Physics · Rotational motion

Parallel-axis theorem for mass moment of inertia

Q: How do I check units and scale for the parallel-axis theorem for mass moment of inertia?

All inertia terms and $Md^2$ have units kg·m². Use metres for d; a distance of 30 cm gives $d^2=0.090\,\mathrm{m^2}$. An area second moment instead shifts by $Ad^2$ and has units m⁴. It cannot be used in the same energy balance or in $L=I\omega$. For two parallel axes A and B neither passing through the centre of mass, refer both to the same centre-of-mass axis: $I_B-I_A=M(d_B^2-d_A^2)$. Do not simply add M times their mutual separation squared. A change in orientation, such as from a rod’s longitudinal axis to a transverse one, also needs a different reference moment.

The moment of inertia about an axis parallel to the center-of-mass axis increases by $Md^2$ .

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Formula

Calculate

I = I_{CM} + M d^{2}

I, $I_{\mathrm{CM}}$ in kg·m²; M in kg; d in m.

Variables & units

Variables & units
Symbol	Meaning	Unit
$I$	Mass moment of inertia about the shifted axis	kg·m²
$ICM$	Mass moment of inertia about the parallel centre-of-mass axis	kg·m²
$M$	Total mass of the same distribution	kg
$d$	Shortest perpendicular distance between the parallel axes	m

LaTeX

I=I_{\mathrm{CM}}+Md^2

QUANTA STUDYquanta-study.de

Horizontal uniform rod of length ℓ with center of mass at its midpoint. Two parallel axes normal to the drawing pass through its left end and center; their separation is d = ℓ/2. — Uniform thin rod: parallel axes through its center of mass and end. With d = ℓ/2, Iend = Mℓ²/12 + Mℓ²/4 = Mℓ²/3.

Parallel axis theorem calculator

Find the moment of inertia about a shifted parallel axis from the centre-of-mass value, mass and axis distance, or any of them in reverse.

Find

Moment about the parallel centre-of-mass axis ICM

Moment of inertia about the parallel axis through the centre of mass, e.g. 0.16 kg·m² for a 3 kg rod 80 cm long.

Total body mass M

Total mass of the body, e.g. 3 kg.

Perpendicular axis separation magnitude d

Perpendicular distance between the two parallel axes, e.g. 0.4 m, half the rod length.

Filled with example values. Type your own and the result updates straight away.

Result: Moment about the shifted axis IA

IA = 0.64 kg·m²

With ICM = 0.16 kg·m², M = 3 kg and d = 0.4 m, IA = 0.64 kg·m².

Working · rearrange for IA, substitute, evaluate

1. Rearrange for IA
$I_A=I_{\mathrm{CM}}+Md^2$
2. Substitute values (in SI units)
$0.16+3\cdot0.4^2$
3. Result
IA = 0.64 kg·m²

Quick tips

The theorem always starts from the centre of mass: if neither axis passes through it, go back to I_CM first and then out to the new axis.
Twice the axis distance, four times the extra term Md². Of all parallel axes, the one through the centre of mass has the smallest moment of inertia.

Assumptions and limits

Same body and mass distribution: parallel axes, reference through the centre of mass. Positive total mass, nonnegative central inertia, shortest perpendicular separation d≥0. No area second moment or axis reorientation. Inverse distance gives no displacement direction.

Derivation & background

The axes must be parallel, and one passes through the center of mass. This is the mass theorem with $Md^2$ , not the area theorem $I_A=I_{A,c}+Ad^2$ used in structural mechanics.

Exam blueprint

Validity range

The same mass distribution and two parallel axes, with the reference axis through the centre of mass. Measure separation perpendicular to the axes. This is the mass theorem in kg·m²; it does not change axis orientation or substitute for an area second moment.

Derivation steps

Expand shifted squared distances; cross terms vanish because the reference axis passes through the centre of mass.

