QiLing Tools

🛞Moment of Inertia Calculator & Rotation Simulator - Parallel Axis, Rolling Race

Four core rotation problems in one tool. Moment of inertia lists I_cm for common bodies about different axes; drag the offset to see the parallel-axis theorem I = I_cm + Md² at work while the body spins about the chosen axis. The composite builder lets you drag point masses, rods, disks, rings, spheres and plates around, move the axis anywhere and get the total I, the centre of mass and each part’s contribution. Rolling race sends a solid sphere, solid cylinder, hollow sphere, hoop or frictionless block down the same incline, with rotating markings so you can see who wins, and a table of accelerations, times and the split between translational and rotational energy — including rolling with slipping when friction is too low. Angular momentum covers a skater pulling in her arms and a person walking on a turntable. Torque & Atwood includes the pulley’s mass, so the two tensions differ, with a table-top version as well.

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How to use Moment of Inertia & Rotational Dynamics

  1. 1Pick a scene: moment of inertia, rolling race, angular momentum or Atwood machine
  2. 2Choose a shape and axis or the racers, and set masses, radii, angle and friction
  3. 3Drag composite parts and the axis, or press play to watch the rolling, spinning or pulley animation
  4. 4Read the moment of inertia, acceleration, times, angular speed, tensions and energy split, with the formulas below

FAQ

Why does a solid sphere always beat a hollow one down a ramp?
Rolling without slipping gives a = g sinθ / (1 + I/mr²). Part of the lost potential energy goes into rotation, and the larger I/mr² is, the bigger that share and the slower the translation. A solid sphere has I/mr² = 2/5, a solid cylinder 1/2, a hollow sphere 2/3 and a thin hoop 1, so the finishing order is always sphere, cylinder, hollow sphere, hoop — whatever their mass or radius. A frictionless block doesn’t rotate at all and is fastest.
How do I use the parallel-axis theorem?
The moment of inertia about any axis equals the moment about a parallel axis through the centre of mass plus the mass times the square of the distance between them: I = I_cm + Md². A thin rod has ML²/12 about its middle; the axis through its end is L/2 away, so I = ML²/12 + M(L/2)² = ML²/3. The theorem only works from the centre-of-mass axis — you can’t shift directly between two axes that both miss the centre of mass.
Why does a skater spin faster with arms pulled in, and where does the energy come from?
Friction from the ice exerts almost no torque, so angular momentum L = Iω is conserved. Pulling the arms in moves mass toward the axis, I falls and ω rises so that I₁ω₁ = I₂ω₂. Rotational kinetic energy L²/2I increases, and the extra energy is the work her muscles do pulling the arms inward against their tendency to keep moving in a wider circle.
Why are the two rope tensions different when the pulley has mass?
Giving a pulley with moment of inertia I an angular acceleration needs a net torque (T₂ − T₁)R = Iα, so the tensions must differ. Combining Newton’s second law for both masses with a = αR gives a = (m₂ − m₁)g / (m₁ + m₂ + I/R²): the pulley acts like an extra mass I/R². The heavier the pulley, the smaller the acceleration and the larger the tension difference.