---
title: "Cord pulled on disk, m = 15 kg, T = 180 N"
subject: "Applied Mechanics (Dynamics)"
chapter: 5
key: "semester_3/applied_mechanics_dynamics:cord-disk-m15-t180-abc"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/cord-disk-m15-t180-abc
site: BCT Engineering
---

# Cord pulled on disk, m = 15 kg, T = 180 N

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

- Asked in: 2074 Asoj, 4 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-abc
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-5

## Question

A cord is wrapped around a homogeneous disk of radius $r = 0.5$ m and mass $m = 15$ kg. If the cord is pulled upward with force $\vec T$ of magnitude 180 N, determine (a) the acceleration of the center of the disk (b) the angular acceleration of the disk (c) 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

Using I = ½mr² = 1.875 kg·m², T−mg = ma gives a = (180−147.15)/15 = 2.19 m/s² upward. Tr = Iα gives α = 180×0.5/1.875 = 48.00 rad/s² CCW. Cord acceleration a_c = 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
$$
\begin{aligned}
\Sigma F_y = m\bar a &: \quad T - W = m\bar a\\
\Sigma M_G = I\alpha &: \quad T r = I\alpha
\end{aligned}
$$

#### Centre of the disk
$$
\bar a = \frac{T - W}{m} = \frac{180 - 147.15}{15} = 2.19\ \text{m/s}^2\ \text{upward}
$$

#### Angular acceleration
$$
\alpha = \frac{Tr}{I} = \frac{180\times0.5}{1.875} = 48.00\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)}
$$

#### Acceleration of the cord
The cord has the same acceleration as the rim point $A$ to which it is attached: the acceleration of $G$ plus the tangential acceleration $\alpha r$ of $A$ relative to $G$ (upward, since $\alpha$ is counter-clockwise and $A$ is on the right):

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

**Answer: (a) $\bar a = 2.19$ m/s² upward; (b) $\alpha = 48.00$ rad/s² counter-clockwise; (c) $a_{cord} = 26.19$ m/s² upward.**

## Check yourself

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

1. Which pair of equations is used for the disk pulled by the cord?
   A. T + W = m a and T r = I alpha
   B. T - W = m a and T r = I alpha
   C. T - W = I alpha and T r = m a
   D. T = m a and W r = I alpha
   Answer: B. Vertical: sum F = m a gives T - W = m a; moments about G: T r = I alpha. (Key point: Equations of motion)

2. For a homogeneous disk of mass 15 kg and radius 0.5 m, I about G is:
   A. 3.75 kg m^2
   B. 7.5 kg m^2
   C. 1.875 kg m^2
   D. 0.5 kg m^2
   Answer: C. I = (1/2) m r^2 = 0.5 x 15 x 0.25 = 1.875 kg m^2. (Key point: Moment of inertia)

3. With T = 180 N and W = 147.15 N, the acceleration of G is:
   A. 2.19 m/s^2 upward
   B. 12 m/s^2 upward
   C. 2.19 m/s^2 downward
   D. 21.9 m/s^2 upward
   Answer: A. a = (180 - 147.15)/15 = 2.19 m/s^2 upward. (Key point: Acceleration of centre)

4. How is the cord's acceleration found?
   A. Same as the acceleration of G
   B. alpha x r only
   C. Acceleration of G plus alpha r, since the cord matches the rim point A
   D. T / m
   Answer: C. The cord moves with rim point A: a_cord = a_G + alpha r = 2.19 + 24 = 26.19 m/s^2. (Key point: Acceleration of cord)

## Close variants set in other exams

- [Cord pulled on disk, angular acceleration and cord](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acord-disk-m15-t180-two-parts) — 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×
- [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 on disk, m = 15 kg, T = 200 N](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acord-disk-m15-t200-impulse-momentum) — Applied Mechanics (Dynamics), chapter 5, asked 2×

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