How this instrument works
Moment of inertia is the rotational counterpart of mass — not a fixed property of an object, but a statement about how far its material sits from a chosen axis. For a single point mass, the formula is as plain as physics gets: I = mr². The r² comes straight out of kinetic energy: a point moving in a circle carries speed v = ωr, so its energy is ½mv² = ½mω²r²; pull the ½ω² back out and what remains, mr², is that mass's own contribution to rotational inertia. Square the radius and you square how much the mass resists a change in spin rate — push it twice as far out and the same mass now takes four times the torque to spin up at a given rate.
Real hardware is never a true point — a bolt, a counterweight, a sensor pod all have their own size — so I = mr² is really the leading term of a larger calculation. Treat a component as a point at its centre of mass and the estimate is good whenever its own dimensions are small next to the radius r; forget that caveat and you undercount, because the object's own moment of inertia about its own centre, I_cm, still adds on top: I_axis = I_cm + mr², the parallel-axis theorem. A fist-sized 5 kg pod mounted metres out is well approximated as a point; the same 5 kg mounted a few centimetres from the axis is not — and that is exactly the small-radius regime this instrument's own numbers sit in.
Spin-test engineers reach for this formula before any hardware gets bolted down. Aerospace and automotive qualification rigs mount a component — a fastener, a bracket, a small assembly — on an arm at a fixed radius and spin it to simulate the centrifugal loads it will meet in service; before a drive motor is even sized, someone has to know how much extra torque that mounted mass demands, and I = mr² supplies the first number in that chain. The mistake this formula invites is treating mass and moment of inertia as roughly interchangeable, since both carry the comforting feel of kilograms. They are not: moving a mass twice as far from the axis has four times the effect of doubling the mass itself, because only mass enters linearly here.
- Enter the object's weight into Point mass — kilograms by default, or switch the unit menu to pounds.
- Enter Radius from the rotation axis — the perpendicular distance from the pivot to the mass, in metres or centimetres.
- Read Moment of inertia, kg·m² — this is the figure a motor or drive shaft must overcome to change that mass's spin rate.
- To feel how sensitive the result is, nudge Radius from the rotation axis and watch Moment of inertia climb with the square, not the difference.
Worked example — a 5 kg mass at 0.3 m on a spin-test arm
A spin-test engineer clamps a 5 kg fixture 0.3 m out from the pivot of a test arm, standing in for a component about to be qualified for centrifugal load. Point mass reads 5 kg, Radius from the rotation axis reads 0.3 m, and Moment of inertia, kg·m² returns I = 5 × 0.3² = 5 × 0.09 = 0.45 kg·m². That figure is what the drive motor has to overcome to bring the arm up to test speed at whatever angular acceleration the qualification procedure specifies, before a single revolution has actually happened.
Move the same fixture out to 0.6 m and inertia does not double to 0.9 kg·m² — it quadruples, to 1.8 kg·m², because doubling the radius squares to four times the effect. Doubling the mass instead, to 10 kg at the original 0.3 m, does exactly what plain intuition expects: inertia doubles cleanly, to 0.9 kg·m². That asymmetry between the two inputs is the entire personality of I = mr², and it is why spin-rig engineers worry far more about how far out a mass sits than about how heavy it is.
Questions
What does 'point mass' actually mean here?
It means all of the object's mass is treated as concentrated at a single location, ignoring the object's own size and shape. That is exact for genuinely small bodies and a good approximation for anything whose own dimensions are small compared with its radius from the axis. Once a component's own bulk becomes comparable to r, its own moment of inertia about its own centre has to be added in through the parallel-axis theorem.
Why does radius matter more than mass in this formula?
Because radius is squared and mass is not. Multiply Point mass by 2 and Moment of inertia, kg·m² exactly doubles — 5 kg at 0.3 m gives 0.45 kg·m², 10 kg at 0.3 m gives 0.9 kg·m². Multiply Radius from the rotation axis by 2 instead and the result quadruples — the same 5 kg moved out to 0.6 m gives 1.8 kg·m², not 0.9. That is why engineers spend more effort moving mass inward on a spinning assembly than trimming its weight.
How is this different from an object's ordinary mass?
Mass measures resistance to a straight-line push and stays fixed no matter how the object is arranged. Moment of inertia measures resistance to a twist about a specific axis, and it changes the moment material moves closer to or further from that axis without the mass itself changing at all. The same 5 kg block reads 0.45 kg·m² at 0.3 m and a completely different number at any other radius — mass alone never says which.
What if my mass isn't really a single point?
Add its own moment of inertia about its own centre of mass, I_cm, to m·r² using the parallel-axis theorem: I_axis = I_cm + m·r². A lookup table or a short integral gives I_cm for common shapes — a solid sphere is two-fifths its mass times its own radius squared, a thin disc is one-half — and this instrument's r² term still correctly accounts for however far that shape's centre sits from the true rotation axis.
Can Point mass or Radius from the rotation axis be zero?
Radius can be zero — a mass sitting exactly on the axis contributes nothing to Moment of inertia, kg·m², because it never has to move as the system turns. Point mass at zero returns zero for the same reason from the other direction: there is no mass left to resist anything. Neither field accepts a negative number, since neither mass nor distance is meaningfully negative in this setup.
Why is this formula the starting point for every other moment-of-inertia case?
Because every rigid body is, mathematically, a collection of point masses. Sum m·r² over every particle in an object — replacing the sum with an integral for a continuous shape — and the result is its full moment of inertia; that is exactly how the standard results for rods, discs and spheres get derived in the first place. Learning I = mr² for one point is what makes those other formulas legible rather than merely memorised.