--- title: "Sphere rolling down incline, friction and velocity" subject: "Applied Mechanics (Dynamics)" chapter: 6 key: "semester_3/applied_mechanics_dynamics:sphere-rolling-incline-80kn" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/sphere-rolling-incline-80kn site: BCT Engineering --- # Sphere rolling down incline, friction and velocity Applied Mechanics (Dynamics) · Chapter 6: Plane motion of rigid bodies: energy and momentum methods - Asked in: 2076 Chaitra, 6 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/applied_mechanics_dynamics/sphere-rolling-incline-80kn - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6 ## Question A sphere of weight 80 KN is released with no initial velocity and rolls without slipping on the incline. Determine: a) the minimum value of the coefficient of friction, b) the velocity of G after the sphere has rolled 15 ft and c) the velocity of G if the sphere were to move 20 ft down a frictionless incline. [Figure: sphere with centre G and contact point C on an incline at $30^\circ$ to the horizontal.] ## Short answer For rolling without slipping, acceleration is (5/7)g sin30 = 11.50 ft/s^2, giving minimum friction coefficient 0.1650. After 15 ft the center velocity is 18.57 ft/s; on a frictionless 20 ft incline it slides with a=g sin30, reaching 25.38 ft/s. ## Answer Solid sphere of weight $W = 80$ kN, rolling without slipping on a $30^\circ$ incline: $\bar I = \tfrac25 mr^2$ and $\bar a = \alpha r$. Distances are in feet, so $g = 32.2$ ft/s². #### (a) Minimum coefficient of friction Take axes along and perpendicular to the incline. Let $F$ be friction up the incline. $$ \begin{aligned} W\sin30^\circ - F &= m\bar a\\ N - W\cos30^\circ &= 0\\ Fr &= \bar I\alpha = \tfrac25 mr^2\frac{\bar a}{r} \;\Rightarrow\; F = \tfrac25 m\bar a \end{aligned} $$ Substituting $F$ in the first equation: $W\sin30^\circ = \tfrac75 m\bar a$, so $$ \bar a = \tfrac57 g\sin30^\circ = \tfrac57(32.2)(0.5) = 11.50\ \text{ft/s}^2 $$ $$ F = \tfrac25 m\bar a = \tfrac27W\sin30^\circ = \tfrac27(80)(0.5) = 11.43\ \text{kN}, \qquad N = W\cos30^\circ = 80\cos30^\circ = 69.28\ \text{kN} $$ For rolling without slipping, $F \le \mu_s N$: $$ \mu_{min} = \frac{F}{N} = \tfrac27\tan30^\circ = 0.1650 $$ #### (b) Velocity of G after rolling 15 ft The acceleration is constant, starting from rest: $$ \bar v^2 = 2\bar a\,s = 2(11.50)(15) \;\Rightarrow\; \bar v = 18.57\ \text{ft/s} \;(= 5.66\ \text{m/s}) $$ (Check by energy: $Ws\sin30^\circ = \tfrac12m\bar v^2 + \tfrac12\bar I\omega^2 = \tfrac12\cdot\tfrac75 m\bar v^2$, giving the same $\bar v$.) #### (c) Frictionless incline, 20 ft The sphere does not rotate; it slides with $a = g\sin30^\circ = 16.1$ ft/s²: $$ \bar v = \sqrt{2(16.1)(20)} = 25.38\ \text{ft/s} \;(= 7.73\ \text{m/s}) $$ **Answer: (a) $\mu_{min} = 0.1650$; (b) $\bar v = 18.57$ ft/s; (c) $\bar v = 25.38$ ft/s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a solid sphere rolling without slipping down the incline, what does the friction equation F r = I alpha give? A. F = (2/5) m a_G, because I = (2/5) m r^2 and a_G = alpha r B. F = m a_G C. F = W sin 30 degrees D. F = 0 Answer: A. Substituting I = (2/5) m r^2 and alpha = a_G / r gives F = (2/5) m a_G. (Key point: Rolling condition and friction) 2. What is the acceleration of G for rolling without slipping? A. g sin 30 degrees = 16.1 ft/s^2 B. (5/7) g sin 30 degrees = 11.50 ft/s^2 C. (2/7) g sin 30 degrees D. (5/7) g cos 30 degrees Answer: B. W sin 30 - F = m a with F = (2/5) m a gives (7/5) m a = W sin 30, so a = (5/7) g sin 30 = 11.50 ft/s^2. (Key point: Acceleration of G) 3. How is the minimum coefficient of friction found, and what is its value? A. mu = tan 30 degrees = 0.577 B. mu_min = F / N = (2/7) tan 30 degrees = 0.165 C. mu_min = (5/7) tan 30 degrees = 0.412 D. mu_min = sin 30 degrees = 0.5 Answer: B. No slipping needs F <= mu N, so mu_min = F / N = (2/7) W sin 30 / (W cos 30) = (2/7) tan 30 = 0.165. (Key point: Minimum friction coefficient) 4. How does the frictionless case (c) differ from rolling? A. The sphere rolls with the same acceleration B. The sphere slides without rotating, with a = g sin 30 degrees, so it is faster (25.38 ft/s after 20 ft) C. The sphere does not move D. The sphere rotates faster Answer: B. With no friction there is no moment about G, so the sphere does not rotate and all the energy goes into translation: v = sqrt(2(16.1)(20)) = 25.38 ft/s. 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