Classical-Mechanics · Unit 21 · Video 5 · Interactive Practice

Two Runs, One Unknown Cancels: Measuring a Rotor's Moment of Inertia

IKey Formulas

FormulaNameWhat you need
mgr−τf=(IR+mr2) α1mgr - \tau_f = (I_R + mr^2)\,\alpha_1 Driven stage (string attached) Rotor Tr−τf=IRα1Tr - \tau_f = I_R\alpha_1 and hanger mg−T=mrα1mg - T = mr\alpha_1 (the string gives a1=rα1a_1 = r\alpha_1); multiply the hanger equation by rr and add, so TrTr cancels
−τf=IR α2-\tau_f = I_R\,\alpha_2 Coasting stage (string detached) Same bearings, same magnitude τf>0\tau_f > 0; with IR>0I_R > 0 this forces α2<0\alpha_2 < 0, a zz-component that keeps its minus sign
IR=mr (g−rα1)α1−α2I_R = \dfrac{mr\,(g - r\alpha_1)}{\alpha_1 - \alpha_2} Friction eliminated (driven minus coasting) Given mm, rr, gg and the two slopes of ω(t)\omega(t); the video's α1=83.5\alpha_1 = 83.5 and α2=−31.2 rad/s2\alpha_2 = -31.2\ \text{rad/s}^2 give 5.3×10−5 kg⋅m25.3 \times 10^{-5}\ \text{kg}\cdot\text{m}^2
τf=−IR α2\tau_f = -I_R\,\alpha_2 Frictional torque (by-product) IRI_R and the coasting slope; the video's rotor gives τf≈1.7×10−3 N⋅m\tau_f \approx 1.7 \times 10^{-3}\ \text{N}\cdot\text{m}, never measured directly

Key Insight: The bearings are the same in both runs, so τf\tau_f is the same number in both equations and subtracting them removes it. Because α2<0\alpha_2 < 0, the denominator α1−α2=∣α1∣+∣α2∣\alpha_1 - \alpha_2 = |\alpha_1| + |\alpha_2| is a sum of magnitudes.

IIVisualization 1 — Two Slopes on One Trace

Each stage's angular acceleration is the slope of its own straight segment; a window across the kink measures neither.

IIIVisualization 2 — One Run Leaves a Line of Answers

The driven run alone fits a whole line of (IR,τf)(I_R, \tau_f) pairs; the coasting run's line crosses it once.

given m = 0.055 kg, r = 0.0127 m, g = 9.8 m/s²; measured α₁ = 83.5 rad/s², α₂ = −31.2 rad/s²

IVVisualization 3 — Change the Friction, Keep the Answer

Tighter or looser bearings change both slopes, yet the two-run formula returns the same IRI_R every time.

VQuiz Questions

Problem 1 · The Closed Form on a New Rotor

Given: a rotor of radius r=20.0 mmr = 20.0\ \text{mm} is driven by a hanging mass m=0.100 kgm = 0.100\ \text{kg} (g=9.8 m/s2g = 9.8\ \text{m/s}^2); the two straight segments of its ω(t)\omega(t) trace have slopes α1=40.0 rad/s2\alpha_1 = 40.0\ \text{rad/s}^2 (string attached) and α2=−10.0 rad/s2\alpha_2 = -10.0\ \text{rad/s}^2 (string detached) — find the rotor's moment of inertia IRI_R.

✅ Correct! mr=2.00×10−3 kg⋅mmr = 2.00 \times 10^{-3}\ \text{kg}\cdot\text{m}, g−rα1=9.00 m/s2g - r\alpha_1 = 9.00\ \text{m/s}^2 and α1−α2=50.0 rad/s2\alpha_1 - \alpha_2 = 50.0\ \text{rad/s}^2, so IR=(2.00×10−3)(9.00)/50.0=3.60×10−4 kg⋅m2I_R = (2.00 \times 10^{-3})(9.00)/50.0 = 3.60 \times 10^{-4}\ \text{kg}\cdot\text{m}^2.
❌ That treats the bearings as frictionless. Dividing by α1\alpha_1 alone is the driven equation with τf=0\tau_f = 0; the coasting run adds ∣α2∣|\alpha_2| to the denominator.
❌ The hanger accelerates, so T≠mgT \ne mg. Its equation mg−T=mrα1mg - T = mr\alpha_1 leaves g−rα1g - r\alpha_1, not gg, in the numerator.
❌ Check the sign of α2\alpha_2. It is a negative zz-component, so α1−α2=40.0+10.0\alpha_1 - \alpha_2 = 40.0 + 10.0, not 40.0−10.040.0 - 10.0.
Show solution

