--- title: "Wheel pulled by cord on inner drum, 50 kg" subject: "Applied Mechanics (Dynamics)" chapter: 5 key: "semester_3/applied_mechanics_dynamics:wheel-inner-drum-50kg" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/wheel-inner-drum-50kg site: BCT Engineering --- # Wheel pulled by cord on inner drum, 50 kg Applied Mechanics (Dynamics) · Chapter 5: Plane Motion of Rigid Bodies: Forces, Moments, and Accelerations - Asked in: 2080 Bhadra, 6 marks - Video: [A cord is wrapped around the inner drum of a wheel and pull horizontally with force 200N](https://www.youtube.com/watch?v=uVJ7zkIVLPU) — Civil Engg Network - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/wheel-inner-drum-50kg - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-5 ## Question A cord wrapped around the inner drum of a wheel and pulled horizontally with a force of 200 N. The wheel has a mass of 50 kg and a radius of gyration of 70 mm. Knowing that the coefficients of friction are $\mu_s = 0.20$ and $\mu_k = 0.15$, determine the acceleration of G and the angular acceleration of the wheel. [Figure: wheel with outer radius 100 mm and inner drum radius 60 mm; 200 N pulled horizontally to the right by the cord on the inner drum; wheel on rough ground.] ## Short answer Rolling assumption fails because required friction exceeds μ_s N. With kinetic friction 73.58 N, a_G = (200−73.58)/50 = 2.53 m/s² right. I = mk² = 0.245 kg·m²; α = (200×0.06−73.58×0.10)/0.245 = 18.95 rad/s² CCW. ## Answer **Assumptions.** The cord leaves the bottom of the inner drum (radius $r=0.060$ m) and pulls horizontally to the right; the wheel (outer radius $R=0.100$ m) rests on rough ground. Take $x$ to the right; the sense of $\alpha$ is stated in each step. The friction force $F$ acts at the contact point. #### Data $m = 50$ kg, $\bar k = 0.070$ m, $P = 200$ N, $\mu_s = 0.20$, $\mu_k = 0.15$ $$ \bar I = m\bar k^2 = 50(0.070)^2 = 0.2450\ \text{kg m}^2, \qquad N = mg = 50\times9.81 = 490.50\ \text{N} $$ #### Step 1: assume rolling without slipping Rolling to the right means the wheel turns clockwise, so $\bar a = \alpha R$ with $\alpha$ clockwise. The pull $P$ (below $G$) gives a counter-clockwise moment $Pr$ about $G$; friction $F$, taken to the left at the contact point, gives a clockwise moment $FR$: $$ \begin{aligned} P - F &= m\bar a\\ FR - Pr &= \bar I\alpha \end{aligned} $$ Eliminating $F$ with $\bar a=\alpha R$: $$ \alpha = \frac{P(R-r)}{m(\bar k^2 + R^2)} = \frac{200(0.100-0.060)}{50(0.0049+0.0100)} = 10.74\ \text{rad/s}^2\ \text{(clockwise)}, \qquad \bar a = \alpha R = 1.07\ \text{m/s}^2 $$ Friction required: $F = P - m\bar a = 200 - 50(1.07) = 146.31$ N. Maximum available: $F_{max} = \mu_s N = 0.20\times490.50 = 98.10$ N. Since $F = 146.31\ \text{N} > F_{max}$, the wheel **slips**. The assumption is wrong, and kinetic friction acts: $F = \mu_k N = 0.15\times490.50 = 73.58$ N. #### Step 2: sliding (kinetic friction) $$ \begin{aligned} \bar a &= \frac{P - F_k}{m} = \frac{200 - 73.58}{50} = 2.53\ \text{m/s}^2\ (\to)\\ \alpha &= \frac{Pr - F_kR}{\bar I} = \frac{200(0.060) - 73.58(0.100)}{0.2450} = 18.95\ \text{rad/s}^2\ \text{(counter-clockwise)} \end{aligned} $$ With $\alpha$ counter-clockwise, the contact point accelerates forward relative to the ground ($\bar a + \alpha R>0$), so it slips forward. This confirms that friction acts to the left. **Answer: $\bar a_G = 2.53$ m/s² to the right; $\alpha = 18.95$ rad/s² counter-clockwise.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the first step in solving this friction problem? A. Assume kinetic friction at once B. Assume rolling without slipping, a = alpha R, and compute the required friction C. Ignore friction D. Assume F = mu_s N Answer: B. Assume rolling, find the friction needed, then compare with the maximum available. (Key point: Rolling assumption) 2. Required friction is 146.31 N while mu_s N = 98.1 N. What follows? A. The wheel rolls without slipping B. The wheel slips, so kinetic friction F = mu_k N acts C. The wheel lifts off D. Friction is zero Answer: B. Required friction exceeds the static maximum, so the wheel slips and F = mu_k N = 73.58 N. (Key point: Slip check) 3. With F_k = 73.58 N, P = 200 N, m = 50 kg, a_G is: A. 2.53 m/s^2 B. 1.07 m/s^2 C. 4.0 m/s^2 D. 5.47 m/s^2 Answer: A. a = (200 - 73.58)/50 = 2.53 m/s^2 to the right. (Key point: Acceleration of G) 4. Which expression gives alpha when sliding? A. (P R - F r)/I B. (P r - F_k R)/I C. P r / I D. (P - F_k) / I Answer: B. Moments about G: P r - F_k R = I alpha, giving 18.95 rad/s^2 counter-clockwise. (Key point: Angular acceleration) ## Close variants set in other exams - [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× ## Related in Applied Mechanics (Dynamics) - [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 = 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 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×