics), chapter 6
- [Rod and spring, angular velocity and reaction](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Arod-spring-pivot-15kg-2-5m-reaction) — Applied Mechanics (Dynamics), chapter 6