Classical-Mechanics · Unit 24 · Video 1 · Interactive Practice

Why the Bottom of a Rolling Wheel Stands Still: Rolling Without Slipping

IKey Formulas

Formula	Name	What it says
$\vec{v}_P = \vec{v}'_P + \vec{V}_{cm},\qquad \vec{v}'_P = R\,\omega_{cm}\,\hat{\boldsymbol{\theta}}$	Law of addition of velocities	Ground-frame velocity of a rim point $P$ = its velocity in the center-of-mass frame (uniform circular motion at tangential speed $R\,\omega_{cm}$ ) plus the velocity of that frame
$\vec{v}_P = (V_{cm} + R\,\omega_{cm}\cos\theta)\,\hat{\mathbf{i}} - R\,\omega_{cm}\sin\theta\,\hat{\mathbf{j}}$	Any rim point, ground frame	$\theta$ measured clockwise from the top of the wheel; uses $\hat{\boldsymbol{\theta}} = \cos\theta\,\hat{\mathbf{i}} - \sin\theta\,\hat{\mathbf{j}}$ . Translation feeds only the $\hat{\mathbf{i}}$ part, rotation feeds both
$\Delta X_{cm} = V_{cm}\,\Delta t,\qquad \Delta s = R\,\Delta\theta = R\,\omega_{cm}\,\Delta t$	Displacements in $\Delta t$	How far the center travels along the ground vs. how much rim passes the bottom of the wheel
$V_{cm} = R\,\omega_{cm}$	Rolling without slipping	$\Delta X_{cm} = \Delta s$ ; $V_{cm} > R\,\omega_{cm}$ is skidding (center outruns rim), $V_{cm} < R\,\omega_{cm}$ is slipping (rim outruns center)

Key Insight: At $\theta = \pi$ the formula gives $\vec{v}_P = (V_{cm} - R\,\omega_{cm})\,\hat{\mathbf{i}}$ : the forward translation and the backward rotation cancel exactly when $V_{cm} = R\,\omega_{cm}$ , so the contact point is momentarily at rest, the top moves at $2V_{cm}$ , and any friction at the contact is static.

IITranslation Plus Rotation at Any Rim Point

Every rim point carries the same forward $\vec{V}_{cm}$ ; the rotation's $\vec{v}'_P$ adds to it at the top and cancels it at the bottom.

\theta

(from top) 135°

R\,\omega_{cm}

(m/s) 2.0

💡 Challenge: with $R\,\omega_{cm} = 4.0\ \text{m/s}$ (slipping, since $V_{cm} = 2.0\ \text{m/s}$ ), find the rim point whose velocity points straight down.

IIIDistance Traveled vs. Rim Passed

Comparing $\Delta X_{cm} = V_{cm}\,\Delta t$ with $\Delta s = R\,\omega_{cm}\,\Delta t$ sorts every wheel into rolling, skidding or slipping.

Case

t

(s) 0.00

IVQuiz Questions

Problem 1 · Angular Speed of a Rolling Tire

Given: A car tire of radius $R = 0.30\ \text{m}$ rolls without slipping while the car moves at $V_{cm} = 12\ \text{m/s}$ — find the angular speed $\omega_{cm}$ .

✅ Correct!

\omega_{cm} = V_{cm}/R = 12/0.30 = 40\ \text{rad/s}

❌ Not quite.

V_{cm}R

has units of

\text{m}^2/\text{s}

, not

\text{rad/s}

. The condition is

V_{cm} = R\,\omega_{cm}

, so divide.

❌ Close, but… that uses the diameter

0.60\ \text{m}

. The tangential speed of the rim is

R\,\omega_{cm}

with the radius.

❌ Not quite.

R/V_{cm}

is upside down — it has units of seconds. Solve

V_{cm} = R\,\omega_{cm}

for

\omega_{cm}

❌ Not quite. Rolling without slipping means

V_{cm} = R\,\omega_{cm}

Show solution

Rolling without slipping means $\Delta X_{cm} = \Delta s$ , i.e. $V_{cm}\,\Delta t = R\,\omega_{cm}\,\Delta t$ , so

\omega_{cm} = \frac{V_{cm}}{R} = \frac{12\ \text{m/s}}{0.30\ \text{m}} = 40\ \text{rad/s}.

In the center-of-mass frame the rim moves at $R\,\omega_{cm} = 12\ \text{m/s}$ — the same speed as the car, which is what lets the contact point stand still.

Problem 2 · Skidding or Slipping?

