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Rotational Motion & Torque

Torque, moment of inertia, and the three-shape spin-up race

A father pushes a playground merry-go-round at its edge — 250 newtons, steady — and the 50-kilogram disk winds up at 6.67 rad/s². Then his kid hops on, just 18 kilograms sitting barely a meter from the center, and the very same push suddenly buys only 4.44 rad/s². Nothing about the push changed. What changed is where the mass sits — and rotation prices location by the square, I = Σmr². That is the lesson separating spinning from sliding. Every straight-line quantity you know has a rotational twin playing by the same grammar: force answers to torque, mass answers to moment of inertia, acceleration answers to angular acceleration — and Newton's second law survives the translation intact: τ = Iα. But the twin has a twist. Mass is a single number; inertia is a whole geography. Take the same five kilograms and the same one-meter radius and build three objects: a solid disk reads I = 2.500 kg·m², a thin ring reads 5.000 — every gram parked at the maximum radius, paying the highest rate — and a solid sphere reads 2.000, its mass huddled near the core. Apply the same 10 N·m and they wind up at 4.000, 2.000, and 5.000 rad/s². Once spinning, they hold energy whether or not they travel anywhere: ½Iω², the same quadratic bill as ½mv² — double the spin, quadruple the joules. In this lab you will flip among the three shapes and catch inertia repricing itself while the scale never moves, push the torque and mass levers against each other until α = τ/I is a reflex, and watch the energy row climb the square of the speed row while the object never leaves its axis.

What you'll be able to do

  • Falsify the same-mass-same-inertia rule with the panel's own numbers at the default state (m = 5.0 kg, R = 1.0 m, τ = 10 N·m): the Solid Disk reads I 2.500 / α 4.000, the Thin Ring reads I 5.000 / α 2.000, the Solid Sphere reads I 2.000 / α 5.000 — identical mass, radius, and torque, with only the mass distribution moved, because 'bodies with more mass concentrated at a greater distance from the axis have greater moments of inertia than bodies of the same mass concentrated near the axis' (OpenStax §10.4, verified); §10.7's merry-go-round is the same lesson with real numbers (α 6.67 → 4.44 rad/s² when the child boards)
  • Price both levers of α = τ/I on the disk: Torque 10 → 20 N·m doubles α 4.000 → 8.000 rad/s²; Mass 5.0 → 10.0 kg halves it back to 2.000; Radius 1.0 → 2.0 m QUARTERS it to 1.000 because the disk's I = ½mR² pays the radius squared (2.500 → 10.000 kg·m²) — the square law the sim's own quiz Q2 traps with 'Doubles', and the exact analog of F = ma with torque, inertia, and angular acceleration in the force, mass, and acceleration seats (§10.7, verified)
  • Keep the spin-up books both ways from rest at defaults (α = 4.000 rad/s²): the ω row climbs linearly with time (ω = αt — ≈4.000 rad/s after one second, 8.000 after two, §10.2 Eq. 10.11) while the KE row climbs quadratically (K = ½Iω²: 20.00 J at ω = 4.000, 80.00 J at ω = 8.000 — four times the energy for twice the speed, §10.4 Eq. 10.18), and the books tie through the work-energy identity τθ = ΔK until the engine's ω = 50 rad/s display cap (registered quirk, not physics)

Formulas

τ=rFsin⁡θ=Iα\tau = rF\sin\theta = I\alpha
Torque (rotational equivalent of F=ma)
I=∑miri2I = \sum m_i r_i^2
Moment of inertia (depends on mass distribution)
Idisk=12mR2I_{\text{disk}} = \tfrac{1}{2}mR^2
Solid disk — mass spread evenly from axis to rim
Iring=mR2I_{\text{ring}} = mR^2
Thin ring — all mass parked at the rim
Isphere=25mR2I_{\text{sphere}} = \tfrac{2}{5}mR^2
Solid sphere — mass huddled toward the core
KErot=12Iω2KE_{rot} = \frac{1}{2}I\omega^2
Rotational kinetic energy

Make a prediction

Default state: Solid Disk, m = 5.0 kg, R = 1.0 m, τ = 10 N·m — the panel reads I = 2.500 kg·m² and α = 4.000 rad/s². Now switch ONLY the object to Thin Ring (same mass, same radius, same torque). What do the Inertia and Angular α rows read?

No grading here — commit to a guess, then scroll down and test it yourself.

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Your prediction

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Answer: I = 5.000 kg·m², α = 2.000 rad/s² — the ring parks ALL its mass at the rim (I = mR² vs ½mR²), inertia doubles, and α = τ/I halves

I_disk = ½mR² = 0.5×5×1 = 2.500 kg·m²; I_ring = mR² = 5×1 = 5.000 kg·m² — exactly double, at identical mass and radius. The textbook states the rule outright: 'Rigid bodies… with more mass concentrated at a greater distance from the axis of rotation have greater moments of inertia than bodies… of the same mass, but concentrated near the axis' (OpenStax UP1 §10.4), and the pricing law behind it is I = Σmr²: the ring's every kilogram sits at the maximum possible radius, paying the highest r² rate on the object. Since α = τ/I (OpenStax §10.7, Newton's second law for rotation), doubling I at fixed τ = 10 N·m halves α from 4.000 to 2.000 rad/s². Option B is the mass-only rule — the instinct the sim's own quiz Q1 is built to catch ('Both accelerate equally' is one of its wrong options): it correctly senses that mass sets inertia but forgets that mass DISTRIBUTION co-signs the bill. Option C's premise is real — the ring IS hollow — but hollowness doesn't delete the mass: all 5 kg are still aboard, parked at the worst possible address. The merry-go-round runs the same lesson in reverse (§10.7 Example 10.16): one child boarding near the center raises I from 56.25 to 84.38 kg·m² and the father's unchanged push drops α from 6.67 to 4.44 rad/s².