Every quantity in the closed form is given or measured:

mr=(0.100)(0.0200)=2.00×10−3 kg⋅mmr = (0.100)(0.0200) = 2.00 \times 10^{-3}\ \text{kg}\cdot\text{m} g−rα1=9.8−(0.0200)(40.0)=9.8−0.800=9.00 m/s2g - r\alpha_1 = 9.8 - (0.0200)(40.0) = 9.8 - 0.800 = 9.00\ \text{m/s}^2 α1−α2=40.0−(−10.0)=50.0 rad/s2\alpha_1 - \alpha_2 = 40.0 - (-10.0) = 50.0\ \text{rad/s}^2 IR=mr (g−rα1)α1−α2=(2.00×10−3)(9.00)50.0=3.60×10−4 kg⋅m2I_R = \frac{mr\,(g - r\alpha_1)}{\alpha_1 - \alpha_2} = \frac{(2.00 \times 10^{-3})(9.00)}{50.0} = 3.60 \times 10^{-4}\ \text{kg}\cdot\text{m}^2

The term rα1=0.800 m/s2r\alpha_1 = 0.800\ \text{m/s}^2 is the hanger's own acceleration, and the negative α2\alpha_2 makes the denominator the sum of the two magnitudes.

Problem 2 · The Coasting Stage on Its Own

Given: the video's rotor has IR=5.32×10−5 kg⋅m2I_R = 5.32 \times 10^{-5}\ \text{kg}\cdot\text{m}^2; after the string detaches from the rim (the hanger was m=0.055 kgm = 0.055\ \text{kg} on radius r=12.7 mmr = 12.7\ \text{mm}) it coasts with α2=−31.2 rad/s2\alpha_2 = -31.2\ \text{rad/s}^2. The bearing torque is −τf k^-\tau_f\,\hat{k} — find the magnitude τf\tau_f.

✅ Correct! τf=−IRα2=−(5.32×10−5)(−31.2)=1.66×10−3 N⋅m\tau_f = -I_R\alpha_2 = -(5.32 \times 10^{-5})(-31.2) = 1.66 \times 10^{-3}\ \text{N}\cdot\text{m}, the video's ≈1.7×10−3\approx 1.7 \times 10^{-3}.
❌ The hanger has gone. Once the string detaches only the rotor spins, so the coasting equation carries IRI_R, not IR+mr2I_R + mr^2; the mr2mr^2 term belongs to the driven stage.
❌ τf\tau_f is a magnitude. The minus sign already sits in −τf k^-\tau_f\,\hat{k}: IRα2=−1.66×10−3 N⋅mI_R\alpha_2 = -1.66 \times 10^{-3}\ \text{N}\cdot\text{m} is the torque's zz-component, which is −τf-\tau_f.
❌ That uses the driven slope. With the string attached the tension torque acts as well; only in the coasting stage is friction the sole torque, so only α2\alpha_2 belongs here.
Show solution

After detachment the only torque about the axis is friction, and only the rotor turns:

−τf=IRα2⟹τf=−IRα2=−(5.32×10−5)(−31.2)=1.66×10−3 N⋅m-\tau_f = I_R\alpha_2 \quad\Longrightarrow\quad \tau_f = -I_R\alpha_2 = -(5.32 \times 10^{-5})(-31.2) = 1.66 \times 10^{-3}\ \text{N}\cdot\text{m}

