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Moment of Inertia Calculator – Rectangle, Circle, Rod, Sphere & All Shapes

📐 Engineering Calculator

Moment of Inertia Calculator

This moment of inertia calculator covers both mass moment of inertia (rotational inertia, kg·m²) for rotation dynamics and area moment of inertia (second moment of area, mm4) for beam bending and structural engineering — for all common shapes including rectangle, circle, rod, sphere, cylinder, ring, cone, and hollow sections. Every result includes the formula and full step-by-step working with numbers substituted.

🌀 Mass Moment of Inertia Rotation • Physics • Dynamics • Flywheels • Shafts kg·m² I = Σmr²
📐 Area Moment of Inertia Beams • Bending • Structural • Section Modulus mm4 / m4 I = ∫y²dA
🌀 Mass Moment of Inertia (Rotational Inertia) — kg·m²

Select a shape to calculate its rotational inertia. Formula: I = Σmr². Results shown in kg·m², kg·cm², g·cm², lb·ft², lb·in².

Iaxis = ½mR²  |  Idiam = ¼mR² + &frac{1}{12}mL²  |  Iedge = &frac32;mR²
axis = spinning like a wheel  |  diam = end-to-end  |  edge = parallel axis theorem
central axis diam R L
I = ⅖mR² = 0.4 × m × R²
About any diameter axis — rotational inertia of a sphere
R axis I = 2/5 mR²
I = ⅔mR²
Hollow sphere / spherical shell — compare: solid = 2/5 mR²
Icentral = mR² (thin hoop)  |  Idiam = ½mR²  |  Iannulus = ½m(R²+r²)
Set inner radius r = 0 for thin hoop. The formula I = mR² applies when all mass is at radius R.
R central axis I = mR²
Icenter = &frac{1}{12}mL²  |  Iend = ⅓mL²
Perpendicular axis through midpoint or one end — moment of inertia of a rod
center axis end axis L
Iz = &frac{1}{12}m(a²+b²)  |  Ix = &frac{1}{12}m(b²+c²)  |  Iy = &frac{1}{12}m(a²+c²)
a = length  |  b = width  |  c = height  |  all axes through center of mass
Iaxis = &frac{3}{10}mR²
About central axis through apex and centre of base — moment of inertia of cone
I = Icm + m × d²
Icm = MOI about centre of mass  |  d = distance between parallel axes  |  Also: Icm = I − md²
CM axis new axis d Icm I = Icm+md²
📐 Area Moment of Inertia (Second Moment of Area) — mm4 / m4

Select a cross-section shape. I = ∫y²dA. Used in beam bending: σ = M×y/I. Section modulus S = I/ymax.

Ix = bh³/12  |  Iy = hb³/12  |  Ibase = bh³/3  |  S = bh²/6
b = width  |  h = height  |  Ix about centroidal horizontal axis  |  area moment of inertia of a rectangle
x (centroid) y base h b
I = πR4/4 = πd4/64  |  J = πR4/2  |  S = πR³/4  |  k = R/2
Solid circle: set inner radius = 0. Hollow circle (pipe): I = π(R4−r4)/4
Ix = (BH³ − bh³)/12  |  Iy = (HB³ − hb³)/12
B,H = outer dims  |  b = B−2t  |  h = H−2t  |  t = wall thickness
Ibase = πR4/8  |  ŷ = 4R/(3π)  |  Icentroid = R4(π/8 − 8/(9π))
Centroid measured from base (diameter) — second moment of area of half circle

Mass Moment of Inertia vs Area Moment of Inertia — What is the Difference?

The symbol I is used for two completely different quantities in engineering. The mass moment of inertia (also called rotational inertia) measures how difficult it is to angularly accelerate a rotating body — it depends on how mass is distributed relative to the rotation axis. The area moment of inertia (also called second moment of area) measures how a cross-section resists bending — it depends on how the cross-sectional area is distributed relative to the bending axis. Always check units: kg·m² means rotational inertia; mm4 or m4 means second moment of area.

PropertyMass Moment of InertiaArea Moment of Inertia
Also calledRotational inertiaSecond moment of area
SymbolI (kg·m²)I (m4 or mm4)
FormulaI = Σmr²I = ∫y²dA
Unitskg·m²m4 or mm4
Used inRotation dynamics, angular accelerationBeam bending, deflection, stress
Relevant equationτ = I × ασ = M × y / I
Example useSpinning flywheel, rotating shaftSteel beam deflection, column buckling
Depends onMass distribution and axis locationCross-section shape and axis location

The polar moment of inertia J = Ix + Iy is the sum of the two area moments of inertia and is used in torsion calculations: τmax = T×R/J. Both area moment of inertia and polar moment of inertia are second moments of area — different axes, same concept.

