Classical-Mechanics · Unit 24 · Video 2 · Interactive Practice

Circle Plus Straight Line: A Bead on a Rolling Wheel and the Curtate Cycloid

IKey Formulas

Formula	Name	What it says
$\theta(t) = \omega_{cm}t = \dfrac{V_{cm}}{R}\,t$	Rolling without slipping	The spoke angle, clockwise from the top, grows steadily
$\vec{r}'_b = b\,\hat{r}, \quad \vec{v}'_b = \dfrac{bV_{cm}}{R}\,\hat{\theta}, \quad \vec{a}'_b = -\dfrac{bV_{cm}^2}{R^2}\,\hat{r}$	Centre-of-mass frame	Uniform circular motion of radius $b$
$x_b = V_{cm}t + b\sin\theta, \quad y_b = R + b\cos\theta$	Ground frame position	$\vec{r}_b = \vec{R}_{cm} + \vec{r}'_b$ : a curtate cycloid for $b \lt R$ , a cycloid for $b = R$
$\vec{v}_b = \vec{V}_{cm} + \vec{v}'_b, \quad \vec{a}_b = \vec{a}'_b$	Adding the motions	$\vec{A}_{cm} = \vec{0}$ , so the acceleration is the same in both frames

Key Insight: Relative to the centre the bead goes round a circle; the centre goes along a straight line. Positions and velocities add, and because the centre does not accelerate, the ground-frame acceleration is the circular one alone: size $bV_{cm}^2/R^2$ , always pointing at the axle.

IIOne Motion, Two Frames

The bead that circles the axle in the centre-of-mass frame traces a wave along the road in the ground frame.

Frame Time t (s) 0.100

IIIVelocities Add, Tip to Tail

The bead's ground velocity is $\vec{V}_{cm}$ plus its circular velocity $\frac{b}{R}V_{cm}\,\hat{\theta}$ about the centre.

💡 Challenge: find the one point of the wheel whose ground-frame velocity is zero, and the point that moves fastest.

IVFrom Curtate Cycloid to Cycloid

Moving the bead out to the rim slows its lowest point to rest and sharpens each trough into a cusp.

Bead distance b / R 0.50

VQuiz Questions

Problem 1 · Speed About the Centre

Given: A wheel of radius $R = 0.30\ \text{m}$ rolls without slipping with $V_{cm} = 6\ \text{m/s}$ . A bead is fixed to a spoke at $b = 0.10\ \text{m}$ from the axle. Find the bead's speed in the centre-of-mass frame.

✅ Correct!

\omega_{cm} = V_{cm}/R = 20\ \text{rad/s}

, and the bead circles at

b\,\omega_{cm} = 2\ \text{m/s}

— the fraction

b/R = \tfrac{1}{3}

of the centre's speed.

❌ That is a rim point's speed. Only a point at

b = R

circles at

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

; the bead is a third of the way out.

❌ Check the units.

bV_{cm}

has units

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

; the speed is

b\,\omega_{cm} = bV_{cm}/R

❌ That is a ground-frame speed.

V_{cm} + bV_{cm}/R = 8\ \text{m/s}

is the bead's speed at the top of its circle as seen from the road; in the centre-of-mass frame the

V_{cm}

term is absent.

❌ Not quite. Find

\omega_{cm}

from rolling without slipping, then multiply by

b

Show solution

Rolling without slipping gives

\omega_{cm} = \frac{V_{cm}}{R} = \frac{6\ \text{m/s}}{0.30\ \text{m}} = 20\ \text{rad/s}.

In the centre-of-mass frame the bead moves on a circle of radius $b$ , so

|\vec{v}'_b| = b\,\omega_{cm} = \frac{bV_{cm}}{R} = (0.10\ \text{m})(20\ \text{rad/s}) = 2\ \text{m/s}.

Problem 2 · Acceleration at the Bottom

Given: The same wheel ( $R = 0.30\ \text{m}$ , $V_{cm} = 6\ \text{m/s}$ , $b = 0.10\ \text{m}$ ). Find the bead's ground-frame acceleration at the instant it is directly below the axle.

✅ Correct!

\vec{A}_{cm} = \vec{0}

, so

\vec{a}_b = \vec{a}'_b

: size

bV_{cm}^2/R^2 = 40\ \text{m/s}^2

, pointing from the bead to the axle — straight up at the bottom.

❌ It is the centre that does not accelerate.

\vec{A}_{cm} = \vec{0}

only removes the frame term; the bead still has its circular acceleration

\vec{a}'_b

about the axle.

❌ That is $V_{cm}^2/R$ . It would be the acceleration of a rim point; the bead circles at radius

b

with speed

bV_{cm}/R

, giving

(bV_{cm}/R)^2/b = bV_{cm}^2/R^2

❌ Wrong speed in the centripetal formula.

v^2/b

with the ground speed

V_{cm}(1 - b/R) = 4\ \text{m/s}

gives

160\ \text{m/s}^2

, but the bead's circle of radius

b

belongs to the centre-of-mass frame, where its speed is

2\ \text{m/s}

❌ Not quite. The ground-frame acceleration equals the centre-of-mass-frame acceleration here.

Show solution

The frames are related by $\vec{a}_b = \vec{a}'_b + \vec{A}_{cm}$ , and $\vec{A}_{cm} = \vec{0}$ because the centre moves at constant velocity. So

\vec{a}_b = -\frac{bV_{cm}^2}{R^2}\,\hat{r}, \qquad |\vec{a}_b| = \frac{(0.10)(6)^2}{(0.30)^2}\ \text{m/s}^2 = \frac{3.6}{0.09}\ \text{m/s}^2 = 40\ \text{m/s}^2.