Positive, as a magnitude must be. The driven stage gives the same number with α1=83.5 rad/s2\alpha_1 = 83.5\ \text{rad/s}^2 and mgr=6.845×10−3 N⋅mmgr = 6.845 \times 10^{-3}\ \text{N}\cdot\text{m}:

τf=mgr−(IR+mr2) α1=6.845×10−3−(6.207×10−5)(83.5)=6.845×10−3−5.183×10−3=1.66×10−3 N⋅m\tau_f = mgr - (I_R + mr^2)\,\alpha_1 = 6.845 \times 10^{-3} - (6.207 \times 10^{-5})(83.5) = 6.845 \times 10^{-3} - 5.183 \times 10^{-3} = 1.66 \times 10^{-3}\ \text{N}\cdot\text{m}

Problem 3 · From the Trace to the Moment of Inertia

Given: a rotor with r=20.0 mmr = 20.0\ \text{mm} is driven by a hanging mass m=0.0600 kgm = 0.0600\ \text{kg} (g=9.8 m/s2g = 9.8\ \text{m/s}^2). Its ω(t)\omega(t) trace rises in a straight line from 00 to 84.0 rad/s84.0\ \text{rad/s} at t=1.20 st = 1.20\ \text{s}, where the string detaches; on the falling segment ω=64.0 rad/s\omega = 64.0\ \text{rad/s} at t=2.00 st = 2.00\ \text{s} and ω=24.0 rad/s\omega = 24.0\ \text{rad/s} at t=3.60 st = 3.60\ \text{s} — find α2\alpha_2, then IRI_R.

What is α₂?

What is I_R?

✅ Correct! α1=70.0\alpha_1 = 70.0 and α2=−25.0 rad/s2\alpha_2 = -25.0\ \text{rad/s}^2, so IR=(1.20×10−3)(8.40)/95.0=1.06×10−4 kg⋅m2I_R = (1.20 \times 10^{-3})(8.40)/95.0 = 1.06 \times 10^{-4}\ \text{kg}\cdot\text{m}^2.
❌ The segment falls, so its slope is negative. Δω=24.0−64.0=−40.0 rad/s\Delta\omega = 24.0 - 64.0 = -40.0\ \text{rad/s} over Δt=1.60 s\Delta t = 1.60\ \text{s}.
❌ The two differences come from different windows. −60.0 rad/s-60.0\ \text{rad/s} is the drop since the kink at 1.20 s1.20\ \text{s}, but 3.60 s3.60\ \text{s} is the time since t=0t = 0; take Δω\Delta\omega and Δt\Delta t between the same two points.
❌ That chord runs from the origin across the kink. It mixes the driven and coasting stages, so its slope is neither α1\alpha_1 nor α2\alpha_2.
❌ Check the denominator's sign. α1−α2=70.0−(−25.0)=95.0 rad/s2\alpha_1 - \alpha_2 = 70.0 - (-25.0) = 95.0\ \text{rad/s}^2, not 45.045.0: the coasting stage adds to it.
❌ Dividing by α1\alpha_1 alone drops the coasting run. That is the frictionless answer; the falling segment shows friction is there.
❌ The numerator needs g−rα1g - r\alpha_1. The hanger accelerates at rα1=1.40 m/s2r\alpha_1 = 1.40\ \text{m/s}^2, so T<mgT < mg and the numerator is mr(8.40)mr(8.40), not mr(9.8)mr(9.8).
❌ Not quite. Take Δω/Δt\Delta\omega / \Delta t between two points on the falling segment.
❌ Not quite. Use IR=mr(g−rα1)/(α1−α2)I_R = mr(g - r\alpha_1)/(\alpha_1 - \alpha_2) with α1=84.0/1.20\alpha_1 = 84.0/1.20.
Show solution

Step 1: the driven slope. The rising segment starts at the origin:

α1=84.0 rad/s1.20 s=70.0 rad/s2\alpha_1 = \frac{84.0\ \text{rad/s}}{1.20\ \text{s}} = 70.0\ \text{rad/s}^2

Step 2: the coasting slope, from two points on the falling segment:

