Classical-Mechanics · Unit 24 · Video 3 · Interactive Practice

One Constraint Per System: Rolling, Yo-Yos and Unwinding Drums

IKey Formulas

Formula	System	What it needs
$\Delta X_{cm} = R\,\Delta\theta \;\Rightarrow\; V_{cm} = R\,\omega_{cm} \;\Rightarrow\; A_{cm} = R\,\alpha_{cm}$	Cylinder rolling without slipping down an incline	Only the radius $R$ of the surface touching the plane; $\beta$ , $M$ , $I_{cm}$ and $h$ do not enter
$b\,\Delta\theta = \Delta l \;\Rightarrow\; b\,\omega_{cm} = V_{cm,y} \;\Rightarrow\; b\,\alpha_{cm} = A_{cm,y}$	Falling yo-yo, string fixed at the top	The axle radius $b$ the string leaves from; the outer radius $R$ plays no part
$\Delta y = R\,\Delta\theta_A + R\,\Delta\theta_B \;\Rightarrow\; a_{A,y} = R\,\alpha_A + R\,\alpha_B$	Drum $A$ hanging on a tape from drum $B$	Both rotations: tape unwinds from $A$ and is fed down from $B$ , and the two lengths add

Key Insight: Each constraint comes from one geometric relation, displacement $=$ radius $\times$ angle, differentiated twice. The radius is that of the rim where the body meets whatever holds it, and no mass, force or moment of inertia appears; once forces and torques enter, the constraint supplies the one equation the dynamics is missing.

IIVisualization 1 — Rolling Down the Incline

The length of plane the cylinder has rolled over always equals the arc $R\,\Delta\theta$ its rim has turned through.

time 0.60 β 30° R (m) 1.20

💡 The live speeds use the solid cylinder's $A_{cm} = \tfrac{2}{3}g\sin\beta$ , which comes from the dynamics of the next chapter; the constraint $A_{cm} = R\,\alpha_{cm}$ holds whatever the value of $A_{cm}$ .

IIIVisualization 2 — Which Radius Unwinds the Yo-Yo?

String peels off the axle, so each drop $\Delta l$ turns the yo-yo through $\Delta\theta = \Delta l / b$ , not $\Delta l / R$ .

drop Δl (cm) 3.0 left axle b (cm) 1.2

IVVisualization 3 — Two Drums, Two Contributions

Drum $A$ drops by the tape unwound from $A$ plus the tape fed down from $B$ : $a_{A,y} = R\,\alpha_A + R\,\alpha_B$ .

time (s) 0.60 α_A (rad/s²) 3.0 α_B (rad/s²) 3.0

VQuiz Questions

Problem 1 · Angular Acceleration on the Incline

Given: a uniform cylinder of radius $R = 0.20\ \text{m}$ rolls without slipping down a plane tilted at $\beta = 25\degree$ ; its center accelerates down the plane at $A_{cm} = 3.0\ \text{m/s}^2$ — find the magnitude of its angular acceleration $\alpha_{cm}$ .

✅ Correct!

A_{cm} = R\,\alpha_{cm}

gives

\alpha_{cm} = A_{cm}/R = 3.0/0.20 = 15\ \text{rad/s}^2

❌ Multiplied instead of divided.

R\,A_{cm} = 0.6

has units

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

; the constraint

A_{cm} = R\,\alpha_{cm}

gives

\alpha_{cm} = A_{cm}/R

❌ That divides by the diameter. The arc traced by a rim point is

R\,\Delta\theta

with the radius

R = 0.20\ \text{m}

, not

2R

❌ The constraint is pure geometry.

\Delta X_{cm} = R\,\Delta\theta

holds along any tilted line, so neither

\beta

nor

M

appears in

A_{cm} = R\,\alpha_{cm}

. They are needed only to find

A_{cm}

itself, which is given.

❌ Not quite. Use

A_{cm} = R\,\alpha_{cm}

with

R = 0.20\ \text{m}

Show solution

Rolling without slipping: in any interval the center moves as far along the plane as a rim point travels along its arc, $\Delta X_{cm} = R\,\Delta\theta$ . Dividing by $\Delta t$ and taking the limit gives $V_{cm} = R\,\omega_{cm}$ ; differentiating once more gives

A_{cm} = R\,\alpha_{cm} \quad\Rightarrow\quad \alpha_{cm} = \frac{A_{cm}}{R} = \frac{3.0\ \text{m/s}^2}{0.20\ \text{m}} = 15\ \text{rad/s}^2.

The tilt $\beta = 25\degree$ is a distractor here: it would matter for computing $A_{cm}$ from forces, but the constraint itself contains no angle and no mass.

Problem 2 · Spin Rate of a Yo-Yo

Given: a yo-yo with axle radius $b = 0.60\ \text{cm}$ and outer radius $R = 3.0\ \text{cm}$ falls on a string whose top is held fixed; at one instant its center moves down at $V_{cm,y} = 1.5\ \text{m/s}$ — find its angular speed $\omega_{cm}$ .

✅ Correct! The string leaves the axle, so

b\,\omega_{cm} = V_{cm,y}

and

\omega_{cm} = 1.5/0.0060 = 250\ \text{rad/s}

❌ That uses the outer radius.

R

is the right radius for a wheel rolling on its rim, but this string unwinds from the axle: the arc that equals the unwound length is

b\,\Delta\theta

❌ $R - b$ has no role here. The unwound string equals the arc on the axle rim, a distance

b

from the center, so divide by

b = 0.0060\ \text{m}

❌ That divides by the axle's diameter. A point where the string leaves is

b

from the center, not

2b

❌ Not quite. Use

b\,\omega_{cm} = V_{cm,y}

with

b

in metres.