1Choose the parallel centre-of-mass axis as z; the target axis passes through (a,b).
2A mass element has target-axis distance squared $r_A^2=(x-a)^2+(y-b)^2$ .
3Expand $r_A^2=x^2+y^2-2ax-2by+a^2+b^2$ .
4Integrate with dm; $\int(x^2+y^2)\,\mathrm dm$ is $I_{\mathrm{CM}}$ .
5The centre-of-mass definition gives $\int x\,\mathrm dm=\int y\,\mathrm dm=0$ , so cross terms vanish.
6Using $\int\mathrm dm=M$ and $d^2=a^2+b^2$ leaves $I_A=I_{\mathrm{CM}}+Md^2$ .

Rearrangements

Moment about the center-of-mass axis

I_{\mathrm{CM}}=I-Md^2

The axes must be parallel, with the reference through the centre of mass. Measure separation perpendicular to the axes. The inverse must give $I_{\mathrm{CM}}\ge0$ ; a negative result signals inconsistent inputs or an inappropriate axis assumption.

Task variant

An object has centre-axis inertia $I_{\mathrm{CM}}$ =0.20 kg·m² and total mass 2.0 kg. A new axis is parallel at perpendicular separation 0.50 m. Find mass inertia about the new axis.

Approach: $I_A=I_{\mathrm{CM}}+Md^2$ .
The addition is $(2.0\,\mathrm{kg})(0.50\,\mathrm m)^2=0.50\,\mathrm{kg\,m^2}$ ; total $I_A=0.70\,\mathrm{kg\,m^2}$ .

Check: the result exceeds centre inertia and has units kg·m².

Common mistakes

Using the simple M d² addition between any two parallel axes.

The reference axis must pass through the centre of mass. Two noncentral parallel axes satisfy $I_B-I_A=M(d_B^2-d_A^2)$ , using both offsets from the same centre axis.

Exam context

Find moments of inertia about shifted axes and for composite rigid bodies.

Formula cluster

Rotational motion

Related relations: traegheitsmoment-punktmassen, rotationsenergie-starrer-koerper, drehmoment.

I = Σ m_ir_i²

Moment of inertia of a particle system

Erot = ½Iω²

Rotational kinetic energy of a rigid body

M = F·r·sin(θ)

Torque

Worked example

For $I_{\mathrm{CM}}=0.20\,\mathrm{kg\,m^2}$ , $M=2.0\,\mathrm{kg}$ and parallel axes separated by $d=0.50\,\mathrm m$ , $I=I_{\mathrm{CM}}+Md^2=0.70\,\mathrm{kg\,m^2}$ . The positive addition $Md^2=0.50\,\mathrm{kg\,m^2}$ confirms greater inertia about this axis.

Applications

Find moments of inertia about shifted axes and for composite rigid bodies.

How do you calculate with Parallel-axis theorem for mass moment of inertia?

Here is how to work through a typical Parallel-axis theorem for mass moment of inertia (I = ICM + Md²) task step by step:

1
Task
An object has centre-axis inertia $I_{\mathrm{CM}}$ =0.20 kg·m² and total mass 2.0 kg. A new axis is parallel at perpendicular separation 0.50 m. Find mass inertia about the new axis.
Solution path
1. Approach: $I_A=I_{\mathrm{CM}}+Md^2$ .
2. The addition is $(2.0\,\mathrm{kg})(0.50\,\mathrm m)^2=0.50\,\mathrm{kg\,m^2}$ ; total $I_A=0.70\,\mathrm{kg\,m^2}$ .
Check: the result exceeds centre inertia and has units kg·m².

Understand and apply the equation

A rod rotates differently about its end than about its midpoint. If inertia about the centre-of-mass axis is known, an added term gives the value for a parallel shifted axis. First check both axes’ location and orientation, then avoid the common mistake of counting kinetic energy twice.

