Rolling without slipping — v = Rω condition: Worked examples

This page is a drill through every case the rolling constraint $v_{c m} = R ω$ can throw at you. We first map out all the scenario types in a matrix, then work an example for each cell. Nothing new is assumed — if a symbol appears, it was built in the parent parent note or below.

Recall The three constraint forms (from the parent)

$x_{c m} = R θ v_{c m} = R ω a_{c m} = R α$ Here $R$ = radius (metres), $θ$ = spin angle in radians, $ω$ = spin rate (rad/s), $α$ = spin acceleration (rad/s²), and the "cm" subscript means "of the centre of mass". These hold only for rolling without slipping.

The scenario matrix

Every problem this topic can throw is one of these cells. The examples below each carry a tag like (Cell B) so you can see the whole space is covered.

Cell	Case class	What makes it special
A	Find $ω$ from $v$ (or reverse)	The plain constraint, one unknown
B	Acceleration form $a = R α$	Differentiated constraint
C	Point velocities (top / bottom / side)	Uses $v_{p o in t} = v_{c m} + ω \times r$
D	Sign / direction of spin	$ω$ can be $+$ or $-$ ; forward vs backward roll
E	Zero / degenerate input	$ω = 0$ , $v = 0$ , or $R \to$ tiny
F	Slipping — constraint FAILS	$v \neq = R ω$ ; must not use the formula
G	Limiting / arbitrary point on rim	Speed of a general point at angle $ϕ$
H	Real-world word problem	Odometer, gear, distance travelled
I	Exam twist	Two wheels / relative spin / mixed units

Example set

Worked example Example 4 — Any point on the rim

(Cell C, G) A ball rolls at $v = 3 m/s$ , radius $R = 0.2 m$ . Find the speed of the point at the side of the wheel (the 3-o'clock point, level with the centre).

Forecast: It's not the top ( $2 v$ ) and not the bottom ( $0$ ). Guess: somewhere in between.

Step 1 — Split into two velocities. Translation gives every point $v = 3 m/s$ forward (horizontal). Rotation gives the side point $R ω = v = 3 m/s$ — but the rotation velocity is perpendicular to the spoke, so at the 3-o'clock point it points straight up (or down). Why this step? $v_{p o in t} = v_{c m} + ω \times r$ : the cross-product $ω \times r$ is always perpendicular to the radius $r$ . Look at the figure — the two arrows are at right angles.

Step 2 — Combine with Pythagoras. They are perpendicular, so $v_{s i d e} = v^{2} + (R ω)^{2} = 3^{2} + 3^{2} = 18 = 4.24 m/s .$ Why this step? Perpendicular vectors add like the two legs of a right triangle.

Verify: Value $4.24$ sits between $0$ (bottom) and $6$ (top) ✓. Since $R ω = v$ here, $v_{s i d e} = v 2 \approx 1.41 v$ — a clean check. $v_{s i d e} \approx 4.24 m/s$

Worked example Example 5 — General angle & the two extreme points

(Cell G) For a wheel rolling at $v$ (so $R ω = v$ ), a point on the rim makes angle $ϕ$ measured from the bottom contact point (going around). Show the speed is $v_{ϕ} = 2 v sin (ϕ /2)$ , then check it at bottom, side, and top.

Forecast: A formula that gives $0$ at the bottom and $2 v$ at the top. Guess whether it grows smoothly.

Step 1 — Why this trick works. In pure rolling the contact point is the instantaneous axis (parent note). So every rim point momentarily rotates about the bottom with speed $= ω \times (distance from contact point)$ . Why this step? Treating the whole wheel as spinning about the fixed contact point turns "translation + rotation" into one simple rotation.

Step 2 — Distance from contact to the point. A point at rim-angle $ϕ$ is a chord away from the contact point; that chord length is $d = 2 R sin (ϕ /2)$ . Why this step? The chord of a circle subtending central angle $ϕ$ has length $2 R sin (ϕ /2)$ — standard circle geometry (see the figure's dashed chord).

