AP Physics C: Mechanics · Unit 6: Energy and Momentum of Rotating Systems · Lesson 6.5

Deep Dive: Rolling

🔬 Deep Dive

This is your textbook for this topic. Take your time. Read it more than once.

6.5.A.1Math

Total Kinetic Energy of a Rolling System

You met this equation back in Lesson 6.1, and it's the foundation of everything in this lesson: a system that both translates and rotates carries both kinds of kinetic energy at once.

Ktot = Ktrans + Krot

💡A rolling wheel, ball, or cylinder is the textbook example — its center of mass moves in a straight (or curved) line while the object itself spins about that center of mass. Neither term alone tells the whole story.

A rolling object carries both kinds of kinetic energy at once. Adjust its mass, center-of-mass speed, rotational inertia, and angular speed to see the split.

m (kg)4

vcm (m/s)3

Icm (kg·m²)1.5

ω (rad/s)6

Ktrans

Krot

Ktrans = ½mv²cm = 18.0 JKrot = ½Icmω² = 27.0 JKtotal = 45.0 J

What's new in this lesson is the special relationship between vcm and ω that a rolling object obeys — which is exactly where we're headed next.

6.5.B.16.5.B.2Concept

Rolling Without Slipping

When a system rolls without slipping, its translational motion and its rotational motion aren't independent — they're locked together by a fixed relationship at every instant:

Δxcm = rΔθ

vcm = rω

acm = rα

🔑These constraints come from a simple geometric fact: the point of the rolling object touching the surface is, at that instant, not sliding — it's momentarily at rest. That's what "without slipping" means.

This has a major energy consequence. Because the contact point isn't sliding, the static friction force acting there does zero work — in the ideal case, rolling without slipping doesn't dissipate any energy at all. That's what lets you safely use conservation of energy on rolling problems.

This is the setup behind the classic race down an incline: release a hoop, a solid disk, and a solid sphere from the same height, and see which reaches the bottom fastest.

Release a hoop, a solid disk, and a solid sphere from rest at the same height on the same incline — the classic race from this unit's suggested activities. Energy conservation (mgh = Ktrans + Krot, with vcm = rω) gives each shape a different speed at the bottom, independent of mass or radius.

height h (m)2

vcm = √(2gh / (1 + c)), where c = Icm / (mR²) — smaller c wins the race

ExampleGuided Example — Racing a Solid Sphere

A solid sphere (Icm = ⅖mR²) is released from rest at the top of a 1.8 m incline and rolls without slipping to the bottom. Find its center-of-mass speed at the bottom.

Step 1 — Set up energy conservation

No energy is lost to friction (rolling without slipping), so mgh = Ktrans + Krot at the bottom.

6.5.C.16.5.C.2Concept⚠ Watch Out

Rolling While Slipping

Everything in the previous section depended on one assumption: no slipping. When a system does slip — think of a car's tires spinning as it peels out, or a bowling ball skidding down the lane before it "grabs" — that clean relationship between vcm and ω breaks down completely.

⚠️While slipping, a system's center-of-mass motion and its rotational motion cannot be directly related. vcm ≠ rω, and you can't use one to find the other — you have to analyze the translational and rotational motion separately, using Newton's second law and its rotational form independently.

The energy story changes too. When a system slips, the point of the surface where kinetic friction acts is actually sliding relative to the ground — so unlike the static-friction case in rolling without slipping, this kinetic friction force does nonzero (negative) work. Slipping dissipates real energy out of the system, typically as heat and sound.

💡This is why spinning tires produce smoke and squealing sound, and rolling-without-slipping wheels don't: it's not the rotation itself that costs energy, it's the sliding. (Rolling friction — the very small energy loss real rolling objects experience even without slipping, due to deformation — is beyond the scope of this course.)

ExampleGuided Example — Peeling Out

A car accelerates hard from a stoplight, its rear tires spinning faster than the car's forward speed divided by the tire radius — classic 'peeling out.' Is this rolling without slipping? What happens to the energy delivered by the engine?

Step 1 — Check the rolling condition

vcm = rω would require the car's speed to exactly match the tire's rim speed. Since the tires are spinning faster than that, vcm ≠ rω — the tires are slipping against the road.

← Back to Lesson 6.5Ready for 6.6? Motion of Orbiting Satellites takes conservation of energy and angular momentum out of the lab and into orbit.