Two parallel axes, one centre of mass

The centre of mass is the object’s mass-weighted mean position. For a uniform thin rod it lies at the midpoint. Shifting to the end axis now requires the same axis orientation, such as perpendicular to the rod in both cases. A longitudinal and a transverse axis are not such a pair.[1]

I_A=I_{\mathrm{CM},\parallel A}+Md^2

M is the total mass of the same object; d is the shortest perpendicular distance between the parallel axes.[1]

The centre of mass fixes the reference axis location, while the target axis fixes its orientation. “Through the centre of mass” alone is insufficient: that axis must be parallel to the new axis. For positive mass, $I_A\ge I_{\mathrm{CM}}$ ; among parallel shifts, the centre-of-mass axis has the smallest mass moment. Translation along the same axis line gives $d=0$ .[1]

The centre axis gives the minimum

QUANTA STUDYquanta-study.de

Data from the fourth exercise. All axes remain parallel; d here is a signed displacement coordinate. Geometric distance is its magnitude. The minimum is at the centre of mass for this family of parallel axes.[1]

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Explore further: Integral proof in centre-of-mass coordinates

Choose the centre-of-mass axis as z. The parallel target axis through (a,b) gives each mass element squared distance $(x-a)^2+(y-b)^2$ . Coordinates x and y are measured from the centre of mass. Expand the squares and integrate each term.[1]

\begin{aligned}I_A&=\int(x^2+y^2)\,\mathrm dm\\&\quad-2a\int x\,\mathrm dm-2b\int y\,\mathrm dm\\&\quad+(a^2+b^2)\int\mathrm dm\end{aligned}

At the centre of mass, the first moments $\int x\,\mathrm dm$ and $\int y\,\mathrm dm$ are zero.[1]

\begin{aligned}\int x\,\mathrm dm&=\int y\,\mathrm dm=0\\d^2&=a^2+b^2\\I_A&=I_{\mathrm{CM}}+Md^2\end{aligned}

The addition depends only on total mass and axis distance. For a reference axis not passing through the centre of mass, cross terms generally remain.[1]

Rearrange and check the geometry

\begin{aligned}&I_{\mathrm{CM}}=I_A-Md^2\\&d=\sqrt{\frac{I_A-I_{\mathrm{CM}}}{M}}\quad(M>0)\end{aligned}

A real distance requires $I_A-I_{\mathrm{CM}}\ge0$ . The square root gives a distance magnitude, not direction.[1]

For two parallel axes A and B neither passing through the centre of mass, refer both to the same centre-of-mass axis: $I_B-I_A=M(d_B^2-d_A^2)$ . Do not simply add M times their mutual separation squared. A change in orientation, such as from a rod’s longitudinal axis to a transverse one, also needs a different reference moment.[1]

The inverse $M=(I_A-I_{\mathrm{CM}})/d^2$ requires $d>0$ and a positive inertia difference. At $d=0$ with equal moments, the equation does not determine mass; different moments about an identical parallel axis are incompatible with this model.[1]

From a rod to a composite rotor

A uniform thin rod

Given: $M=3.0\,\mathrm{kg}$ and length $\ell=0.80\,\mathrm m$ ; the target axis passes through an end perpendicular to the rod. The known parallel centre-of-mass axis has $I_{\mathrm{CM}}=\frac{M\ell^2}{12}$ .
Find: Find inertia about the end.

Approach: A uniform rod’s centre of mass is at its midpoint. Its end axis stays parallel to the centre axis, with half the rod length between them.
$\begin{aligned}I_A&=I_{\mathrm{CM}}+Md^2\\I_{\mathrm{CM}}&=\frac{M\ell^2}{12}\\d&=\frac\ell2\end{aligned}$
Substitution: Use the specified masses and perpendicular axis distances in SI units.
$\begin{aligned}I_{\mathrm{CM}}&=\frac{3.0(0.80)^2}{12}=0.160\,\mathrm{kg\,m^2}\\{} d&=0.40\,\mathrm m\end{aligned}$
Add the shift contribution.
$I_{\mathrm{end}}=0.160+3.0(0.40)^2=0.640\,\mathrm{kg\,m^2}$

Result: The end axis has $I=0.640\,\mathrm{kg\,m^2}$ , four times the centre-of-mass value.

Check: Independent integral: $I_A=\int_0^\ell x^2\frac M\ell\,\mathrm dx=\frac{M\ell^2}{3}=0.640\,\mathrm{kg\,m^2}$ . Both axes have the same orientation; d is not the full rod length.