Step 3 — Multiply by $ω$ . $v_{ϕ} = ω d = ω \cdot 2 R sin (ϕ /2) = 2 (R ω) sin (ϕ /2) = 2 v sin (ϕ /2)$ . Why this step? Speed about a pivot $=$ (angular rate) $\times$ (distance to pivot), and $R ω = v$ .

Verify (all three cases):

Bottom $ϕ = 0$ : $2 v sin 0 = 0$ ✓ (Cell E degenerate: contact at rest).
Side $ϕ = 9 0^{\circ}$ : $2 v sin 4 5^{\circ} = 2 v \cdot 0.707 = 1.41 v$ ✓ (matches Example 4).
Top $ϕ = 18 0^{\circ}$ : $2 v sin 9 0^{\circ} = 2 v$ ✓. $v_{ϕ} = 2 v sin (ϕ /2)$

Worked example Example 6 — Direction / sign of spin

(Cell D) A wheel rolls to the left at $v = 2 m/s$ , $R = 0.25 m$ . Give $ω$ with its correct sign (take counter-clockwise as $+$ ) and the velocity of the top point (magnitude and direction).

Forecast: Rolling left ⇒ the wheel spins the opposite way from rolling right. Guess the sign flips.

Step 1 — Magnitude from the constraint. $∣ ω ∣ = v / R = 2/0.25 = 8 rad/s$ . Why this step? The constraint links magnitudes; sign comes from geometry, not from the equation.

Step 2 — Fix the sign. Rolling left, the top of the wheel moves left while the bottom (contact) stays put — that is clockwise, which is the negative sense. So $ω = - 8 rad/s$ . Why this step? A wheel rolling in the direction its top leans; leftward motion ⇒ clockwise spin ⇒ negative by our convention.

Step 3 — Top point velocity. Top speed $= 2 v = 4 m/s$ , pointing left (same direction as the roll). Why this step? The top always moves in the direction of travel at twice centre speed, regardless of sign convention.

Verify: $∣ R ω ∣ = 0.25 \times 8 = 2 = v$ ✓. Sign check: if it rolled right we'd get $ω = + 8$ ; leftward flips it to $- 8$ ✓. $ω = - 8 rad/s, v_{t o p} = 4 m/s left$

Worked example Example 7 — Degenerate inputs

(Cell E) Two quick sub-cases. (a) A wheel spins with $ω = 0$ while its centre glides forward at $v = 1 m/s$ on frictionless ice — is this rolling without slipping? (b) A wheel is pinned at its centre and spun at $ω = 10 rad/s$ , $R = 0.1 m$ , but the centre cannot move ( $v = 0$ ). What is the contact-point speed?

Forecast: Both violate $v = R ω$ . Guess which slides forward and which slides backward.

Step 1 — Test (a) against the constraint. $R ω = R \cdot 0 = 0 \neq = v = 1$ . Constraint fails ⇒ it is sliding, not rolling. Why this step? Rolling without slipping requires $v = R ω$ ; if it fails, there is slipping. The contact point moves forward at $v - R ω = 1 - 0 = 1 m/s$ — pure skid.

Step 2 — Test (b). $R ω = 0.1 \times 10 = 1 m/s$ , but $v = 0$ , so $v \neq = R ω$ . The contact-point velocity $= v - R ω = 0 - 1 = - 1 m/s$ (backward). This is a spinning tyre on ice: rubber flies backward past the ground. Why this step? Same test; here the contact point slips backward — opposite of (a).

Verify: (a) contact slips $+ 1$ m/s forward; (b) contact slips $- 1$ m/s (backward). Neither is $0$ , so neither is true rolling ✓. This is exactly the parent's " $v = R ω$ is NOT always true" warning. $(a) slip + 1 m/s, (b) slip - 1 m/s$

Worked example Example 8 — Real-world word problem

(Cell H) A bike wheel of radius $R = 0.34 m$ makes $N = 1000$ full turns without slipping. How far did the bike travel, and if it took $t = 200 s$ , what was the average $v$ and $ω$ ?

