Physics I: Mechanics › Rotation › full formula sheet

vcm = ωR  ·  K = ½mvcm² + ½Iω²

Say it: “rolling without slipping means the center moves at omega R; the total kinetic energy is translation of the center plus rotation about the center”

Rolling without slipping

A wheel is two motions in a trench coat — gliding forward and spinning — locked together by one constraint.

Notation on this page: vcm is the center-of-mass speed; R the wheel’s radius; I its moment of inertia about the central axis. “Without slipping” means the contact point doesn’t slide.

Before this lesson: Rotational kinetic energy, Moment of inertia

Where it comes from

Watch a rolling wheel’s contact point: at the instant it touches the ground, it is at rest — the wheel lays itself down like tape unrolling. If the contact slid, that’d be slipping (like a car spinning its tires). No slip means the ground-speed of the contact point is zero:

Before reading on: the center moves forward at vcm; the rim spins backward relative to the center at ωR. For the contact point to be instantaneously at rest, what must relate vcm and ωR?
contact velocity
=
vcm − ωR = 0
Forward trip motion minus backward rim motion. No slip ⇒ they cancel exactly.
⇒
vcm = ωR
The constraint. Translation and rotation aren’t independent — one number determines the other.

Same result from arc length: rolling through angle Δθ unrolls arc RΔθ onto the ground, so the center advances Δx = RΔθ — differentiate and vcm = Rω again. Two pictures, one lock.

Derivation

Part 1: the no-slip constraint, from the contact point’s velocity. Part 2: the energy split — a rolling body’s kinetic energy is translation of the CM plus rotation about the CM.

vcontact
=
vcm + (−ωR)
Step 1 — add the motions. The contact point rides the center forward at vcm, while rotation drags it backward at rim speed ωR. (Top of the wheel: vcm + ωR = 2vcm — the top moves twice as fast as the center!)
no slip:
⇒
vcm − ωR = 0  ⇒  vcm = ωR
Step 2 — impose the constraint. “Without slipping” means vcontact = 0. Solve: vcm = ωR. ∎ (constraint)
K
=
½mvcm² + ½Icmω²
Step 3 — split the energy. Any rigid motion = translation of the CM + rotation about the CM (König’s theorem). The CM carries ½mvcm²; the spin about it carries ½Iω². They add — no cross term. ∎ (energy)

Why I about the CM here, not about the contact point? The split is defined that way: translation of the CM plus rotation about the CM. (You can also write K = ½Icontactω² via the parallel-axis theorem — same number, one term. Both are correct; the split form is more useful.)

How to use it

The procedure, every time:

  1. Confirm “rolls without slipping.” Those exact words (or “rolls” in a physics problem) license vcm = ωR. Slipping/skidding → stop, different problem.
  2. Kill one variable: use vcm = ωR to eliminate whichever of v, ω you don’t know.
  3. Write both KE terms: K = ½mvcm² + ½Iω². Forgetting the spin term is the #1 rolling error.
  4. Energy problems: mgh = ½mv² + ½Iω². With I = cMR² and ω = v/R this becomes mgh = ½mv²(1 + c).
v = √(2gh / (1 + c))the downhill-race formula: I = cMR² — smaller c winsSay it: “the race speed is the square root of two g h over one plus c”
Common mistake: using v = ωR with the diameter, or forgetting ω must be in rad/s. R is the radius; ω is radians per second — both, every time.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: how fast is the wheel spinning?

  1. Identify. Bike wheel rolling without slipping: R = 0.35 m, vcm = 7 m/s.
  2. Constraint. ω = v/R = 7/0.35 = 20 rad/s.
  3. Read it. About 191 rpm — and the top of the tire moves at 2v = 14 m/s while the contact patch is instantaneously at rest.
Common mistake: ω = v×R (multiplying). Dimensional check kills it: (m/s)×m = m²/s ≠ 1/s. Division gives 1/s ✓.
Your turn — Wheel R = 0.4 m rolling at v = 8 m/s. ω = ?

Answer: 20 rad/s. ω = 8/0.4 = 20 rad/s.

Example 2 — both KE terms: a rolling hoop

  1. Setup. Hoop M = 2 kg, R = 0.3 m, rolling at v = 4 m/s. I = MR² = 2 × 0.09 = 0.18 kg·m².
  2. ω. ω = 4/0.3 ≈ 13.33 rad/s.
  3. Translation. ½mv² = 0.5 × 2 × 16 = 16 J.
  4. Rotation. ½Iω² = 0.5 × 0.18 × 177.8 = 16 J — equal! (For a hoop, c = 1, so the split is always 50/50.)
  5. Total. K = 32 J.
Common mistake: reporting 16 J — the translation alone. A rolling object’s spin energy is real energy; dropping it halves the answer for a hoop.
Your turn — Solid disk M = 4 kg, R = 0.25 m rolling at v = 6 m/s. Total K = ?