Mass Moment of Inertia Formulas — Reference Table

All formulas below give rotational inertia I in kg·m² when mass is in kg and dimensions in metres. The fundamental formula is I = Σmr² — each element of mass multiplied by the square of its distance from the axis. Note that I = mR² appears for the thin hoop, I = mR² is the baseline maximum for a given mass and radius.

ShapeAxisFormulaNotes
Solid cylinder / diskCentral (spinning)I = ½mR²Wheel-like spin
Solid cylinderDiameter (end-to-end)I = ¼mR² + &frac{1}{12}mL²Tumbling axis
Disk about edgeTangent to rimI = &frac32;mR²Parallel axis theorem
Solid sphereAny diameterI = ⅖mR²Rotational inertia of a sphere
Hollow sphere (shell)Any diameterI = ⅔mR²Higher than solid sphere
Thin hoop / ringCentral axis (⊥ to plane)I = mR²All mass at radius R; thin hoop moment of inertia
Thin hoopDiameter axisI = ½mR²In-plane axis
Thick ring / annulusCentral axisI = ½m(R²+r²)Inertia of a ring with wall thickness
Thin rodThrough center (⊥)I = &frac{1}{12}mL²Moment of inertia of a rod about center
Thin rodThrough end (⊥)I = ⅓mL²Moment of inertia of a rod about end
Rectangular boxz-axis through centerI = &frac{1}{12}m(a²+b²)Moment of inertia of a box
Rectangular boxx-axis through centerI = &frac{1}{12}m(b²+c²)Rotational inertia of a cube when a=b=c
Solid coneCentral axisI = &frac{3}{10}mR²Moment of inertia of cone; about base axis
Thin plate (a×b)Through center parallel to bI = &frac{1}{12}ma²Moment of inertia of a plate

Area Moment of Inertia Formulas — Reference Table

These formulas give the second moment of area I in mm4 when dimensions are in mm. The formula I = bh³/12 is fundamental — it appears for rectangles about their centroidal axis. The section modulus S = I / ymax is used in beam design to find maximum bending stress: σmax = M/S.

Cross-sectionAxisFormulaSection Modulus S
Rectangle (b×h)Centroidal x (horiz)I = bh³/12S = bh²/6
Rectangle (b×h)BaseI = bh³/3—
Rectangle (b×h)Centroidal y (vert)I = hb³/12S = hb²/6
Square (a×a)CentroidalI = a4/12S = a³/6
Hollow rectangleCentroidal xI = (BH³−bh³)/12S = I/(H/2)
Circle (radius R)Any diameterI = πR4/4S = πR³/4
Circle (diameter d)Any diameterI = πd4/64S = πd³/32
Hollow circle / pipeAny diameterI = π(R4−r4)/4S = I/R
Semicircle (radius R)Base (diameter)I = πR4/8—
SemicircleCentroidal (∥ to base)I = R4(π/8−8/(9π))S = I/(R−ŷ)
Triangle (b×h)BaseI = bh³/12—
Triangle (b×h)CentroidalI = bh³/36—

Parallel Axis Theorem — How to Find Moment of Inertia About Any Axis

The parallel axis theorem allows you to find the moment of inertia about any axis if you know the moment of inertia about a parallel axis through the centroid (or centre of mass for mass MOI). Both the mass and area versions follow the same principle.

I = Icm + m × d²   (Mass MOI)
Icm = MOI about centre of mass  |  m = total mass  |  d = perpendicular distance between axes
I = Icentroid + A × d²   (Area MOI)
Icentroid = second moment about centroidal axis  |  A = cross-sectional area  |  d = distance between axes

Worked Example 1 — Rod About End From Center

  1. I about center: Icm = (1/12)mL²
  2. Distance from center to end: d = L/2
  3. Iend = (1/12)mL² + m(L/2)² = (1/12)mL² + (1/4)mL²
  4. Iend = mL²(1/12 + 3/12) = mL²(4/12) = (1/3)mL² ✓

Worked Example 2 — Disk About Edge

  1. I about center (spinning): Icm = (1/2)mR²
  2. Distance from center to rim: d = R
  3. Iedge = (1/2)mR² + mR² = (3/2)mR² ✓
  4. Example: m=2kg, R=0.1m → Iedge = 1.5×2×0.01 = 0.03 kg·m²

Worked Example 3 — Rectangle About Base From Centroid

  1. I about centroid: Ix = bh³/12
  2. Area A = b×h, distance from centroid to base: d = h/2
  3. Ibase = bh³/12 + (bh)(h/2)² = bh³/12 + bh³/4 = bh³/12 + 3bh³/12 = bh³/3 ✓

Section Modulus Calculator — Rectangle Circle and Hollow Sections

Section modulus S = I / ymax, where I is the area moment of inertia and ymax is the distance from the centroid to the extreme (outermost) fibre. Units are mm³ or m³ — one order of dimension lower than I (mm4). The section modulus is the single most important property for beam design because maximum bending stress equals σmax = M / S, where M is the applied bending moment.

