Classical-Mechanics · Unit 25 · Video 4 · Interactive Practice

Three Equations, Three Unknowns: The Falling Yo-Yo's Tension, Acceleration and Final Spin

IKey Formulas

Formula	Name	What it says
$mg - T = ma_y$ , $\ bT = I_{\text{cm}}\alpha_z$ , $\ a_y = b\,\alpha_z$	The three equations	Newton along $\hat{j}$ (down), torque about the center of mass, unwinding constraint; unknowns $T$ , $a_y$ , $\alpha_z$
$T = \dfrac{mg}{1 + 2b^2/R^2}$	String tension	With $I_{\text{cm}} = \tfrac{1}{2}mR^2$ ; always less than the weight
$\alpha_z = \dfrac{2bg}{R^2 + 2b^2}$ , $\ a_y = b\,\alpha_z = \dfrac{g}{1 + R^2/(2b^2)}$	Angular and linear acceleration	Both constant and positive: down, and clockwise ( $\hat{k}$ into the page); $a_y < g$
$\omega_f = \sqrt{\dfrac{4gl}{R^2 + 2b^2}}$	Final spin	Energy conservation after the string length $l$ unwinds; the tension does no work

Key Insight: The axle radius $b$ enters twice, as the moment arm of the tension and as the radius in the constraint; the spool radius $R$ enters only through $I_{\text{cm}}$ . For $R = 3b$ : $T = \tfrac{9}{11}mg$ and $a_y = \tfrac{2}{11}g$ .

IIVisualization 1 — One Tension Satisfies All Three Equations

For a guessed tension, Newton's law and torque-plus-constraint each predict an acceleration; they agree at one $T$ only.

Spool : axle Guess T / mg 1.00

IIIVisualization 2 — Thinner Axle: Slower Fall, Faster Final Spin

On the same spool, shrinking the axle radius $b$ lengthens the fall yet raises $\omega_f = \sqrt{4gl/(R^2 + 2b^2)}$ .

Yo-yo B axle Time t 0.000 s

💡 As $b \to 0$ the fall time grows without bound while $\omega_f$ approaches $\sqrt{4gl}/R \approx 162\ \text{rad/s}$ for this spool.

IVVisualization 3 — Energy Bookkeeping on the Way Down

With $U = -mgy$ and $E_i = 0$ , the lost potential energy splits between translation and spin in the ratio $2b^2 : R^2$ .

Axle Depth y 0.30 m

VQuiz Questions

Problem 1 · Tension for a Fatter Axle

Given: A yo-yo of mass $m$ has axle radius $b$ and spool radius $R = 2b$ , with $I_{\text{cm}} = \tfrac{1}{2}mR^2$ . It is released from rest with the top of its vertical string held fixed. Find the tension $T$ .

✅ Correct!

2b^2/R^2 = 2b^2/(4b^2) = \tfrac{1}{2}

, so

T = mg/(1 + \tfrac{1}{2}) = \tfrac{2}{3}mg

❌ Close, but…

\tfrac{1}{3}

a_y/g

for this yo-yo, not

T/mg

. Newton's law then gives

T = mg - ma_y = mg - \tfrac{1}{3}mg

❌ Not quite. That comes from writing the constraint as

a_y = R\alpha_z

. The string unwinds from the axle, so the constraint radius is

b

a_y = b\,\alpha_z

❌ Not quite.

T = mg

leaves zero net force, so

a_y = 0

; but the torque

bT

would still give

\alpha_z > 0

, and the constraint

a_y = b\,\alpha_z

fails.

❌ Not quite. Combine

mg - T = ma_y

bT = I_{\text{cm}}\alpha_z

and

a_y = b\,\alpha_z

Show solution

Substitute the constraint into Newton's law, then $\alpha_z = bT/I_{\text{cm}}$ :

mg - T = mb\,\alpha_z = \frac{mb^2\,T}{I_{\text{cm}}} \quad\Rightarrow\quad T = \frac{mg}{1 + mb^2/I_{\text{cm}}} = \frac{mg}{1 + 2b^2/R^2}

With $R = 2b$ : $2b^2/R^2 = 2b^2/(4b^2) = \tfrac{1}{2}$ , so

T = \frac{mg}{1 + \tfrac{1}{2}} = \tfrac{2}{3}mg

Check: $a_y = g - T/m = \tfrac{1}{3}g$ , and $b\,\alpha_z = b^2T/I_{\text{cm}} = \tfrac{2b^2}{R^2}\cdot\tfrac{2}{3}g = \tfrac{1}{2}\cdot\tfrac{2}{3}g = \tfrac{1}{3}g$ ✓

Problem 2 · Which Radius Unwinds?

Given: The video's yo-yo: axle radius $b = 1.0\ \text{cm}$ , spool radius $R = 3.0\ \text{cm}$ , $I_{\text{cm}} = \tfrac{1}{2}mR^2$ , $g = 9.8\ \text{m/s}^2$ . Find the magnitude of the acceleration of its center of mass.

✅ Correct!

R = 3b

gives

a_y = g/(1 + 9/2) = \tfrac{2}{11}g \approx 1.78\ \text{m/s}^2

❌ Close, but…

3.92 = \tfrac{2}{5}g

comes from the constraint

a_y = R\,\alpha_z

. The string leaves the axle, so the unwound length per radian is

b

, not

R

❌ Not quite.

2.18 = 2b^2g/R^2

drops the

2b^2

from the denominator:

a_y = \dfrac{2b^2 g}{R^2 + 2b^2}

❌ Not quite.