Answer: 108 J. ω = 6/0.25 = 24 rad/s; I = 0.5 × 4 × 0.0625 = 0.125; K = 0.5 × 4 × 36 + 0.5 × 0.125 × 576 = 72 + 36 = 108 J. (Disk c = 1/2: spin gets 1/3 of the total.)

Before reading on: hoop (c=1), disk (c=1/2), sphere (c=2/5) race from height h. Rank them before you compute — which shape’s small c pays off?

Example 3 — the famous downhill race

  1. Energy. mgh = ½mv²(1 + c) ⇒ v = √(2gh/(1+c)). Mass and radius cancel — only the shape matters.
  2. h = 2 m. 2gh = 39.2.
  3. Hoop (c=1): v = √(39.2/2) = √19.6 ≈ 4.43 m/s.
  4. Disk (c=1/2): v = √(39.2/1.5) = √26.13 ≈ 5.11 m/s.
  5. Sphere (c=2/5): v = √(39.2/1.4) = √28 ≈ 5.29 m/s — wins 🏆.
  6. The lesson. Smaller c → less energy trapped in spin → more for speed. A marble beats a tractor tire, every time.
Common mistake: “heavier wins” or “bigger wins.” M and R both cancel in v = √(2gh/(1+c)). The race is shape-only — the most counterintuitive result in the unit.
Your turn — Same three racers, h = 3 m. All three speeds?

Answer: hoop 5.42, disk 6.26, sphere 6.48 m/s. 2gh = 58.8: hoop √(58.8/2) = √29.4 = 5.42; disk √(58.8/1.5) = √39.2 = 6.26; sphere √(58.8/1.4) = √42 = 6.48 m/s. Sphere still wins.

Example 4 — car wheel rpm from road speed

  1. Setup. Wheel R = 0.33 m, car at v = 15 m/s (~34 mph).
  2. ω. ω = 15/0.33 ≈ 45.45 rad/s.
  3. rpm. 45.45 × 60/(2π) = 2727/6.283 ≈ 434 rpm.
  4. Read it. Your wheels spin hundreds of rpm at city speeds — and the no-slip constraint is what your speedometer is secretly measuring.
Common mistake: rpm = ω × 60 (forgetting /2π) — giving 2727 rpm, 2π× too big. rpm = ω × 60/(2π).
Your turn — Wheel R = 0.3 m at v = 20 m/s. rpm = ?

Answer: ≈ 637 rpm. ω = 20/0.3 = 66.67 rad/s; rpm = 66.67 × 60/6.283 = 636.6 ≈ 637 rpm.

Memorization tips

  • Say it aloud: “v equals omega R — no slip, no slide.”
  • The tape picture: a rolling wheel unrolls like tape — arc RΔθ laid down as the center advances. If you see the tape, you own v = ωR.
  • Two-term KE: trip (½mv²) + spin (½Iω²). Write both terms before you compute either.
  • Race formula: v = √(2gh/(1+c)) with c = 1, 1/2, 2/5. Smaller c wins; M and R cancel — say “shape only” when you use it.
  • Top = 2v, contact = 0: the velocity profile (top twice the center, contact at rest) is the quickest no-slip self-check.
  • “Rolls” = no slip in physics problems unless slipping is stated. The magic words license the constraint.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and rolling is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What does “rolling without slipping” mean?

The contact point is instantaneously at rest relative to the ground — the wheel unrolls like tape, laying down arc length R·Δθ as it advances. This gives the constraint vcm = ωR.

Why is total KE ½mv² + ½Iω²?

A rolling body’s motion splits into translation of the center of mass plus rotation about the center of mass. Each carries its own kinetic energy, and they add: ½m·vcm² for the trip, ½Iω² for the spin.

Which wins a race rolling downhill: hoop, disk, or sphere?

The solid sphere, then the disk, then the hoop. With I = c·M·R², energy gives v = √(2gh/(1+c)): smaller c means less energy locked in spin, so more goes to speed. Sphere c = 2/5 wins.

Does mass or radius affect who wins the race?

No — both cancel out. v = √(2gh/(1+c)) depends only on the shape factor c and the height. A tiny marble beats a huge hoop.

What if the wheel is slipping?

Then vcm ≠ ωR: the contact point slides, kinetic friction dissipates energy, and the simple energy split needs a friction work term. The formulas on this page assume no slip.

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