At the bottom, $\theta = \pi$ and $\hat{r}$ points down, so $-\hat{r}$ points straight up: $\vec{a}_b = 40\ \text{m/s}^2\ \hat{j}$ . This upward acceleration turns the path back up out of each trough.

Problem 3 · The Video's Wheel at a Quarter Turn

Given: $V_{cm} = 5\ \text{m/s}$ , $R = 0.25\ \text{m}$ , $b = 0.125\ \text{m}$ , with the bead at the top at $t = 0$ and the ground origin directly below the starting centre. Find the bead's ground-frame position and velocity at $t = \pi/40\ \text{s}$ .

Position $(x_b, y_b)$ in metres?

Velocity $(v_{b,x}, v_{b,y})$ in m/s?

✅ Correct! A quarter turn puts the bead level with the axle and ahead of it, moving forward with the centre and downward around its circle.

❌ The angle has moved on.

(V_{cm}t,\ R + b)

keeps the bead at the top; at this time

\theta = \pi/2

, so

\sin\theta = 1

and

\cos\theta = 0

❌ Check the direction of $\theta$ .

\theta

is measured clockwise from the top and the wheel turns clockwise, so at

\theta = \pi/2

the bead is ahead of the axle:

x_b = V_{cm}t + b

❌ Check the position.

\theta = (V_{cm}/R)\,t

, then

x_b = V_{cm}t + b\sin\theta

and

y_b = R + b\cos\theta

❌ That is the speed at the top.

V_{cm}(1 + b/R)

holds only at

\theta = 0

; here

\cos\theta = 0

and

\sin\theta = 1

❌ Check the sign of $v_{b,y}$ .

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

; at

\theta = \pi/2

it points straight down, the way a clockwise wheel carries the front of its rim.

❌ That is the centre-of-mass-frame velocity. In the ground frame add

\vec{V}_{cm} = 5\ \text{m/s}\ \hat{i}

❌ Check the velocity.

\vec{v}_b = \vec{V}_{cm} + \frac{b}{R}V_{cm}(\cos\theta\,\hat{i} - \sin\theta\,\hat{j})

Show solution

Angle: $\omega_{cm} = V_{cm}/R = 20\ \text{rad/s}$ , so $\theta = 20 \cdot \pi/40 = \pi/2$ .

Position:

x_b = V_{cm}t + b\sin\theta = 5 \cdot \frac{\pi}{40} + 0.125 = 0.393 + 0.125 = 0.518\ \text{m}

y_b = R + b\cos\theta = 0.25 + 0 = 0.250\ \text{m}

Velocity: with $\frac{b}{R}V_{cm} = 2.5\ \text{m/s}$ ,

v_{b,x} = V_{cm} + \frac{b}{R}V_{cm}\cos\theta = 5 + 0 = 5\ \text{m/s}

v_{b,y} = -\frac{b}{R}V_{cm}\sin\theta = -2.5\ \text{m/s}

The speed is $\sqrt{5^2 + 2.5^2} \approx 5.59\ \text{m/s}$ , between the trough's $2.5\ \text{m/s}$ and the crest's $7.5\ \text{m/s}$ .

Problem 4 · A Pebble in the Tread

Given: A car tyre of radius $R = 0.30\ \text{m}$ rolls without slipping at $V_{cm} = 30\ \text{m/s}$ with a pebble wedged in its tread ( $b = R$ ). Find the pebble's greatest ground-frame speed and its acceleration at the instant it touches the road.

Greatest ground-frame speed?

Acceleration at the contact instant?

✅ Correct! At the top the pebble moves at

2V_{cm} = 60\ \text{m/s}

; at the cusp it stops for an instant yet accelerates at

V_{cm}^2/R = 3000\ \text{m/s}^2 \approx 300g

straight up.

❌ That is its speed about the axle. In the ground frame add

V_{cm}

: at the top both velocities point forward.

❌ That is the speed level with the axle. At

\theta = \pi/2

the two velocities are perpendicular; they line up at

\theta = 0

, the top.

❌ Check the top speed.

v_{\text{top}} = V_{cm}(1 + b/R)

with

b = R

❌ Zero velocity is not zero acceleration. The pebble's acceleration is its circular one,

-\frac{bV_{cm}^2}{R^2}\hat{r}

, at every instant — including the instant it is at rest.

❌ Wrong speed in $v^2/R$ . The circle of radius

R

is the centre-of-mass-frame path, where the pebble's speed is

V_{cm}

, not the ground-frame top speed

2V_{cm}

❌ Check the direction. The acceleration points from the pebble to the axle, and at the contact point the axle is directly above.

❌ Check the acceleration. Put

b = R

into

|\vec{a}_b| = bV_{cm}^2/R^2

Show solution

Top speed: $v_{b,x} = V_{cm}\big(1 + \frac{b}{R}\cos\theta\big)$ and $v_{b,y} = -\frac{b}{R}V_{cm}\sin\theta$ give, with $b = R$ ,

|\vec{v}_b|^2 = V_{cm}^2\big[(1 + \cos\theta)^2 + \sin^2\theta\big] = 2V_{cm}^2(1 + \cos\theta),

largest at $\theta = 0$ : $|\vec{v}_b| = 2V_{cm} = 60\ \text{m/s}$ .

Contact acceleration: $\vec{A}_{cm} = \vec{0}$ , so $\vec{a}_b = -\frac{bV_{cm}^2}{R^2}\hat{r}$ . At the bottom ( $\theta = \pi$ ) with $b = R$ :

\vec{a}_{\text{contact}} = \frac{V_{cm}^2}{R}\,\hat{j} = \frac{(30)^2}{0.30}\ \text{m/s}^2\ \hat{j} = 3000\ \text{m/s}^2\ \hat{j}.

The pebble comes down into the cusp, stops for an instant, and is pulled straight up again.

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