Same body, different axis.

Pick a body and an axis; I = Mk² is built and compared, with parallel-axis shift and same-torque spin-up race showing larger I turns slower.

Shift a disc's axis parallel to its central axis until its moment of inertia doubles (d = R/√2), then find which body has I = MR² about its central axis (ring).

Treating I as a fixed property of a body regardless of axis.

• Each particle of a body rotating with ω has speed ωrᵢ, so the kinetic energy is K = ½(Σmᵢrᵢ²)ω² = ½Iω², where I = Σmᵢrᵢ² is the moment of inertia about the axis and rᵢ is the perpendicular distance from the axis.
• Comparing ½Iω² with ½mv² shows that I plays the role of mass in rotation: it measures a body's resistance to a change in its rotation (rotational inertia).
• A thin ring of mass M and radius R about the axis through its centre, perpendicular to its plane: I = MR², since all its mass is at R. Two masses M/2 at the ends of a light rod of length l, about a perpendicular axis through the middle: I = Ml²/4.
• Standard results: ring about a diameter MR²/2; thin rod about a perpendicular axis at the midpoint ML²/12; disc about the perpendicular central axis MR²/2 and about a diameter MR²/4; hollow cylinder about its axis MR²; solid cylinder about its axis MR²/2; solid sphere about a diameter 2MR²/5.
• Unlike mass, I is not fixed for a body: it depends on the mass, the shape and size, how the mass is spread about the axis, and the position and orientation of the axis.
• Radius of gyration k: I = Mk². It is the distance from the axis at which a point mass equal to the whole mass would have the same I. Rod about its middle: k = L/√12; disc about a diameter: k = R/2.
• Dimensions of I are M L², SI unit kg m².
• A flywheel is a disc of large moment of inertia on an engine's shaft. It resists sudden changes in speed, so a vehicle's speed changes gradually and the ride is smooth.

Teaching this idea? Put the simulation on your own page or LMS with this code.