Rotational Inertia
▶︎ Watch it animatedinteractive step-through · ~3 min · optionalRotational inertia I measures how hard it is to change a body's spin about a chosen axis. You build it piece by piece: I = Σ miri2 for point masses, or I = ∫ r2 dm for a continuous body, where every bit of mass is weighted by the square of its distance r from the axis. That r2 weighting is the whole story: mass far from the axis counts far more than mass near it. To move to a parallel axis a distance d from the center of mass, the parallel-axis theorem adds one term: I = Icm + Md2.
Setting up the integral is where rotational inertia goes wrong. Dropping the r2 weighting, or forgetting that dm = λ dx or σ dA, so the changing r can't be pulled outside. Misusing the parallel-axis theorem — subtracting the shift, starting from a non-center axis, or picking the wrong distance d. And treating I as a fixed property of the body, as if equal mass meant equal inertia or all the mass acted at the center. Keep the r2 weighting, add Md2 only from the center-of-mass axis, and ask which axis every time.
The work
3 ways in · any order
Lesson
Rotational Inertia
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Builds rotational inertia as I = ∫ r2 dm: the r2 weighting, choosing the mass element dm = λ dx or σ dA, and the parallel-axis theorem I = Icm + Md2. Shows why pulling r outside the integral overcounts, why the shift always adds, and why equal mass can give very different inertia. Closes with a ten-scenario skill check.
Diagnostic
10-item topic check
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Ten items on the three misconceptions for Topic 5.4: setting up the ∫ r2 dm integral wrong, misusing the parallel-axis theorem I = Icm + Md2, and treating rotational inertia as axis-independent or mass-only. Take it cold to see what is still tangled, or after the lesson to confirm it is not.
Targeted Practice
Drill a single misconception
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Pick one misconception you keep missing and drill it on its own. The round adapts: two correct in a row clears it and you move on.