Section Modulus Formulas — Common Cross-Sections

  • Rectangle: Sx = bh²/6  |  Sy = hb²/6
  • Circle (diameter d): S = πd³/32
  • Hollow circle: S = π(D4−d4)/(32D)
  • Square tube (outer B×H, inner b×h): S = (BH³−bh³)/(6H)

Elastic vs Plastic section modulus: The elastic section modulus Se = I/ymax assumes linear stress distribution (used for serviceability). The plastic section modulus Sp assumes full yielding across the section (used for ultimate strength design). Sp > Se always.

Worked Example — Maximum Bending Stress in Rectangular Beam

Rectangular beam: b=50mm, h=100mm, M=5 kN·m

  1. Ix = bh³/12 = 50×100³/12 = 4,166,667 mm4
  2. ymax = h/2 = 50 mm
  3. Sx = I/ymax = 4,166,667/50 = 83,333 mm³
  4. σmax = M/S = 5,000,000 N·mm / 83,333 mm³ = 60 N/mm² = 60 MPa

Moment of Inertia of a Rod

The moment of inertia of a rod describes the rotational inertia of a thin uniform rod rotating about an axis perpendicular to its length. There are two standard cases, and the parallel axis theorem connects them. Note that rotation about the rod's own length axis gives I ≈ 0 for a thin rod (negligible radius).

Case 1 — Axis Through Centre (Perpendicular)

Icenter = &frac{1}{12}mL²
Perpendicular axis through the midpoint of the rod

This is the minimum moment of inertia for a rod rotating perpendicular to its length. The mass is distributed symmetrically about the centre, so distances range from 0 (at center) to L/2 (at ends), giving I = (1/12)mL².

Case 2 — Axis Through One End (Perpendicular)

Iend = ⅓mL²
Perpendicular axis through one end — exactly 4× the center case (not 3×)

The moment of inertia of a rod about one end is exactly 4 times larger than (1/12)mL² — wait, (1/3)/(1/12) = 4. Yes: rotating about the end is 4× harder than rotating about the center because all mass is now at distances from 0 to L. Via parallel axis theorem: Iend = (1/12)mL² + m(L/2)² = (1/12)mL² + (3/12)mL² = (4/12)mL² = (1/3)mL².

Worked Example 1: m=0.8kg, L=1.2m

  1. L² = 1.2² = 1.44 m²
  2. Icenter = (1/12) × 0.8 × 1.44 = 0.096 kg·m²
  3. Iend = (1/3) × 0.8 × 1.44 = 0.384 kg·m²
  4. Check: 0.096 + 0.8×(0.6)² = 0.096 + 0.288 = 0.384 ✓

Worked Example 2: Baton (m=0.2kg, L=0.5m)

  1. Icenter = (1/12) × 0.2 × 0.25 = 0.00417 kg·m²
  2. Iend = (1/3) × 0.2 × 0.25 = 0.01667 kg·m²
  3. Torque needed for α = 10 rad/s² about center: τ = 0.00417×10 = 0.0417 N·m

Moment of Inertia of a Rectangle

The moment of inertia of a rectangle is an area moment of inertia (second moment of area) used in structural engineering to quantify a rectangular beam's resistance to bending. The formula is I = bh³/12 about the centroidal horizontal axis, where b is width and h is height (depth). Note that height h appears cubed — this means doubling the depth multiplies bending resistance by 8×, while doubling the width only doubles it. This is why beams are always oriented with their greater dimension vertical.

