Classical mechanicsLesson 1 of 1 in the course Angular Momentum

Rolling Motion: No-Slip Constraints, Inclines, and Energy

5 flashcards · 5 quiz questions · 5 match pairs · 5 fill-in sentences · English

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Summary

Rolling motion combines translation of an object's center of mass with rotation about that center. This lesson focuses on the special case in which the object rolls without slipping, and on the relationships that connect its linear and angular motion.

A wheel rolling without slipping moves forward while rotating. The point of the wheel touching the ground is momentarily at rest relative to the ground. This does not mean that the wheel is not moving; rather, the point of contact has zero instantaneous velocity because the wheel's forward motion and rotational motion cancel there.

All flashcards

No-slip rolling
The contact point is instantaneously at rest relative to the surface, linking translation and rotation.
Rolling acceleration constraint
For constant radius and no slipping, the magnitudes satisfy aCM=Rαa_{CM}=R\alpha.
Acceleration of a rolling object on an incline
With no slipping, aCM=mgsin⁡θm+ICM/r2a_{CM}=\frac{mg\sin\theta}{m+I_{CM}/r^2}.
Kinetic friction during slipping
Its magnitude is fk=μkNf_k=\mu_kN, and it opposes relative sliding at the contact.
Mechanical energy of a rolling object
It includes translational kinetic, rotational kinetic, and gravitational potential energy.

Quiz questions

  1. A solid cylinder rolls without slipping down an incline at angle θ\theta. What is its center-of-mass acceleration?Answer: aCM=23gsin⁡θa_{CM}=\frac{2}{3}g\sin\theta.
  2. What minimum coefficient of static friction allows a solid cylinder to roll without slipping down an incline at angle θ\theta?Answer: μs≥13tan⁡θ\mu_s\geq\frac{1}{3}\tan\theta.
  3. A cylinder slides down an incline at angle θ\theta while kinetic friction acts up the slope. What is its translational acceleration?Answer: aCM=g(sin⁡θ−μkcos⁡θ)a_{CM}=g(\sin\theta-\mu_k\cos\theta).
  4. A solid cylinder of radius rr slips down an incline at angle θ\theta under kinetic friction. What is its angular acceleration?Answer: α=2μkgcos⁡θr\alpha=\frac{2\mu_k g\cos\theta}{r}.
  5. An object with moment of inertia ICM=kmr2I_{CM}=kmr^2 starts from rest and rolls without slipping through a vertical drop hh. What is its final center-of-mass speed, neglecting dissipative effects?Answer: vCM=2gh1+kv_{CM}=\sqrt{\frac{2gh}{1+k}}.

Based on University Physics Volume 1 by William Moebs, Samuel J. Ling, Jeff Sanny (OpenStax), licensed CC BY 4.0. Adapted into a summary, flashcards and quizzes.

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