Two small disks, two axes

Given: Two uniform thin disks each have $m=0.50\,\mathrm{kg}$ and $R=0.10\,\mathrm m$ . Their centres lie at $x=\pm0.30\,\mathrm m$ on a massless frame. All axes are perpendicular to the disk plane.
Find: Find I about the combined centre-of-mass axis and a parallel axis shifted by 0.20 m along +x.

Approach: Each disk retains intrinsic inertia about its parallel central axis. First add both disk contributions about the combined centre of mass, then shift the total mass.
$\begin{aligned}I_{\mathrm{CM}}&=\sum_{j=1}^2\left(\frac12mR^2+md_j^2\right)\\I_A&=I_{\mathrm{CM}}+Md^2\end{aligned}$
Substitution: Use the specified masses and perpendicular axis distances in SI units.
$I_{\mathrm{CM}}=2\left[\frac12(0.50)(0.10)^2+0.50(0.30)^2\right]=0.095\,\mathrm{kg\,m^2}$
Shift the whole object using its total mass 1.0 kg.
$I_A=0.095+(1.0)(0.20)^2=0.135\,\mathrm{kg\,m^2}$

Result: $I_{\mathrm{CM}}=0.095\,\mathrm{kg\,m^2}$ and $I_A=0.135\,\mathrm{kg\,m^2}$ . Here the disks should not be treated as particles without intrinsic contributions.

Check: Direct distances to the target axis are 0.10 and 0.50 m. Thus $I_A=(2\cdot0.0025+0.50\cdot0.10^2+0.50\cdot0.50^2)\,\mathrm{kg\,m^2}=0.135\,\mathrm{kg\,m^2}$ . This verifies the whole-body shift.

Explore further: Composite objects and removed material

\begin{aligned}&I_A=\sum_j\left(I_{\mathrm{CM},j}+m_jd_j^2\right)\\&x_{\mathrm{CM}}=\frac{\sum_jm_jx_j}{\sum_jm_j}\end{aligned}

Every component moment must use parallel axes and distance to the common target axis.[1]

For an asymmetric arrangement, the combined centre of mass is not automatically at the geometric midpoint. Find its location first if you need the combined $I_{\mathrm{CM}}$ . A hole can be subtracted as missing material when it has the same assumed density; total mass and possibly centre of mass change. Negative bookkeeping contributions represent removed material, not physically negative mass.[1]

Moment of inertia of a system of particles sums contributions from the actual mass distribution to the chosen axis. For the extended components considered here, retain intrinsic inertia as well; shifting the reference does not automatically turn a small disk into a point mass.[1]

Use the same axis in energy and dynamics

\frac12I_A\omega^2=\frac12I_{\mathrm{CM}}\omega^2+\frac12M(d\omega)^2

The same energy in two descriptions: rotation about A, or intrinsic rotation plus centre-of-mass motion. Do not then add translational energy a second time.[2]

Rotational energy of a rigid body requires inertia about the actual axis used. The parallel-axis theorem supplies that axial value; it creates no extra energy term beyond the already equivalent descriptions.[2]

Five exercises with solutions

Calculate · Foundation

An object with $I_{\mathrm{CM}}=0.18\,\mathrm{kg\,m^2}$ and $M=1.6\,\mathrm{kg}$ has a parallel axis 0.25 m away. Find IA.

Show hint

Add $Md^2$ , not Md.

Learning goal: Add the quadratic offset contribution to centre-of-mass inertia.

Show solution and calculation

Approach: The reference is the parallel centre-of-mass axis, so $I_A=I_{\mathrm{CM}}+Md^2$ .
$Md^2=1.6\cdot0.25^2\,\mathrm{kg\,m^2}=0.10\,\mathrm{kg\,m^2}$ .
Thus $I_A=I_{\mathrm{CM}}+Md^2=0.28\,\mathrm{kg\,m^2}$ .
It exceeds $I_{\mathrm{CM}}$ , as required for a centre-of-mass reference axis.

I_A=0.18+1.6(0.25)^2=0.28\,\mathrm{kg\,m^2}

Calculate · Application

$I_A=0.50\,\mathrm{kg\,m^2}$ , $M=2.0\,\mathrm{kg}$ and $d=0.40\,\mathrm m$ to a parallel centre-of-mass axis. Find $I_{\mathrm{CM}}$ .

