SCIENCE · INVESTIGATE AND EXPLAIN

Integrate sliding motion to its first no-slip event

Evolve translation and rotation separately while kinetic friction acts.

Learning outcome: Analyze forward slip, backward slip and an already rolling case. Predict the friction direction, solve both motion equations and locate the first event before checking. Reconcile kinetic-energy loss with the contact-slip integral, and explain why center kinetic energy alone is not a dissipation measurement.

Starting recap

State the reference axis and system, distinguish time and angle integrals, and verify an accessible and supported motion branch.

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Try

With v0=3 and omega0=0, u0=3, tstar=.5 s, vf=2 m/s and omega_f=4 rad/s. Initial K=9 J and final K=6 J. The lost 3 J also equals 4×3×.5/2. For v0=1,omega0=8, slip is backward; friction accelerates the center while reducing spin.

Open the experiment

Predict

While the body is slipping, can vCM=R omega be imposed throughout the motion?

Your prediction and reason

Experiment

Analyze forward slip, backward slip and an already rolling case. Predict the friction direction, solve both motion equations and locate the first event before checking. Reconcile kinetic-energy loss with the contact-slip integral, and explain why center kinetic energy alone is not a dissipation measurement.

Match a kinetic-sliding branch to first no-slip motion

Worked teaching model · synthetic values · Solid cylinder M=2 kg,R=0.5 m,I=.5MR² on horizontal stationary surface,mu_k=0.2,g=10. Rightward v and clockwise omega positive. Slip velocity at contact is v−Romega. Kinetic friction stops at first zero slip; no other horizontal force or rolling resistance.

Initial center speed: 3 m/s Initial clockwise angular speed: 0 rad/s

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Contact slip approaching first rolling

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Contact slip approaching first rolling: numerical coordinates

time (s)	contact slip (m/s)
0	3
0.0125	2.925
0.025	2.85
0.0375	2.775
0.05	2.7
0.0625	2.625
0.075	2.55
0.0875	2.475
0.1	2.4
0.1125	2.325
0.125	2.25
0.1375	2.175
0.15	2.1
0.1625	2.025
0.175	1.95
0.1875	1.875
0.2	1.8
0.2125	1.725
0.225	1.65
0.2375	1.575
0.25	1.5
0.2625	1.425
0.275	1.35
0.2875	1.275
0.3	1.2
0.3125	1.125
0.325	1.05
0.3375	0.975
0.35	0.9
0.3625	0.825
0.375	0.75
0.3875	0.675
0.4	0.6
0.4125	0.525
0.425	0.45
0.4375	0.375
0.45	0.3
0.4625	0.225
0.475	0.15
0.4875	0.075
0.5	0

Calculated values under the declared model
Quantity	Value	Unit
Initial contact slip velocity	3	m/s
Kinetic-friction branch duration	0.5	s
Center velocity at first rolling	2	m/s
Angular velocity at first rolling	4	rad/s
Final slip residual	0	m/s
Initial kinetic energy	9	J
Final kinetic energy	6	J
Dissipated kinetic energy	3	J

Kinetic friction changes translation and rotation until the first zero of contact slip.

For nonzero slip, aCM=−sign(slip)mu g and alpha=sign(slip)mu g/(beta R), with clockwise omega positive. Thus slip changes at −sign(slip)6 m/s². At the first zero switch to ideal rolling without rolling resistance; do not continue kinetic friction into a fictitious oscillation of slip.

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Kinetic friction changes translation and rotation until the first zero of contact slip.

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Understand

Diagnose relative motion at the contact

Rightward center speed v and clockwise angular speed omega are positive. For radius R, the contact slip speed relative to a stationary horizontal surface is u=v−R omega. Its sign sets the direction of kinetic friction. A rotating body can slip forward, slip backward or already roll without slipping; v alone does not distinguish these cases.

Integrate the two equations separately

The solid cylinder has M=2 kg, R=.5 m, I=.25 kg m², mu_k=.2 and g=10. If s=sign(u0), friction is −s4 N. Thus dv/dt=−s2 m/s² and clockwise domega/dt=s8 rad/s². Before the first zero of slip, v=v0−s2t and omega=omega0+s8t. Consequently u=u0−s6t and the first no-slip time is |u0|/6.

Change branches at the event

At the first u=0, reconnect v=R omega. With no other horizontal force, no axle torque and no rolling resistance, zero friction is sufficient afterward and both speeds remain constant. Continuing the old kinetic-friction direction beyond this event would create an unphysical overshoot. If u0=0, the system starts on the no-slip branch and there is no sliding interval.

Audit the dissipated energy

K=Mv²/2+Iomega²/2=v²+.125omega² for the stated parameters. During slip the total mechanical power delivered by friction is −|f_k||u|, so the energy lost is integral |f_k||u|dt. For constant deceleration of slip this is a triangle area in time: 4|u0|tstar/2. Friction can increase rotational kinetic energy while decreasing the total; looking at one component alone misses dissipation.

Worked original case

Investigation

Public numerical transfer practice

Try a new practice case

For the taught cylinder, v0=3 m/s and omega0=0. What is the first no-slip time in seconds?

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax angular momentum. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.

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Public practice checks a specific claim. It does not establish independent mastery.

What additional information would be needed if friction varied with slip speed?

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Previous: Derive and test a no-slip rolling constraint Next: Reconcile orbital energy with angular momentum

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