---
title: "Cord pulled on disk, m = 30 kg, T = 200 N"
subject: "Applied Mechanics (Dynamics)"
chapter: 5
key: "semester_3/applied_mechanics_dynamics:cord-disk-m30-t200"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/cord-disk-m30-t200
site: BCT Engineering
---

# Cord pulled on disk, m = 30 kg, T = 200 N

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

- Asked in: 2073 Shrawan, 6 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-m30-t200
- 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 $m = 30$ kg as shown in figure below. If the cord is pulled upward with a force $T$ of magnitude 200 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

I=3.75 kg·m². T (200 N) < weight, so a = (200−294.3)/30 = −3.14 m/s² downward. α = 200×0.5/3.75 = 26.67 rad/s² CCW. Cord a_c = −3.14 + 13.33 = 10.19 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 pull $T$, acting 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
$m=30$ kg, $r=0.5$ m, $T=200$ N, $g = 9.81$ m/s²

$$
W = mg = 30\times9.81 = 294.30\ \text{N}, \qquad I = \tfrac12 mr^2 = \tfrac12(30)(0.5)^2 = 3.75\ \text{kg m}^2
$$

#### 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
$$
\bar a = \frac{T - W}{m} = \frac{200 - 294.30}{30} = -3.14\ \text{m/s}^2, \quad\text{i.e. } 3.14\ \text{m/s}^2\ \text{downward}
$$

The pull is less than the weight, so the centre accelerates downward.

#### Angular acceleration
$$
\alpha = \frac{Tr}{I} = \frac{200\times0.5}{3.75} = 26.67\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)}
$$

#### 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):

$$
a_{cord} = \bar a_y + \alpha r = (-3.14) + 26.67\times0.5 = 10.19\ \text{m/s}^2\ \uparrow
$$

**Answer: (a) acceleration of the centre $\bar a = 3.14$ m/s² downward; (b) angular acceleration $\alpha = 26.67$ rad/s² counter-clockwise; (c) acceleration of the cord $= 10.19$ m/s² upward.**

## Check yourself

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

1. For m = 30 kg and T = 200 N, why does the centre accelerate downward?
   A. T is larger than W
   B. T is smaller than W = 294.3 N, so net force is downward
   C. The disk rotates counter-clockwise
   D. The cord is pulled at the rim
   Answer: B. T - W = 200 - 294.3 is negative, so a_G = 3.14 m/s^2 downward. (Key point: Direction of G acceleration)

2. With I = 3.75 kg m^2, the angular acceleration is:
   A. 26.67 rad/s^2
   B. 53.3 rad/s^2
   C. 13.3 rad/s^2
   D. 40 rad/s^2
   Answer: A. alpha = 200 x 0.5 / 3.75 = 26.67 rad/s^2. (Key point: Angular acceleration)

3. How is the downward a_G combined with alpha r to get the cord acceleration?
   A. 3.14 + 13.33
   B. 13.33 - 3.14, taking upward positive
   C. 13.33 x 3.14
   D. Only alpha r is used
   Answer: B. a_cord = (-3.14) + 26.67 x 0.5 = 10.19 m/s^2 upward. (Key point: Sign in cord acceleration)

4. The acceleration of the cord is:
   A. 16.47 m/s^2 upward
   B. 10.19 m/s^2 upward
   C. 13.33 m/s^2 upward
   D. 3.14 m/s^2 downward
   Answer: B. -3.14 + 13.33 = 10.19 m/s^2 upward. (Key point: Cord 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, 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 = 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×

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