Moment of inertia calculator
Calculate the mass and mass moment of inertia of homogeneous bodies from their dimensions and density: solid cylinder, hollow cylinder, cuboid, sphere, cone and ring of circular cross-section. The tool returns the inertia about the rotational axis and, for cylinders and cuboids, the transverse axes as well. In addition you can apply the parallel-axis (Steiner) theorem, reduce a load inertia through a gear stage and reduce a linear mass to the rotational axis, live with every input.
Moment of inertia calculator
Model: homogeneous, ideal basic bodies with constant density, inertia about the centroidal principal axes. The ring of circular cross-section is approximated as a thin ring. Determine composite bodies by adding the individual inertias (with Steiner where needed).
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Formulas and fundamentals
The mass moment of inertia J describes a body's resistance to angular acceleration and is the rotational counterpart of mass: M = J·α, analogous to F = m·a. It depends not only on the mass but quadratically on the distance of the mass elements from the axis, J = ∫ r² dm. Closed-form expressions exist for the common basic bodies about their symmetry axis Z. In every case the mass follows from m = ρ·V with the density ρ (steel about 7860 kg/m³) and the geometric volume.
About the rotational or symmetry axis: solid cylinder J_z = ½·m·r², hollow cylinder J_z = ½·m·(R²+r²), sphere J = 2/5·m·r², cone J_z = 3/10·m·R² (base radius R), cuboid J_z = m·(a²+b²)/12 with the edges a and b perpendicular to the axis. For cylinders and cuboids the transverse axis through the centre of mass also matters: solid cylinder J_x = m·(3r²+h²)/12, cuboid J_x = m·(b²+c²)/12. A thin ring of circular cross-section (mean diameter D_m, wire diameter d) has m = ρ·(π²/4)·D_m·d² and approximately J_z = ½·m·D_m².
Two conversions are central in drive engineering. The parallel-axis (Steiner) theorem shifts the reference axis by the distance s: J = J_eigen + m·s²; the inertia is minimal only about the centre of mass. For motor sizing all inertias are reduced to the motor shaft: a load inertia behind a gear stage with ratio i acts only as J_red = J_last/i², a linearly moving mass m as J_red = m·(K_VA/2π)², where K_VA is the linear travel per revolution (screw lead or π·d for a wheel or pulley).
Worked example
A solid steel shaft of radius r = 0.05 m and length h = 1.3 m has, at ρ = 7860 kg/m³, a mass m = ρ·π·r²·h = 80.25 kg. About its longitudinal axis J_z = ½·m·r² = 0.1003 kg·m², whereas about a transverse axis through the middle J_x = m·(3r²+h²)/12 = 11.35 kg·m² – more than a hundred times larger, because the length enters quadratically.
If a load with J_last = 10 kg·m² sits behind this shaft together with a gearbox of ratio i = 5, only J_red = 10/5² = 0.4 kg·m² acts at the motor shaft. A mass of 200 kg moved through a screw with K_VA = 0.01 m/rev reduces to J = 200·(0.01/2π)² = 5.07·10⁻⁴ kg·m².
For a rotational axis offset by s = 2 m from the centre of mass of a body with J_eigen = 1.5 kg·m² and m = 2 kg, the parallel-axis theorem gives J = 1.5 + 2·2² = 9.5 kg·m². The Steiner term m·s² clearly dominates the intrinsic part here – a typical effect for eccentrically mounted masses.
Frequently asked questions
What is the difference between J about the rotational and the transverse axis?
The rotational or symmetry axis Z is the longitudinal axis of a cylinder or the axis of symmetry. About it the moment of inertia is usually smallest. The transverse axis X is perpendicular to it and passes through the centre of mass; for long, slender bodies J_x is much larger because of the quadratic length term. Drives usually care about the rotational axis, pendulum or tilting motions about the transverse axis.
When do I need the parallel-axis (Steiner) theorem?
Whenever the axis of rotation does not pass through the centre of mass. J = J_eigen + m·s² adds the term m·s² for the axis distance s to the centroidal moment of inertia. Because s enters quadratically, this extra term can quickly exceed the intrinsic part. The moment of inertia is minimal only about the centre of mass.
How do I reduce a load inertia to the motor shaft?
Through a gear stage with ratio i the relation is J_red = J_last/i². A larger ratio reduces the reflected inertia quadratically and thus dynamically decouples the load from the motor. This is why gearboxes greatly improve the acceleration capability of servo axes.
How does a linearly moving mass enter?
A mass m moved linearly through a screw, pinion or pulley acts on the rotational axis as J = m·(K_VA/2π)². K_VA is the linear travel per revolution: the lead for a screw, π·d for a wheel or pulley. The factor K_VA/2π is the effective radius at which the mass appears rotationally.
Which density should I use?
About 7860 kg/m³ for steel, 7900 for stainless steel, 7200 for grey cast iron, 2700 for aluminium, 8500 for brass, 8900 for copper, 4500 for titanium and roughly 1400 kg/m³ for engineering plastics. The tool offers these values for selection, but the density can also be entered freely. Since J scales linearly with mass and thus with density, a wrong density has a directly proportional effect.
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