--- title: "Cord on disk, m = 15 kg, T = 200 N" subject: "Applied Mechanics (Dynamics)" chapter: 5 key: "semester_3/applied_mechanics_dynamics:cord-disk-m15-t200-impulse-momentum" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/cord-disk-m15-t200-impulse-momentum site: BCT Engineering --- # Cord on disk, m = 15 kg, T = 200 N Applied Mechanics (Dynamics) · Chapter 5: Plane Motion of Rigid Bodies: Forces, Moments, and Accelerations - Asked in: 2079 Bhadra, 4+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-m15-t200-impulse-momentum - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-5 ## Question Explain the principle of impulse and momentum for the plane motion of rigid body. 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 a force $\vec T$ 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 The impulse–momentum principle equates changes in linear momentum m v̄ and angular momentum Īω to the impulses of external forces and moments. For the disk: I=1.875 kg·m²; a = (200−147.15)/15 = 3.52 m/s² up; α = 200×0.5/1.875 = 53.33 rad/s² CCW; cord a_c = 3.52+26.67 = 30.19 m/s² up. ## Answer **Principle of impulse and momentum for plane motion of a rigid body:** the momenta of all particles of the body at time $t_1$, plus the impulses of the external forces acting between $t_1$ and $t_2$, equal the momenta at time $t_2$. The momenta of the particles reduce to a linear momentum vector $m\bar{\vec v}$ at the mass centre and an angular momentum $\bar I\omega$ about it: $$ \begin{aligned} m\bar{\vec v}_1 + \sum\int \vec F\,dt &= m\bar{\vec v}_2\\ \bar I\omega_1 + \sum\int M_G\,dt &= \bar I\omega_2 \end{aligned} $$ The moment equation may also be written about a fixed point $O$ (or about any point, with the appropriate momentum terms): $H_{O1} + \sum\int M_O\,dt = H_{O2}$. In scalar form the principle gives three equations (two for the linear part, one for the rotational part). It is the best method when forces are given as functions of time, or when a time interval (rather than a distance) is involved, as in impact problems. ```text (momenta)1 + (impulses) = (momenta)2 m v1 .--. F dt m v2 .--. ---> (G) I w1 + ------> = ---> (G) I w2 '--' '--' ``` #### Numerical part 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, $T=200$ 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 $$ \bar a = \frac{T - W}{m} = \frac{200 - 147.15}{15} = 3.52\ \text{m/s}^2\ \text{upward} $$ #### Angular acceleration $$ \alpha = \frac{Tr}{I} = \frac{200\times0.5}{1.875} = 53.33\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)} $$ #### 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): $$ a_{cord} = \bar a_y + \alpha r = (+3.52) + 53.33\times0.5 = 30.19\ \text{m/s}^2\ \uparrow $$ **Answer: (a) $\bar a = 3.52$ m/s² upward; (b) $\alpha = 53.33$ rad/s² counter-clockwise; (c) $a_{cord} = 30.19$ m/s² upward.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The principle of impulse and momentum for a rigid body states that: A. Initial momenta plus impulses of external forces equal final momenta B. Initial momenta equal final momenta, always C. Work done equals change in kinetic energy D. Impulse equals force times distance Answer: A. Momenta at t1 plus impulses of external forces between t1 and t2 equal momenta at t2. (Key point: Impulse-momentum principle) 2. The momenta of the particles of a body in plane motion reduce to: A. m v at G only B. I omega only C. m v at G and a couple I omega D. m v at O and m a at G Answer: C. They are equivalent to linear momentum m v at G plus angular momentum I omega. (Key point: Momenta of the body) 3. For T = 200 N, m = 15 kg, W = 147.15 N, a_G is: A. 13.3 m/s^2 B. 3.52 m/s^2 upward C. 9.81 m/s^2 D. 53.3 m/s^2 Answer: B. a = (200 - 147.15)/15 = 3.52 m/s^2 upward. (Key point: Acceleration of centre) 4. Using I = 1.875 kg m^2, r = 0.5 m, T = 200 N, alpha is: A. 53.33 rad/s^2 B. 106.7 rad/s^2 C. 26.67 rad/s^2 D. 3.52 rad/s^2 Answer: A. alpha = T r / I = 100/1.875 = 53.33 rad/s^2. (Key point: Angular acceleration) ## 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× ## Related in Applied Mechanics (Dynamics) - [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, 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× - [Wheel pulled by cord on inner drum, 50 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Awheel-inner-drum-50kg) — Applied Mechanics (Dynamics), chapter 5, asked 2× - [Wheel pulled by cord on inner drum, 45 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Awheel-inner-drum-45kg) — Applied Mechanics (Dynamics), chapter 5, asked 2× - [Angular momentum of rigid body in plane motion](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aangular-momentum-rigid-body-plane) — Applied Mechanics (Dynamics), chapter 5 ## Related in other subjects - [Impulse-momentum principle; blocks on incline](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aimpulse-momentum-blocks-a-b-incline-pulley) — Applied Mechanics, chapter 9