Quiz (0/4)

A disk (I = 2 kg·m²) has a net torque of 8 N·m applied. What is its angular acceleration?

A skater spins at 2 rad/s with I = 4 kg·m². They pull in arms to I = 1 kg·m². New ω?

Why does a longer wrench make it easier to loosen a bolt?

A solid disk (I = ½MR²) and a hollow ring (I = MR²) of equal mass and radius start from rest on an incline. Which reaches the bottom first?

You can now

  • Falsify the same-mass-same-inertia rule with the panel's own numbers at the default state (m = 5.0 kg, R = 1.0 m, τ = 10 N·m): the Solid Disk reads I 2.500 / α 4.000, the Thin Ring reads I 5.000 / α 2.000, the Solid Sphere reads I 2.000 / α 5.000 — identical mass, radius, and torque, with only the mass distribution moved, because 'bodies with more mass concentrated at a greater distance from the axis have greater moments of inertia than bodies of the same mass concentrated near the axis' (OpenStax §10.4, verified); §10.7's merry-go-round is the same lesson with real numbers (α 6.67 → 4.44 rad/s² when the child boards)
  • Price both levers of α = τ/I on the disk: Torque 10 → 20 N·m doubles α 4.000 → 8.000 rad/s²; Mass 5.0 → 10.0 kg halves it back to 2.000; Radius 1.0 → 2.0 m QUARTERS it to 1.000 because the disk's I = ½mR² pays the radius squared (2.500 → 10.000 kg·m²) — the square law the sim's own quiz Q2 traps with 'Doubles', and the exact analog of F = ma with torque, inertia, and angular acceleration in the force, mass, and acceleration seats (§10.7, verified)
  • Keep the spin-up books both ways from rest at defaults (α = 4.000 rad/s²): the ω row climbs linearly with time (ω = αt — ≈4.000 rad/s after one second, 8.000 after two, §10.2 Eq. 10.11) while the KE row climbs quadratically (K = ½Iω²: 20.00 J at ω = 4.000, 80.00 J at ω = 8.000 — four times the energy for twice the speed, §10.4 Eq. 10.18), and the books tie through the work-energy identity τθ = ΔK until the engine's ω = 50 rad/s display cap (registered quirk, not physics)

What you'll learn

  • Moment of Inertia. Moment of inertia is the rotational equivalent of mass — it measures how hard it is to change an object's rotation. Unlike mass, it depends on how the mass is distributed relative to the axis. Moving mass farther from the axis dramatically increases the moment of inertia.
  • The Translation-Rotation Dictionary. Every linear quantity has a rotational twin: force becomes torque, mass becomes moment of inertia, acceleration becomes angular acceleration. Newton's 2nd law translates word for word — F = ma becomes τ = Iα. In this lab all three twins are live: the Torque slider plays force, the shape/mass/radius controls set I, and the α row answers.
  • Torque: The Rotational Force. Torque is what causes angular acceleration, just as force causes linear acceleration. It depends on three factors: the magnitude of the force, the distance from the pivot (lever arm), and the angle of application. This is why longer wrenches make bolts easier to turn.
  • Rotational Kinetic Energy. A spinning object carries kinetic energy even though its center of mass never moves. As the applied torque spins the object up, the work it does — torque times angle turned — lands in the KE_rot row: ω climbs linearly with time, but the energy climbs with ω², which is why the KE row races ahead of the ω row.

Step-by-step

  1. Pick a shape — Solid Disk, Thin Ring, or Solid Sphere — and read its moment of inertia on the panel.
  2. Drag the Torque slider and watch α = τ/I respond; drag Mass or Radius and watch I re-price (any slider change restarts the spin from rest).
  3. Let it run and watch ω climb linearly while rotational KE climbs with ω².
  4. Use the speed buttons to slow the clock without changing the physics.

Key formulas

  • τ=rFsin⁡θ=Iα\tau = rF\sin\theta = I\alphaTorque (rotational equivalent of F=ma)
  • I=∑miri2I = \sum m_i r_i^2Moment of inertia (depends on mass distribution)
  • Idisk=12mR2I_{\text{disk}} = \tfrac{1}{2}mR^2Solid disk — mass spread evenly from axis to rim
  • Iring=mR2I_{\text{ring}} = mR^2Thin ring — all mass parked at the rim
  • Isphere=25mR2I_{\text{sphere}} = \tfrac{2}{5}mR^2Solid sphere — mass huddled toward the core
  • KErot=12Iω2KE_{rot} = \frac{1}{2}I\omega^2Rotational kinetic energy

Frequently asked questions

A disk (I = 2 kg·m²) has a net torque of 8 N·m applied. What is its angular acceleration?
The correct answer is: 4 rad/s². Α = τ/I.
A skater spins at 2 rad/s with I = 4 kg·m². They pull in arms to I = 1 kg·m². New ω?
The correct answer is: 8 rad/s. L = Iω is conserved: I₁ω₁ = I₂ω₂.
Why does a longer wrench make it easier to loosen a bolt?
The correct answer is: It increases the lever arm, producing more torque for the same force. Τ = rF — larger r means larger torque for same force.
A solid disk (I = ½MR²) and a hollow ring (I = MR²) of equal mass and radius start from rest on an incline. Which reaches the bottom first?
The correct answer is: The solid disk (lower I means more translational KE). Lower I → more of PE converts to translational KE. Use energy conservation.