α2=24.0−64.03.60−2.00=−40.0 rad/s1.60 s=−25.0 rad/s2\alpha_2 = \frac{24.0 - 64.0}{3.60 - 2.00} = \frac{-40.0\ \text{rad/s}}{1.60\ \text{s}} = -25.0\ \text{rad/s}^2

Step 3: the closed form.

mr=(0.0600)(0.0200)=1.20×10−3 kg⋅m,g−rα1=9.8−1.40=8.40 m/s2mr = (0.0600)(0.0200) = 1.20 \times 10^{-3}\ \text{kg}\cdot\text{m}, \qquad g - r\alpha_1 = 9.8 - 1.40 = 8.40\ \text{m/s}^2 IR=(1.20×10−3)(8.40)70.0−(−25.0)=1.008×10−295.0=1.06×10−4 kg⋅m2I_R = \frac{(1.20 \times 10^{-3})(8.40)}{70.0 - (-25.0)} = \frac{1.008 \times 10^{-2}}{95.0} = 1.06 \times 10^{-4}\ \text{kg}\cdot\text{m}^2

As a by-product, τf=−IRα2=(1.061×10−4)(25.0)=2.65×10−3 N⋅m\tau_f = -I_R\alpha_2 = (1.061 \times 10^{-4})(25.0) = 2.65 \times 10^{-3}\ \text{N}\cdot\text{m}.

Problem 4 · The Same Trick on a Track

Given: a cart of unknown mass MM on a level track is pulled by a string over a light pulley to a hanging mass m=0.050 kgm = 0.050\ \text{kg} (g=9.8 m/s2g = 9.8\ \text{m/s}^2); the track exerts a constant, unknown friction force ff that is the same in both stages. While the mass falls the cart speeds up at a1=0.70 m/s2a_1 = 0.70\ \text{m/s}^2; after the mass lands and the string goes slack the cart slows at a2=−0.35 m/s2a_2 = -0.35\ \text{m/s}^2 — find MM.

✅ Correct! Subtracting the two runs removes ff exactly as it removed τf\tau_f: M=m(g−a1)/(a1−a2)=(0.050)(9.10)/1.05=0.433 kgM = m(g - a_1)/(a_1 - a_2) = (0.050)(9.10)/1.05 = 0.433\ \text{kg}.
❌ That is the driven run alone with f=0f = 0. The slack-string stage shows friction is present, and its equation −f=Ma2-f = Ma_2 is what removes ff.
❌ Check the sign of a2a_2. The cart is slowing, so a2<0a_2 < 0 and a1−a2=1.05 m/s2a_1 - a_2 = 1.05\ \text{m/s}^2, a sum of magnitudes.
❌ The hanging mass accelerates too. Setting T=mgT = mg drops its ma1ma_1; the driven equation is mg−f=(M+m)a1mg - f = (M + m)a_1.
Show solution

Driven stage (string taut, both bodies at a1a_1): cart T−f=Ma1T - f = Ma_1, hanger mg−T=ma1mg - T = ma_1. Adding them cancels the unmeasured tension:

mg−f=(M+m) a1mg - f = (M + m)\,a_1

Slack stage: friction is the only horizontal force on the cart:

−f=Ma2-f = Ma_2

Subtract the second from the first, so ff cancels:

mg=(M+m) a1−Ma2⟹M=m (g−a1)a1−a2=(0.050)(9.8−0.70)0.70−(−0.35)=0.4551.05=0.433 kgmg = (M + m)\,a_1 - Ma_2 \quad\Longrightarrow\quad M = \frac{m\,(g - a_1)}{a_1 - a_2} = \frac{(0.050)(9.8 - 0.70)}{0.70 - (-0.35)} = \frac{0.455}{1.05} = 0.433\ \text{kg}

This is the rotor's closed form with rα→ar\alpha \to a and IR/r2→MI_R/r^2 \to M. The by-product is f=−Ma2=(0.433)(0.35)=0.152 Nf = -Ma_2 = (0.433)(0.35) = 0.152\ \text{N}.

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