Show hint

Subtract the addition when reversing the shift.

Learning goal: Recover centre-of-mass inertia from an offset axis.

Show solution and calculation

Approach: Returning to the parallel centre-of-mass axis requires subtracting the shift contribution.
$I_{\mathrm{CM}}=I_A-Md^2=(0.50-2.0\cdot0.40^2)\,\mathrm{kg\,m^2}=0.18\,\mathrm{kg\,m^2}$ .

Check: $(0.18+0.32)\,\mathrm{kg\,m^2}=0.50\,\mathrm{kg\,m^2}$ . A negative result would signal inconsistent data or an unsuitable axis assumption.

I_{\mathrm{CM}}=0.50-2.0(0.40)^2=0.18\,\mathrm{kg\,m^2}

Error analysis · Application

For $M=2.0\,\mathrm{kg}$ and $I_{\mathrm{CM}}=0.10\,\mathrm{kg\,m^2}$ , parallel axes A and B lie 0.20 m and 0.50 m from the same centre-of-mass axis on the same side. Find IA and IB.

Show hint

Shift each axis separately from the centre of mass.

Learning goal: Compare two noncentral parallel axes through their common centre-of-mass axis.

Show solution and calculation

Approach: Both axes share the same known centre reference; calculate $I_A=I_{\mathrm{CM}}+Md_A^2$ and $I_B=I_{\mathrm{CM}}+Md_B^2$ separately.
$I_A=0.18\,\mathrm{kg\,m^2}$ and $I_B=0.60\,\mathrm{kg\,m^2}$ .
Their mutual separation is 0.30 m; $I_A+M(0.30\,\mathrm m)^2=0.36\,\mathrm{kg\,m^2}$ would be wrong.
Correctly, $I_B-I_A=M(0.50^2-0.20^2)\,\mathrm{m^2}=0.42\,\mathrm{kg\,m^2}$ .

I_B-I_A=M(d_B^2-d_A^2)=0.42\,\mathrm{kg\,m^2}

Interpret · Application

A rigid 2.0 kg body has $I_{\mathrm{CM}}=0.12\,\mathrm{kg\,m^2}$ about a specified centre-of-mass axis. A graph plots I for parallel axes against signed displacement d to the right or left. Find I at $d=-0.30\,\mathrm m$ , 0 and +0.30 m. What shape and minimum must the curve have?

Show hint

Axis offset enters quadratically. Axes stay parallel; only their position changes.

Learning goal: Infer symmetry of an inertia–offset graph from the parallel-axis theorem.

Show solution and calculation

Approach: use the stated centre-of-mass axis as the reference and add the quadratic offset contribution.
Both offsets ±0.30 m give $I=0.30\,\mathrm{kg\,m^2}$ ; the centre axis gives 0.12 kg·m².
The curve is an upward parabola with its minimum at $d=0$ .

Check: the added term is never negative; displacement cannot lower inertia below the centre value for this family of parallel axes.

\begin{aligned}I(d)&=I_{\mathrm{CM}}+Md^2\\I(\pm0.30\,\mathrm m)&=0.12+2.0(0.30)^2=0.30\,\mathrm{kg\,m^2}\\I(0)&=0.12\,\mathrm{kg\,m^2}\end{aligned}

[2][1]

Error analysis · Application

A massless rigid rod carries two point masses of 0.50 kg each at $x=\pm0.40\,\mathrm m$ , with centre of mass at the origin. Axis A is the x-axis along the rod; axis B is the z-axis through the same origin. A solution uses $d=0$ in the parallel-axis theorem and states $I_B=I_A$ . Check using the point masses’ perpendicular distances to each axis.

Show hint

The axes intersect but are not parallel. Use the definition of mass moment of inertia separately for each axis.

Learning goal: Avoid replacing axis orientation by separation alone.

Show solution and calculation

Approach: perpendicular distances are zero for both point masses about the x-axis, but 0.40 m about the z-axis.
Thus $I_A=0$ and $I_B=0.16\,\mathrm{kg\,m^2}$ .
The parallel-axis theorem does not apply to this axis pair.