Given: A wheel of radius $R = 0.40\ \text{m}$ has $V_{cm} = 6.0\ \text{m/s}$ to the right ( $+\hat{\mathbf{i}}$ ) and turns clockwise at $\omega_{cm} = 10\ \text{rad/s}$ — find the ground-frame velocity of the contact point and name the motion.

Velocity of the contact point ( $\theta = \pi$ )?

What is the wheel doing?

✅ Correct!

R\,\omega_{cm} = 4.0\ \text{m/s} < V_{cm}

, so the contact point slides forward at

2.0\ \text{m/s}

: the center outruns the rim — skidding.

❌ Not quite. The contact point is at rest only if

V_{cm} = R\,\omega_{cm}

. Here

R\,\omega_{cm} = 0.40 \times 10 = 4.0\ \text{m/s} \neq 6.0\ \text{m/s}

❌ Check the sign. At the bottom the rotation contributes

-R\,\omega_{cm}\,\hat{\mathbf{i}} = -4.0\,\hat{\mathbf{i}}

and the translation

+6.0\,\hat{\mathbf{i}}

; the sum is positive.

❌ Not quite.

V_{cm} + R\,\omega_{cm}

is the velocity of the top (

\cos 0 = 1

). At the bottom

\cos\pi = -1

, so the speeds subtract.

❌ Close, but the names are swapped. Slipping means the rim outruns the center (

V_{cm} < R\,\omega_{cm}

). Here the center outruns the rim.

❌ Not quite. Use

\vec{v}_P(\theta = \pi) = (V_{cm} - R\,\omega_{cm})\,\hat{\mathbf{i}}

❌ Not quite. Compare

V_{cm} = 6.0\ \text{m/s}

with

R\,\omega_{cm} = 4.0\ \text{m/s}

Show solution

Tangential speed of the rim in the center-of-mass frame: $R\,\omega_{cm} = (0.40\ \text{m})(10\ \text{rad/s}) = 4.0\ \text{m/s}$ .

At the bottom, $\theta = \pi$ : $\cos\pi = -1$ , $\sin\pi = 0$ , so

\vec{v}_P(\theta = \pi) = (V_{cm} - R\,\omega_{cm})\,\hat{\mathbf{i}} = (6.0 - 4.0)\,\hat{\mathbf{i}} = 2.0\,\hat{\mathbf{i}}\ \text{m/s}.

The contact point slides forward over the ground. Since $V_{cm} > R\,\omega_{cm}$ , in any interval $\Delta X_{cm} > \Delta s$ : the center travels farther than the rim turns, which is skidding.

Problem 3 · A Rim Point Partway Round

Given: A wheel rolls without slipping to the right with $V_{cm} = 4.0\ \text{m/s}$ . Point $P$ on the rim is at $\theta = 60\degree$ , measured clockwise from the top — find its ground-frame velocity and speed.

Velocity $\vec{v}_P$ ?

Speed $|\vec{v}_P|$ ?

✅ Correct!

\vec{v}_P = (6.0\,\hat{\mathbf{i}} - 3.46\,\hat{\mathbf{j}})\ \text{m/s}

, speed

\sqrt{48} \approx 6.93\ \text{m/s}

— between

0

at the bottom and

2V_{cm} = 8.0\ \text{m/s}

at the top.

❌ Check sine and cosine.

\theta

is measured from the vertical, so

\hat{\boldsymbol{\theta}} = \cos\theta\,\hat{\mathbf{i}} - \sin\theta\,\hat{\mathbf{j}}

: the

\hat{\mathbf{i}}

part of the rotation carries

\cos\theta

❌ Check the sign of the $\hat{\mathbf{j}}$ part. On the front of a clockwise-turning wheel the rim moves down:

-R\,\omega_{cm}\sin\theta\,\hat{\mathbf{j}}

❌ Check the $\hat{\mathbf{i}}$ part. It is

V_{cm} + R\,\omega_{cm}\cos\theta = 4.0 + 4.0\cos 60\degree = 6.0

; the minus sign belongs only to the

\hat{\mathbf{j}}

term.

❌ Not quite.

V_{cm} + R\,\omega_{cm} = 8.0

adds the two speeds as if the vectors were parallel — true only at the top. Add the vectors, then take the magnitude.

❌ Not quite. Use

\vec{v}_P = (V_{cm} + R\,\omega_{cm}\cos\theta)\,\hat{\mathbf{i}} - R\,\omega_{cm}\sin\theta\,\hat{\mathbf{j}}

with

R\,\omega_{cm} = V_{cm}

❌ Not quite.

|\vec{v}_P| = \sqrt{v_x^2 + v_y^2}

with the components you found.

