---
title: "Cord pulled on disk, m = 20 kg, F = 250 N"
subject: "Applied Mechanics (Dynamics)"
chapter: 5
key: "semester_3/applied_mechanics_dynamics:cord-disk-m20-f250"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/cord-disk-m20-f250
site: BCT Engineering
---

# Cord pulled on disk, m = 20 kg, F = 250 N

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

- Asked in: 2074 Chaitra, 6 marks
- 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-m20-f250
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-5

## Question

A cord is wrapped around a homogenous disk of radius $r = 0.5$ m and mass 20 kg. If the cord is pulled upward with a force of magnitude $F = 250$ N, determine (a) the angular acceleration of the disk, (b) the acceleration of the disk and (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 F.]

## Short answer

I=½(20)(0.5²)=2.5 kg·m²; T−mg gives a = (250−196.2)/20 = 2.69 m/s² up. Tr = Iα gives α = 250×0.5/2.5 = 50.00 rad/s² CCW; cord a_c = a + αr = 27.69 m/s² up.

## 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 $T$ (the given force $F$), 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
$m=20$ kg, $r=0.5$ m, $T=250$ N (the force called $F$ in the question), $g = 9.81$ m/s²

$$
W = mg = 20\times9.81 = 196.20\ \text{N}, \qquad I = \tfrac12 mr^2 = \tfrac12(20)(0.5)^2 = 2.5\ \text{kg m}^2
$$

#### 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
$$
\bar a = \frac{T - W}{m} = \frac{250 - 196.20}{20} = 2.69\ \text{m/s}^2\ \text{upward}
$$

#### Angular acceleration
$$
\alpha = \frac{Tr}{I} = \frac{250\times0.5}{2.5} = 50.00\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)}
$$

#### 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):

$$
a_{cord} = \bar a_y + \alpha r = (+2.69) + 50.00\times0.5 = 27.69\ \text{m/s}^2\ \uparrow
$$

**Answer: (a) angular acceleration $\alpha = 50.00$ rad/s² counter-clockwise; (b) acceleration of the disk $\bar a = 2.69$ m/s² upward; (c) acceleration of the cord $= 27.69$ m/s² upward.**

## Check yourself

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

1. For the 20 kg, 0.5 m radius homogeneous disk, I about G is:
   A. 5.0 kg m^2
   B. 2.5 kg m^2
   C. 10 kg m^2
   D. 1.25 kg m^2
   Answer: B. I = 0.5 x 20 x 0.25 = 2.5 kg m^2. (Key point: Moment of inertia)

2. With F = 250 N, the angular acceleration is:
   A. 100 rad/s^2
   B. 25 rad/s^2
   C. 50 rad/s^2
   D. 12.5 rad/s^2
   Answer: C. alpha = F r / I = 250 x 0.5 / 2.5 = 50 rad/s^2. (Key point: Angular acceleration)

3. Using F - W = m a with W = 196.2 N, the acceleration of the centre is:
   A. 12.5 m/s^2
   B. 2.69 m/s^2 upward
   C. 2.69 m/s^2 downward
   D. 9.81 m/s^2 upward
   Answer: B. a = (250 - 196.2)/20 = 2.69 m/s^2 upward. (Key point: Acceleration of disk)

4. The acceleration of the cord is:
   A. 27.69 m/s^2 upward
   B. 25 m/s^2 upward
   C. 2.69 m/s^2 upward
   D. 22.31 m/s^2 upward
   Answer: A. a_cord = a_G + alpha r = 2.69 + 25 = 27.69 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 = 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 = 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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