Check: sharing a centre of mass does not specify orientation. Zero inertia here belongs to ideal point masses on a massless line, not a rod with finite cross-section.

\begin{aligned}I_A&=\sum_i m_i r_{\perp,A,i}^2=0\\I_B&=2(0.50\,\mathrm{kg})(0.40\,\mathrm m)^2=0.16\,\mathrm{kg\,m^2}\\\hat{\vec e}_A\cdot\hat{\vec e}_B&=0\end{aligned}

[1][3]

What the theorem does not replace

For nonparallel axes, use direct calculation or an inertia tensor with transformed directions. Even a correct separation cannot rescue a wrongly oriented centre-of-mass axis. Record shape, density assumption and axis direction for every tabulated value before adding the shift term.[3]

Key points

Check that the axes are parallel and the reference axis passes through the centre of mass. Add $Md^2$ using total mass and perpendicular axis separation. Composite objects retain intrinsic moments; avoid counting rotational and translational energy twice.

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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October 2, 2026Linked mass distribution and rotational energy and added conditions of the mass inverse; preserved geometry, examples, exercises and figure.
October 1, 2026Clarified the entry and learning transitions; checked and refined existing exercise solutions, typeset mathematical prose and added an original quantitative article figure.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Parallel-axis theorem for mass moment of inertia. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/massentraegheitsmoment-parallelachsensatz

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Frequently asked questions about Parallel-axis theorem for mass moment of inertia

What does the parallel-axis theorem for mass moment of inertia describe physically?+

M is the total mass of the same object; d is the shortest perpendicular distance between the parallel axes. The centre of mass fixes the reference axis location, while the target axis fixes its orientation. “Through the centre of mass” alone is insufficient: that axis must be parallel to the new axis. For positive mass, $I_A\ge I_{\mathrm{CM}}$ ; among parallel shifts, the centre-of-mass axis has the smallest mass moment. Translation along the same axis line gives $d=0$ . The addition depends only on total mass and axis distance. For a reference axis not passing through the centre of mass, cross terms generally remain.

How can I check a calculation involving the parallel-axis theorem for mass moment of inertia?+

The target axis is shifted 0.20 m along +x from the combined centre-of-mass axis. Two uniform thin disks each have $m=0.50\,\mathrm{kg}$ and $R=0.10\,\mathrm m$ . Their centres lie at $x=\pm0.30\,\mathrm m$ on a massless frame. All axes are perpendicular to the disk plane. Each disk contributes its intrinsic $\frac12mR^2$ plus $md^2$ . Shift the whole object using its total mass 1.0 kg. $I_{\mathrm{CM}}=0.095\,\mathrm{kg\,m^2}$ and $I_A=0.135\,\mathrm{kg\,m^2}$ . Here the disks should not be treated as particles without intrinsic contributions. Direct distances to the target axis are 0.10 and 0.50 m: $(2\cdot0.0025+0.50\cdot0.10^2+0.50\cdot0.50^2)\,\mathrm{kg\,m^2}=0.135\,\mathrm{kg\,m^2}$ . This verifies the whole-body shift. Every component moment must use parallel axes and distance to the common target axis.

What assumptions and limits apply to the parallel-axis theorem for mass moment of inertia?+

The parallel-axis theorem is a geometric identity for the given mass distribution. It requires neither a particular rotation rate nor equilibrium. Its use in rigid-body dynamics nevertheless requires an appropriate description of that distribution during the motion. If liquid moves within a tank, centre of mass and inertia must be recalculated; the theorem does not supply one permanently constant rotor value. For nonparallel axes, use direct calculation or an inertia tensor with transformed directions. Even a correct separation cannot rescue a wrongly oriented centre-of-mass axis. Record shape, density assumption and axis direction for every tabulated value before adding the shift term.

How do I check units and scale for the parallel-axis theorem for mass moment of inertia?+

All inertia terms and $Md^2$ have units kg·m². Use metres for d; a distance of 30 cm gives $d^2=0.090\,\mathrm{m^2}$ . An area second moment instead shifts by $Ad^2$ and has units m⁴. It cannot be used in the same energy balance or in $L=I\omega$ . For two parallel axes A and B neither passing through the centre of mass, refer both to the same centre-of-mass axis: $I_B-I_A=M(d_B^2-d_A^2)$ . Do not simply add M times their mutual separation squared. A change in orientation, such as from a rod’s longitudinal axis to a transverse one, also needs a different reference moment.