Show solution

As the yo-yo turns through $\Delta\theta$ , a point on the axle rim sweeps the arc $b\,\Delta\theta$ , which equals the length $\Delta l$ of string unwound. The top of the string is fixed, so the center drops by that same $\Delta l$ . Dividing by $\Delta t$ and taking the limit:

b\,\omega_{cm} = V_{cm,y} \quad\Rightarrow\quad \omega_{cm} = \frac{V_{cm,y}}{b} = \frac{1.5\ \text{m/s}}{0.0060\ \text{m}} = 250\ \text{rad/s}.

The outer radius $R = 3.0\ \text{cm}$ enters the moment of inertia, and therefore the dynamics, but not this constraint.

Problem 3 · Two Drums from Rest

Given: drum $B$ turns on a fixed axle and drum $A$ hangs below it on a tape unwinding from the rim of each; both have radius $R = 0.25\ \text{m}$ and start at rest, with constant angular accelerations $\alpha_A = 4.0\ \text{rad/s}^2$ and $\alpha_B = 6.0\ \text{rad/s}^2$ — find the acceleration of drum $A$ 's center and how far it has dropped after $t = 2.0\ \text{s}$ .

What is $a_{A,y}$ ?

How far has $A$ dropped at $t = 2.0\ \text{s}$ ?

✅ Correct!

a_{A,y} = 0.25(4.0 + 6.0) = 2.5\ \text{m/s}^2

and

\Delta y = \tfrac{1}{2}(2.5)(2.0)^2 = 5.0\ \text{m}

, which is

2.0\ \text{m}

unwound from

A

plus

3.0\ \text{m}

fed from

B

❌ The two contributions add, not subtract. Both rotations lengthen the hanging stretch of tape, so

a_{A,y} = R\,\alpha_A + R\,\alpha_B

❌ That keeps only one drum's term.

A

drops by tape unwound from itself and tape fed down from

B

; each rotation contributes its own

R\,\alpha

❌ Missing the factor $\tfrac{1}{2}$ . From rest with constant acceleration,

\Delta y = \tfrac{1}{2}a_{A,y}t^2

, not

a_{A,y}t^2

❌ Check $a_{A,y}$ . Differentiate

\Delta y = R\,\Delta\theta_A + R\,\Delta\theta_B

twice.

❌ Check the drop. Use

\Delta y = \tfrac{1}{2}a_{A,y}t^2

with both drums' contribution in

a_{A,y}

Show solution

Step 1: the constraint. The center of $A$ drops by the tape unwound from both drums, $\Delta y = R\,\Delta\theta_A + R\,\Delta\theta_B$ ; differentiating twice,

a_{A,y} = R\,\alpha_A + R\,\alpha_B = 0.25(4.0) + 0.25(6.0) = 1.0 + 1.5 = 2.5\ \text{m/s}^2.

Step 2: the drop. Starting from rest with constant acceleration,

\Delta y = \tfrac{1}{2}a_{A,y}t^2 = \tfrac{1}{2}(2.5)(2.0)^2 = 5.0\ \text{m}.

Check by pieces: $\Delta\theta_A = \tfrac{1}{2}(4.0)(2.0)^2 = 8.0\ \text{rad}$ gives $R\,\Delta\theta_A = 2.0\ \text{m}$ ; $\Delta\theta_B = \tfrac{1}{2}(6.0)(2.0)^2 = 12\ \text{rad}$ gives $R\,\Delta\theta_B = 3.0\ \text{m}$ ; the sum is $5.0\ \text{m}$ .

Problem 4 · A Yo-Yo on a Moving Hand

Given: a yo-yo with axle radius $b = 1.0\ \text{cm}$ hangs from a string whose top is held in a hand. The hand accelerates upward at $2.0\ \text{m/s}^2$ while the yo-yo's center accelerates downward at $3.0\ \text{m/s}^2$ — find the magnitude of the yo-yo's angular acceleration.

✅ Correct! The string between hand and axle lengthens with acceleration

3.0 + 2.0 = 5.0\ \text{m/s}^2

, so

\alpha_{cm} = 5.0/0.010 = 500\ \text{rad/s}^2

— like drum

A

, whose tape is itself moving.

❌ That treats the top as fixed.

b\,\alpha_{cm} = A_{cm,y}

assumes the string's top is at rest. Here the hand moves away too, and that adds string to unwind, just as drum

B

adds tape for drum

A

❌ Check the signs. Taking down as positive, the hand's acceleration is

-2.0\ \text{m/s}^2

; the string length grows with acceleration

3.0 - (-2.0) = 5.0\ \text{m/s}^2

, because the two ends move apart.

❌ That keeps only the hand's term. The center's own drop also unwinds string; both separations add.

❌ Not quite. The unwound length is the hanging string length

l = y_{cm} - y_{hand}

; differentiate it twice.

Show solution

Point the $y$ -axis down. The hanging string runs from the hand to the axle, so its length is $l = y_{cm} - y_{hand}$ , and every bit of growth in $l$ is string unwound from the axle: $\Delta l = b\,\Delta\theta$ . Differentiating twice,

b\,\alpha_{cm} = \frac{d^2 l}{dt^2} = A_{cm,y} - a_{hand,y} = 3.0 - (-2.0) = 5.0\ \text{m/s}^2.

\alpha_{cm} = \frac{5.0\ \text{m/s}^2}{0.010\ \text{m}} = 500\ \text{rad/s}^2.

With the hand still ( $a_{hand,y} = 0$ ) this reduces to the video's $b\,\alpha_{cm} = A_{cm,y}$ . A moving support adds its own term, exactly as the moving tape from drum $B$ adds $R\,\alpha_B$ to drum $A$ 's acceleration.

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