Ix = bh³/12  |  Iy = hb³/12  |  Ibase = bh³/3
b = width  |  h = height (depth)  |  x = centroidal horizontal axis  |  y = centroidal vertical axis

Example 1: 200×400mm Beam (b=200mm, h=400mm)

  1. Ix = 200×400³/12 = 200×64,000,000/12 = 1,066,666,667 mm4 = 1.067×10? mm4
  2. Sx = Ix/(h/2) = 1,066,666,667/200 = 5,333,333 mm³
  3. A = 200×400 = 80,000 mm²
  4. kx = √(I/A) = √(1,066,666,667/80,000) = 115.5 mm

Example 2: Square 150×150mm

  1. Ix = Iy = 150×150³/12 = 1504/12 = 42,187,500 mm4
  2. S = I/(h/2) = 42,187,500/75 = 562,500 mm³

Example 3: I about base (b=100mm, h=200mm)

  1. Icentroid = bh³/12 = 100×200³/12 = 66,666,667 mm4
  2. Ibase = bh³/3 = 100×200³/3 = 266,666,667 mm4 (= 4 × Icentroid)
  3. Check via parallel axis: 66,666,667 + (100×200)×100² = 66,666,667 + 200,000,000 = 266,666,667 ✓

Worked Examples

1. Moment of Inertia of a Solid Cylinder About Its Axis

A solid cylinder spinning like a wheel uses I = (1/2)mR². For m=5kg, R=0.15m: I = 0.5×5×0.15² = 0.5×5×0.0225 = 0.05625 kg·m². Angular acceleration with 2 N·m torque: α = τ/I = 2/0.05625 = 35.6 rad/s².

2. Moment of Inertia of a Disk About Its Edge

Disk about its edge uses the parallel axis theorem: Iedge = Icenter + mR² = (1/2)mR² + mR² = (3/2)mR². For m=2kg, R=0.1m: Iedge = 1.5×2×0.01 = 0.03 kg·m². This is exactly 3× the spinning (central axis) value of 0.01 kg·m².

3. Moment of Inertia of a Rod — Center and End

m=0.8kg, L=1.2m: Icenter = (1/12)×0.8×1.44 = 0.096 kg·m². Iend = (1/3)×0.8×1.44 = 0.384 kg·m². Verification: 0.096 + 0.8×0.36 = 0.096 + 0.288 = 0.384 ✓. The end value is 4× the center value.

4. Rotational Inertia of a Sphere

Solid sphere: I = (2/5)mR². For m=3kg, R=0.08m: I = 0.4×3×0.0064 = 0.00768 kg·m². Hollow sphere same mass and radius: I = (2/3)×3×0.0064 = 0.01280 kg·m² — 67% higher because all mass is at the outer surface.

5. Moment of Inertia of a Rectangle for Beam Bending

b=100mm, h=200mm: Ix = 100×200³/12 = 66,666,667 mm4. Sx = bh²/6 = 100×40,000/6 = 666,667 mm³. For M=10 kN·m: σmax = 10×106/666,667 = 15 MPa.

6. Area Moment of Inertia of a Circular Cross Section

Solid circle d=50mm (R=25mm): I = π×254/4 = π×390,625/4 = 306,796 mm4. J = 2I = 613,592 mm4. S = I/R = 306,796/25 = 12,272 mm³. Radius of gyration k = R/2 = 12.5mm.

7. Using the Parallel Axis Theorem Step by Step

Find I of rectangle (100×200mm) about its base: Icentroid = 100×200³/12 = 66,666,667 mm4. Area A = 20,000 mm². Distance centroid to base d = 100mm. Ibase = 66,666,667 + 20,000×100² = 66,666,667 + 200,000,000 = 266,666,667 mm4.

8. Section Modulus for a Rectangular Beam

b=50mm, h=150mm: Sx = bh²/6 = 50×150²/6 = 50×22,500/6 = 187,500 mm³. Maximum allowable moment for σallow=120MPa: M = S×σ = 187,500×120 = 22,500,000 N·mm = 22.5 kN·m.

9. Moment of Inertia of a Hollow Sphere

Hollow sphere (spherical shell): I = (2/3)mR². For m=2kg, R=0.1m: I = (2/3)×2×0.01 = 0.01333 kg·m². Compare solid sphere same mass/radius: I = (2/5)×2×0.01 = 0.008 kg·m². The hollow sphere has 67% more rotational inertia — its mass is all at maximum radius.

10. Moment of Inertia of a Thin Hoop About Its Diameter

Thin hoop about diameter axis: I = (1/2)mR². For m=0.5kg, R=0.2m: I = 0.5×0.5×0.04 = 0.01 kg·m². Compare central axis (perpendicular to plane): I = mR² = 0.5×0.04 = 0.02 kg·m². The in-plane (diameter) rotation is half as resistant as the out-of-plane (central axis) rotation.

