---
title: "Cord pulled on disk, angular acceleration and cord"
subject: "Applied Mechanics (Dynamics)"
chapter: 5
key: "semester_3/applied_mechanics_dynamics:cord-disk-m15-t180-two-parts"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/cord-disk-m15-t180-two-parts
site: BCT Engineering
---

# Cord pulled on disk, angular acceleration and cord

Applied Mechanics (Dynamics) · Chapter 5: Plane Motion of Rigid Bodies: Forces, Moments, and Accelerations

- Asked in: 2081 Bhadra, 5 marks
- Video: [Motion Developed During Pulling of a Cord Wrapped around a Disk](https://www.youtube.com/watch?v=YauaqYYY4xU) — Indra Lal Subedi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/cord-disk-m15-t180-two-parts
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-5

## Question

A cord is wrapped around a homogeneous disk with a radius of $r = 0.5$ m and a mass $m = 15$ kg. If the cord is pulled upward with a force $T$ of magnitude 180 N, determine, (i) The angular acceleration of the disk. (ii) The acceleration of the cord. [Figure: disk with centre G, radius 0.5 m, cord wrapped on the rim and pulled up at A with force T.]

## Short answer

With Tr = Iα and I = ½mr², α = 180×0.5/1.875 = 48.00 rad/s² CCW. Disk centre accelerates at a = (180−147.15)/15 = 2.19 m/s² up, so cord accelerates at a + αr = 26.19 m/s² upward.

## Answer

Take upward as positive for translation. The disk is homogeneous, so $I = \tfrac12 mr^2$. The forces on the disk are its weight $W=mg$ (down, at $G$) and the cord tension $T$ (up, at the rim, a distance $r$ from $G$). The cord leaves the rim tangentially, so $T$ has moment $Tr$ about $G$.

```text
        T ^
          | A
      .---+---.
     /    G    \
     \         /
      '-------'
          | W
          v
```

#### Data
$m=15$ kg, $r=0.5$ m, $T=180$ N, $g = 9.81$ m/s²

$$
W = mg = 15\times9.81 = 147.15\ \text{N}, \qquad I = \tfrac12 mr^2 = \tfrac12(15)(0.5)^2 = 1.875\ \text{kg m}^2
$$

#### Equations of motion
$$
\begin{aligned}
\Sigma F_y = m\bar a &: \quad T - W = m\bar a\\
\Sigma M_G = I\alpha &: \quad T r = I\alpha
\end{aligned}
$$

#### Angular acceleration
$$
\alpha = \frac{Tr}{I} = \frac{180\times0.5}{1.875} = 48.00\ \text{rad/s}^2\ \text{(counter-clockwise)}
$$

#### Acceleration of the cord
First, the centre: $\bar a = (T-W)/m = (180 - 147.15)/15 = 2.19$ m/s² upward. Cord acceleration = acceleration of rim point $A$:

$$
a_{cord} = \bar a + \alpha r = 2.19 + 48.00\times0.5 = 26.19\ \text{m/s}^2\ \uparrow
$$

**Answer: (i) $\alpha = 48.00$ rad/s² counter-clockwise; (ii) $a_{cord} = 26.19$ m/s² upward.**

## Check yourself

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

1. Which equation gives the angular acceleration of the disk?
   A. T - W = I alpha
   B. T r = I alpha about G
   C. W r = I alpha
   D. T r = m alpha
   Answer: B. Only T has a moment about G; sum M_G = I alpha gives T r = I alpha. (Key point: Moment equation)

2. With T = 180 N, r = 0.5 m and I = 1.875 kg m^2, alpha is:
   A. 96 rad/s^2
   B. 24 rad/s^2
   C. 48 rad/s^2
   D. 12 rad/s^2
   Answer: C. alpha = 180 x 0.5 / 1.875 = 48 rad/s^2, counter-clockwise. (Key point: Value of alpha)

3. The cord's acceleration equals:
   A. alpha r alone
   B. The acceleration of G plus alpha r of the rim point
   C. g
   D. The acceleration of G minus g
   Answer: B. The cord moves with rim point A, so a = a_G + alpha r. (Key point: Cord acceleration method)

4. With a_G = 2.19 m/s^2 up and alpha r = 24 m/s^2, the cord acceleration is:
   A. 21.81 m/s^2
   B. 24 m/s^2
   C. 26.19 m/s^2
   D. 48 m/s^2
   Answer: C. 2.19 + 24 = 26.19 m/s^2 upward. (Key point: Cord acceleration value)

## Close variants set in other exams

- [Cord pulled on disk, m = 15 kg, T = 180 N](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acord-disk-m15-t180-abc) — Applied Mechanics (Dynamics), chapter 5, asked 5×
- [Cord pulled on disk, m = 20 kg, F = 250 N](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acord-disk-m20-f250) — Applied Mechanics (Dynamics), chapter 5, asked 4×
- [Cord pulled on disk, m = 30 kg, T = 200 N](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acord-disk-m30-t200) — Applied Mechanics (Dynamics), chapter 5, asked 4×

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