--- title: "Cylinder rolling down an incline" subject: "Engineering Mechanics" chapter: 6 key: "semester_3/engineering_mechanics:cylinder-rolling-down-incline" source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/cylinder-rolling-down-incline site: BCT Engineering --- # Cylinder rolling down an incline Engineering Mechanics · Chapter 6: Plane Kinetics of Rigid Bodies: Force, Mass and Acceleration - Asked in: year not given, 8 marks - Video: [Physics 13.1 Moment of Inertia Application (3 of 11) Solid Cylinder Rolling Down an Incline](https://www.youtube.com/watch?v=cPMSMDSY-rc) — Michel van Biezen - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/cylinder-rolling-down-incline - Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-6 ## Question A solid homogeneous cylinder of mass 20 kg and radius 0.3 m rolls without slipping down an incline of 30° to the horizontal, starting from rest. Determine (a) the linear acceleration of its centre, (b) the angular acceleration, (c) the friction force and the normal reaction, and (d) the minimum coefficient of static friction needed to prevent slipping. ## Short answer Using ΣF along slope, ΣF perpendicular, ΣM_G = I_G α, and rolling condition a = α r gives a = (2/3)g sin30° = 3.27 m/s², α = 10.90 rad/s², friction 32.7 N, normal reaction 169.9 N. Minimum coefficient of static friction is 0.192. ## Answer **Given:** $m = 20$ kg, $r = 0.3$ m, $\theta = 30^\circ$, rolling without slipping. $I_G = \tfrac12 mr^2 = 0.90$ kg·m². ```text \ \ O (G) (o)--> a \ F (friction up slope) \ \ 30 deg ``` #### Equations of motion Let $F$ be the friction force up the slope at the contact point, $N$ the normal reaction, and $a$ the acceleration of G down the slope. Along the slope: $mg\sin\theta - F = ma$ (1) Perpendicular to the slope: $N = mg\cos\theta$ (2) Moments about G: $F r = I_G\alpha$ (3) Rolling without slipping: $a = \alpha r$ (4) From (3) and (4): $F = \dfrac{I_G a}{r^2} = \tfrac12 ma$. Put in (1): $$ mg\sin\theta = ma + \tfrac12 ma = \tfrac32 ma \;\Rightarrow\; a = \tfrac23 g\sin\theta $$ #### (a) Linear acceleration $$ a = \tfrac23(9.81)(0.5) = 3.27\ \text{m/s}^2 $$ #### (b) Angular acceleration $$ \alpha = \frac{a}{r} = \frac{3.27}{0.3} = 10.90\ \text{rad/s}^2 $$ #### (c) Friction and normal reaction $$ F = \tfrac12 ma = \tfrac12(20)(3.27) = 32.7\ \text{N} \qquad N = 20(9.81)\cos30^\circ = 169.9\ \text{N} $$ #### (d) Minimum coefficient of friction For no slipping $F \le \mu_s N$: $$ \mu_{min} = \frac{F}{N} = \frac{32.7}{169.9} = 0.1925 \;\left(= \tfrac13\tan30^\circ\right) $$ **Answer: a = 3.27 m/s²; α = 10.90 rad/s²; F = 32.7 N; N = 169.9 N; μ_min = 0.192.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which equation relates friction to angular acceleration for the cylinder? A. F r = I_G alpha B. F r = m a C. mg sin(theta) r = I_G alpha D. N r = I_G alpha Answer: A. Only friction has a moment about G (weight and N pass through G). (Key point: Moment equation about G) 2. What is the acceleration of the centre of a solid cylinder rolling without slipping down a 30° incline? A. g sin 30° = 4.91 m/s^2 B. (2/3) g sin 30° = 3.27 m/s^2 C. (1/2) g sin 30° = 2.45 m/s^2 D. (3/2) g sin 30° = 7.36 m/s^2 Answer: B. mg sin(theta) = ma + (1/2)ma, so a = (2/3) g sin(theta). (Key point: Acceleration of a rolling cylinder) 3. What is the friction force F, and in which direction does it act? A. mg sin(theta) = 98.1 N down the slope B. (1/2) m a = 32.7 N up the slope C. mu N = 169.9 N up the slope D. (1/2) m a = 32.7 N down the slope Answer: B. Friction opposes sliding, acts up the slope, and from F = I_G a/r^2 equals ma/2. (Key point: Friction force) 4. How is the minimum coefficient of static friction found, and what is its value? A. F/N = 0.192, which equals (1/3) tan 30° B. tan 30° = 0.577 C. N/F = 5.2 D. (2/3) tan 30° = 0.385 Answer: A. No slip needs F <= mu N, so mu_min = 32.7/169.9 = 0.192. 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