Forecast: Each turn lays out one circumference. Guess a distance in the low thousands of metres.

Step 1 — Angle in radians. $θ = N \cdot 2 π = 1000 \times 2 π = 6283.2 rad$ . Why this step? The constraint $x_{c m} = R θ$ needs $θ$ in radians; one turn is $2 π$ rad.

Step 2 — Distance. $x = R θ = 0.34 \times 6283.2 = 2136.3 m$ . Why this step? No-slip means ground distance = total rim unrolled.

Step 3 — Averages. $v = x / t = 2136.3/200 = 10.68 m/s$ ; $ω = v / R = 10.68/0.34 = 31.4 rad/s$ . Why this step? Average speed is distance over time; then the constraint gives $ω$ .

Verify: Cross-check via circumference: $x = N \cdot 2 π R = 1000 \times 2 π \times 0.34 = 2136.3 m$ ✓. And $ω = θ / t = 6283.2/200 = 31.4 rad/s$ matches ✓. $x \approx 2136 m, v \approx 10.7 m/s, ω \approx 31.4 rad/s$

Worked example Example 9 — Exam twist: two coupled wheels

(Cell I) A big wheel ( $R_{1} = 0.4 m$ ) and a small wheel ( $R_{2} = 0.1 m$ ) are on the same axle rolling without slipping on the same road (a toy-train double wheel). If the small wheel is the one actually on the track and the axle centre moves at $v = 1.2 m/s$ , find the common $ω$ and the speed of the top of the big wheel.

Forecast: Same axle ⇒ same $ω$ . The big rim, spinning at that $ω$ , will overshoot the ground — it must be above the track. Guess whether its top exceeds $2 v$ .

Step 1 — Which wheel sets $ω$ ? Only the contact wheel (radius $R_{2}$ ) obeys $v = R_{2} ω$ with the ground. So $ω = v / R_{2} = 1.2/0.1 = 12 rad/s$ . Why this step? The no-slip constraint applies to the wheel touching the surface, not to a rim that isn't in contact.

Step 2 — Top of the big wheel. Its top speed $= v + R_{1} ω = 1.2 + 0.4 \times 12 = 1.2 + 4.8 = 6.0 m/s$ . Why this step? Top point = translation $v$ + rotation $R_{1} ω$ (both forward). Note $R_{1} ω = 4.8 \neq = v$ , so this is NOT the simple " $2 v$ " case — the big rim is not the rolling one.

Verify: Sanity: the big wheel's bottom would be $v - R_{1} ω = 1.2 - 4.8 = - 3.6 m/s$ (backward) — which is fine because that point is above the track and free to move. $ω = 12 rad/s, v_{t o p, bi g} = 6.0 m/s$

Recall Coverage check — did we hit every cell?

Cell A: Ex 1, 2 ::: plain constraint both directions Cell B: Ex 3 ::: acceleration form $a = R α$ Cell C: Ex 4 ::: side-point velocity Cell D: Ex 6 ::: sign of $ω$ (leftward roll) Cell E: Ex 5 (bottom), Ex 7 ::: degenerate / zero inputs Cell F: Ex 7 ::: slipping, constraint fails Cell G: Ex 5 ::: general rim angle $2 v sin (ϕ /2)$ Cell H: Ex 8 ::: odometer word problem Cell I: Ex 9 ::: two coupled wheels

Connections

Instantaneous axis of rotation — the pivot trick used in Example 5
Angular velocity and angular acceleration — signs of $ω$ in Example 6
Static vs kinetic friction — slipping cases (Example 7) involve kinetic friction
Kinetic energy of rolling bodies — next step once point speeds are understood
Rolling down an incline — where $a = R α$ (Example 3) gets used
Moment of inertia — needed to turn these kinematics into dynamics