When should I combine the parallel-axis theorem for mass moment of inertia with another relation?+

A door rotates about a hinge line near its edge, and a pendulum about its suspension. Both equations of motion need inertia about that actual axis. The centre-of-mass moment alone underestimates resistance to angular acceleration. For pure fixed rotation, $E=\frac12I_A\omega^2$ ; alternatively, add centre-of-mass translation and intrinsic rotation. The same energy in two descriptions: rotation about A, or intrinsic rotation plus centre-of-mass motion. Do not then add translational energy a second time. For an asymmetric arrangement, the combined centre of mass is not automatically at the geometric midpoint. Find its location first if you need the combined $I_{\mathrm{CM}}$ . A hole can be subtracted as missing material when it has the same assumed density; total mass and possibly centre of mass change. Negative bookkeeping contributions represent removed material, not physically negative mass.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §10.5: Calculating Moments of Inertia

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

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Symbol

Meaning

Unit

I

Mass moment of inertia about the shifted axis

kg·m²

ICM

Mass moment of inertia about the parallel centre-of-mass axis

kg·m²

M

Total mass of the same distribution

d

Shortest perpendicular distance between the parallel axes

Exam blueprint

Validity range

Derivation steps

Expand shifted squared distances; cross terms vanish because the reference axis passes through the centre of mass.

1Choose the parallel centre-of-mass axis as z; the target axis passes through (a,b).
2A mass element has target-axis distance squared $r_A^2=(x-a)^2+(y-b)^2$ .
3Expand $r_A^2=x^2+y^2-2ax-2by+a^2+b^2$ .
4Integrate with dm; $\int(x^2+y^2)\,\mathrm dm$ is $I_{\mathrm{CM}}$ .
5The centre-of-mass definition gives $\int x\,\mathrm dm=\int y\,\mathrm dm=0$ , so cross terms vanish.
6Using $\int\mathrm dm=M$ and $d^2=a^2+b^2$ leaves $I_A=I_{\mathrm{CM}}+Md^2$ .

How do you calculate with Parallel-axis theorem for mass moment of inertia?

Here is how to work through a typical Parallel-axis theorem for mass moment of inertia (I = ICM + Md²) task step by step:

Task

An object has centre-axis inertia $I_{\mathrm{CM}}$ =0.20 kg·m² and total mass 2.0 kg. A new axis is parallel at perpendicular separation 0.50 m. Find mass inertia about the new axis.

Solution path

Approach: $I_A=I_{\mathrm{CM}}+Md^2$ .
The addition is $(2.0\,\mathrm{kg})(0.50\,\mathrm m)^2=0.50\,\mathrm{kg\,m^2}$ ; total $I_A=0.70\,\mathrm{kg\,m^2}$ .

Check: the result exceeds centre inertia and has units kg·m².

Two parallel axes, one centre of mass

I_A=I_{\mathrm{CM},\parallel A}+Md^2

M is the total mass of the same object; d is the shortest perpendicular distance between the parallel axes.[1]

The centre axis gives the minimum

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Explore further: Integral proof in centre-of-mass coordinates

\begin{aligned}I_A&=\int(x^2+y^2)\,\mathrm dm\\&\quad-2a\int x\,\mathrm dm-2b\int y\,\mathrm dm\\&\quad+(a^2+b^2)\int\mathrm dm\end{aligned}

At the centre of mass, the first moments $\int x\,\mathrm dm$ and $\int y\,\mathrm dm$ are zero.[1]

\begin{aligned}\int x\,\mathrm dm&=\int y\,\mathrm dm=0\\d^2&=a^2+b^2\\I_A&=I_{\mathrm{CM}}+Md^2\end{aligned}

The addition depends only on total mass and axis distance. For a reference axis not passing through the centre of mass, cross terms generally remain.[1]