6.53 = \tfrac{2}{3}g

is the answer for a string wound on the rim (

b = R

); here the thin axle makes the yo-yo fall far more slowly.

❌ Not quite. Use

a_y = g/\bigl(1 + R^2/(2b^2)\bigr)

Show solution

From the three equations, $a_y = b\,\alpha_z = \dfrac{2b^2 g}{R^2 + 2b^2} = \dfrac{g}{1 + R^2/(2b^2)}$ . With $R/b = 3$ :

\frac{R^2}{2b^2} = \frac{9}{2}, \qquad a_y = \frac{g}{1 + \tfrac{9}{2}} = \frac{2}{11}g = \frac{2}{11}(9.8) \approx 1.78\ \text{m/s}^2

Only the ratio $R/b$ matters, and the radius in the constraint is the axle radius $b$ , where the string leaves the yo-yo.

Problem 3 · Final Spin by Energy

Given: The same yo-yo ( $b = 1.0\ \text{cm}$ , $R = 3.0\ \text{cm}$ , $I_{\text{cm}} = \tfrac{1}{2}mR^2$ ) starts from rest and unwinds a string of length $l = 0.80\ \text{m}$ ; $g = 9.8\ \text{m/s}^2$ , no air resistance. Find its angular speed $\omega_f$ at the bottom.

✅ Correct!

\omega_f = \sqrt{4gl/(R^2 + 2b^2)} = \sqrt{31.36/(1.1\times10^{-3})} \approx 169\ \text{rad/s}

❌ Not quite. That sets

\tfrac{1}{2}mv_f^2 = mgl

and ignores the rotational kinetic energy, which here takes

\tfrac{9}{11}

of the energy.

❌ Close, but… That keeps only

\tfrac{1}{4}mR^2\omega_f^2

. The center also moves, adding

\tfrac{1}{2}mb^2\omega_f^2

❌ Not quite. That uses

v_f = R\,\omega_f

. The point at rest is where the string leaves the axle, so

v_f = b\,\omega_f

❌ Not quite. Set

\bigl(\tfrac{1}{2}mb^2 + \tfrac{1}{4}mR^2\bigr)\omega_f^2 = mgl

Show solution

The tension acts at a point momentarily at rest, so it does no work and $E_f = E_i = 0$ . With $v_f = b\,\omega_f$ and $I_{\text{cm}} = \tfrac{1}{2}mR^2$ :

\tfrac{1}{2}mb^2\omega_f^2 + \tfrac{1}{4}mR^2\omega_f^2 - mgl = 0 \quad\Rightarrow\quad \omega_f = \sqrt{\frac{4gl}{R^2 + 2b^2}}

R^2 + 2b^2 = 9.0\times10^{-4} + 2.0\times10^{-4} = 1.1\times10^{-3}\ \text{m}^2, \qquad 4gl = 4(9.8)(0.80) = 31.36\ \text{m}^2/\text{s}^2

\omega_f = \sqrt{\frac{31.36}{1.1\times10^{-3}}} = \sqrt{2.851\times10^{4}} \approx 169\ \text{rad/s}

Kinematic check: $a_y = \tfrac{2}{11}(9.8) \approx 1.782\ \text{m/s}^2$ , $\Delta t = \sqrt{2l/a_y} \approx 0.948\ \text{s}$ , $\alpha_z = 2bg/(R^2 + 2b^2) \approx 178.2\ \text{rad/s}^2$ , and $\alpha_z\,\Delta t \approx 169\ \text{rad/s}$ ✓

Problem 4 · Same Method, Rolling Cylinder

Given: A uniform solid cylinder ( $I_{\text{cm}} = \tfrac{1}{2}mR^2$ ) rolls without slipping down a $\theta = 30\degree$ incline from rest. Static friction $f$ acts up the slope at the contact point, a distance $R$ from the center. Find the acceleration $a$ of the center and the friction force $f$ .

What is the acceleration of the center?

What is the friction force?

✅ Correct! Friction plays the string's role with

b = R

a = \tfrac{2}{3}g\sin 30\degree = \tfrac{1}{3}g

and

f = \tfrac{1}{2}ma = \tfrac{1}{6}mg

❌ Check the acceleration.

\tfrac{1}{2}g = g\sin 30\degree

ignores friction,

\tfrac{2}{3}g

drops the

\sin 30\degree

, and

\tfrac{1}{4}g

belongs to a hoop (

I_{\text{cm}} = mR^2

). Solve

mg\sin\theta - f = ma

fR = I_{\text{cm}}\alpha

a = R\alpha

❌ Check the friction. The torque equation with the constraint gives

f = I_{\text{cm}}a/R^2 = \tfrac{1}{2}ma

; friction cannot be zero, since it alone supplies the torque about the center.

Show solution

Take the down-slope direction positive. The three equations, exactly as for the yo-yo:

mg\sin\theta - f = ma, \qquad fR = \tfrac{1}{2}mR^2\,\alpha, \qquad a = R\,\alpha

The second and third give $f = \tfrac{1}{2}ma$ . Then $mg\sin\theta = \tfrac{3}{2}ma$ , so

a = \tfrac{2}{3}g\sin\theta = \tfrac{2}{3}g\cdot\tfrac{1}{2} = \tfrac{1}{3}g \approx 3.27\ \text{m/s}^2, \qquad f = \tfrac{1}{2}ma = \tfrac{1}{6}mg

This is the yo-yo result with $b = R$ and $mg$ replaced by $mg\sin\theta$ : $f = \dfrac{mg\sin\theta}{1 + 2R^2/R^2} = \tfrac{1}{3}mg\sin\theta$ .

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