Frequently Asked Questions

What is the moment of inertia formula for a rectangle? ▼
The area moment of inertia of a rectangle: Ix = bh³/12 (about centroidal x-axis), Iy = hb³/12 (about centroidal y-axis), Ibase = bh³/3 (about base). Section modulus S = bh²/6. For b=100mm, h=200mm: Ix = 100×200³/12 = 66,666,667 mm4 = 66.67×106 mm4.
What is the difference between mass and area moment of inertia? ▼
Mass moment of inertia (rotational inertia, kg·m², I=Σmr²) resists angular acceleration in rotating systems (τ=I×α). Area moment of inertia (second moment of area, mm4, I=∫y²dA) resists bending in beams (σ=M×y/I). Same symbol I, completely different physical quantities — always check units.
What is the moment of inertia of a rod about its center? ▼
I = mL²/12. For m=1kg, L=1m: I = 1/12 = 0.0833 kg·m². About one end: I = mL²/3 = 4 times larger. These are perpendicular axes — about the rod's own length axis, I≈0 for a thin rod.
What is the moment of inertia of a solid sphere? ▼
I = (2/5)mR² = 0.4mR² about any diameter axis. For m=1kg, R=0.1m: I = 0.4×1×0.01 = 0.004 kg·m². Hollow sphere (spherical shell): I = (2/3)mR² — 67% higher because all mass is at the maximum radius R.
How do you find the moment of inertia of a disk about its edge? ▼
Use the parallel axis theorem: Iedge = Icenter + mR² = (1/2)mR² + mR² = (3/2)mR². For m=2kg, R=0.1m: Iedge = 1.5×2×0.01 = 0.03 kg·m². The disk-about-edge moment of inertia is 3 times the spinning (central axis) value.
What is the parallel axis theorem? ▼
I = Icm + md² (mass MOI) or I = Icentroid + Ad² (area MOI). It lets you find I about any axis parallel to a centroidal axis. The centroidal I must be known first, then add md² or Ad² for the offset. Example: rod about end = Icenter + m(L/2)² = (1/12)mL² + (1/4)mL² = (1/3)mL².
What is section modulus and how is it calculated? ▼
S = I/ymax (mm³). Maximum bending stress σ = M/S. Rectangle: S = bh²/6. Circle: S = πd³/32. For b=100mm, h=200mm: S = 100×200²/6 = 666,667 mm³. With M=10 kN·m: σmax = 10×106/666,667 = 15 MPa.
What is the moment of inertia of a thin hoop? ▼
Thin hoop central axis (perpendicular to plane): I = mR². This is the maximum possible I for given mass and radius — all mass is at the maximum distance R. About diameter axis: I = (1/2)mR². Thick ring (annulus): I = (1/2)m(R²+r²) about central axis. For m=0.5kg, R=0.2m: I = 0.5×0.04 = 0.02 kg·m².
What is the polar moment of inertia? ▼
J = Ix + Iy — sum of two area moments of inertia, about the z-axis perpendicular to the cross-section. Units mm4. Circle: J = πR4/2 = πd4/32. Rectangle: J = bh(h²+b²)/12. Used in torsion: τmax = T×R/J. Higher J means less shear stress for the same torque.
How does moment of inertia affect rotation? ▼
Newton's second law for rotation: τ = I×α. Higher I requires more torque for the same angular acceleration. Rotational KE = (1/2)Iω². Angular momentum L = I×ω. Flywheels and gyroscopes deliberately maximise I by placing mass at large radius. Ice skaters pull arms in to reduce I and spin faster (conservation of angular momentum).

Related Calculators

🌀 Mass MOI Quick Reference
Cyl
I = ½mR² (central axis)
Sph
I = ⅖mR² (solid sphere)
HSp
I = ⅔mR² (hollow sphere)
Hoop
I = mR² (thin hoop central)
Rod
I = mL²/12 (center)
Rod
I = mL²/3 (end)
Box
I = m(a²+b²)/12
PAT
I = Icm + md²
📐 Area MOI Quick Reference
Rect
I = bh³/12 (centroid)
Rect
I = bh³/3 (base)
Circ
I = πR4/4 = πd4/64
Pol
J = πR4/2 = πd4/32
Sec
S = I / ymax = bh²/6
PAT
I = Ic + Ad²
📏 Unit Conversions
kg·m²
1 kg·m² = 10,000 kg·cm² = 107 g·cm²
lb·ft²
1 kg·m² = 23.730 lb·ft² = 3417 lb·in²
mm4
1 mm4 = 10?4 cm4 = 10?¹² m4
in4
1 in4 = 416,231 mm4 = 41.623 cm4
mm³
1 mm³ = 0.001 cm³ = 10?? m³
⚖ Key Equations
τ = I × α
Torque = MOI × angular accel.
KE = ½ Iω²
Rotational kinetic energy
σ = My/I
Bending stress
σmax = M/S
Max stress via section modulus
τ = TR/J
Torsional shear stress

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