Show solution

Rolling without slipping: $R\,\omega_{cm} = V_{cm} = 4.0\ \text{m/s}$ . With $\cos 60\degree = \tfrac{1}{2}$ and $\sin 60\degree = \tfrac{\sqrt{3}}{2}$ :

\vec{v}_P = \left(4.0 + 4.0 \cdot \tfrac{1}{2}\right)\hat{\mathbf{i}} - 4.0 \cdot \tfrac{\sqrt{3}}{2}\,\hat{\mathbf{j}} = (6.0\,\hat{\mathbf{i}} - 2\sqrt{3}\,\hat{\mathbf{j}})\ \text{m/s} \approx (6.0\,\hat{\mathbf{i}} - 3.46\,\hat{\mathbf{j}})\ \text{m/s}.

|\vec{v}_P| = \sqrt{6.0^2 + (2\sqrt{3})^2} = \sqrt{36 + 12} = \sqrt{48} \approx 6.93\ \text{m/s}.

Check: the same formula gives $0$ at $\theta = \pi$ and $2V_{cm} = 8.0\ \text{m/s}$ at $\theta = 0$ ; $6.93\ \text{m/s}$ lies between, as it should for a point on the upper front of the rim.

Problem 4 · Drive Wheel on Ice

Given: A car's drive wheel, radius $R = 0.35\ \text{m}$ , spins clockwise at $\omega_{cm} = 30\ \text{rad/s}$ on ice while the car creeps forward at $V_{cm} = 2.5\ \text{m/s}$ . Over $\Delta t = 0.20\ \text{s}$ — find how much more rim passes the bottom than the distance the wheel travels, and the velocity of the contact point.

$\Delta s - \Delta X_{cm}$ ?

Velocity of the contact point?

✅ Correct! The rim laid past the bottom exceeds the distance traveled by

1.6\ \text{m}

, and the tread scrapes backward over the ice at

8.0\ \text{m/s}

— slipping.

❌ Not quite.

2.1\ \text{m}

\Delta s

alone. Subtract the

\Delta X_{cm} = V_{cm}\,\Delta t

the wheel actually traveled.

❌ Not quite.

0.50\ \text{m}

\Delta X_{cm}

alone, the distance traveled. You need the difference

\Delta s - \Delta X_{cm}

❌ Not quite. That adds the two displacements. The question compares them: subtract.

❌ Check the sign.

V_{cm} - R\,\omega_{cm} = 2.5 - 10.5

is negative: the rim outruns the center, so the bottom of the tread moves backward.

❌ Not quite.

V_{cm} + R\,\omega_{cm} = 13\ \text{m/s}

is the velocity of the top of the wheel. At the bottom the rotation term enters with

\cos\pi = -1

❌ Not quite. The contact point is at rest only when

V_{cm} = R\,\omega_{cm}

; here

R\,\omega_{cm} = 10.5\ \text{m/s}

far exceeds

V_{cm}

❌ Not quite. Compute

\Delta s = R\,\omega_{cm}\,\Delta t

and

\Delta X_{cm} = V_{cm}\,\Delta t

separately, then subtract.

❌ Not quite. Use

\vec{v}_P(\theta = \pi) = (V_{cm} - R\,\omega_{cm})\,\hat{\mathbf{i}}

Show solution

Tangential speed of the rim: $R\,\omega_{cm} = (0.35\ \text{m})(30\ \text{rad/s}) = 10.5\ \text{m/s}$ .

\Delta X_{cm} = V_{cm}\,\Delta t = (2.5)(0.20) = 0.50\ \text{m}, \qquad \Delta s = R\,\omega_{cm}\,\Delta t = (10.5)(0.20) = 2.1\ \text{m}

\Delta s - \Delta X_{cm} = 2.1 - 0.50 = 1.6\ \text{m}.

Since $\Delta X_{cm} < \Delta s$ (equivalently $V_{cm} < R\,\omega_{cm}$ ), the rim outruns the center: the wheel is slipping. The contact point moves at

\vec{v}_P(\theta = \pi) = (V_{cm} - R\,\omega_{cm})\,\hat{\mathbf{i}} = (2.5 - 10.5)\,\hat{\mathbf{i}} = -8.0\,\hat{\mathbf{i}}\ \text{m/s},

so the tread rubs backward over the ice — the friction there is kinetic, not static.

Solved: 0 / 4