Rearrange and check the geometry

\begin{aligned}&I_{\mathrm{CM}}=I_A-Md^2\\&d=\sqrt{\frac{I_A-I_{\mathrm{CM}}}{M}}\quad(M>0)\end{aligned}

A real distance requires $I_A-I_{\mathrm{CM}}\ge0$ . The square root gives a distance magnitude, not direction.[1]

From a rod to a composite rotor

A uniform thin rod

Given: $M=3.0\,\mathrm{kg}$ and length $\ell=0.80\,\mathrm m$ ; the target axis passes through an end perpendicular to the rod. The known parallel centre-of-mass axis has $I_{\mathrm{CM}}=\frac{M\ell^2}{12}$ .
Find: Find inertia about the end.

Approach: A uniform rod’s centre of mass is at its midpoint. Its end axis stays parallel to the centre axis, with half the rod length between them.
$\begin{aligned}I_A&=I_{\mathrm{CM}}+Md^2\\I_{\mathrm{CM}}&=\frac{M\ell^2}{12}\\d&=\frac\ell2\end{aligned}$
Substitution: Use the specified masses and perpendicular axis distances in SI units.
$\begin{aligned}I_{\mathrm{CM}}&=\frac{3.0(0.80)^2}{12}=0.160\,\mathrm{kg\,m^2}\\{} d&=0.40\,\mathrm m\end{aligned}$
Add the shift contribution.
$I_{\mathrm{end}}=0.160+3.0(0.40)^2=0.640\,\mathrm{kg\,m^2}$

Result: The end axis has $I=0.640\,\mathrm{kg\,m^2}$ , four times the centre-of-mass value.

Check: Independent integral: $I_A=\int_0^\ell x^2\frac M\ell\,\mathrm dx=\frac{M\ell^2}{3}=0.640\,\mathrm{kg\,m^2}$ . Both axes have the same orientation; d is not the full rod length.

Two small disks, two axes

Given: Two uniform thin disks each have $m=0.50\,\mathrm{kg}$ and $R=0.10\,\mathrm m$ . Their centres lie at $x=\pm0.30\,\mathrm m$ on a massless frame. All axes are perpendicular to the disk plane.
Find: Find I about the combined centre-of-mass axis and a parallel axis shifted by 0.20 m along +x.

Approach: Each disk retains intrinsic inertia about its parallel central axis. First add both disk contributions about the combined centre of mass, then shift the total mass.
$\begin{aligned}I_{\mathrm{CM}}&=\sum_{j=1}^2\left(\frac12mR^2+md_j^2\right)\\I_A&=I_{\mathrm{CM}}+Md^2\end{aligned}$
Substitution: Use the specified masses and perpendicular axis distances in SI units.
$I_{\mathrm{CM}}=2\left[\frac12(0.50)(0.10)^2+0.50(0.30)^2\right]=0.095\,\mathrm{kg\,m^2}$
Shift the whole object using its total mass 1.0 kg.
$I_A=0.095+(1.0)(0.20)^2=0.135\,\mathrm{kg\,m^2}$

Result: $I_{\mathrm{CM}}=0.095\,\mathrm{kg\,m^2}$ and $I_A=0.135\,\mathrm{kg\,m^2}$ . Here the disks should not be treated as particles without intrinsic contributions.

Explore further: Composite objects and removed material

\begin{aligned}&I_A=\sum_j\left(I_{\mathrm{CM},j}+m_jd_j^2\right)\\&x_{\mathrm{CM}}=\frac{\sum_jm_jx_j}{\sum_jm_j}\end{aligned}

Every component moment must use parallel axes and distance to the common target axis.[1]

Use the same axis in energy and dynamics

\frac12I_A\omega^2=\frac12I_{\mathrm{CM}}\omega^2+\frac12M(d\omega)^2

The same energy in two descriptions: rotation about A, or intrinsic rotation plus centre-of-mass motion. Do not then add translational energy a second time.[2]

Five exercises with solutions

Calculate · Foundation

An object with $I_{\mathrm{CM}}=0.18\,\mathrm{kg\,m^2}$ and $M=1.6\,\mathrm{kg}$ has a parallel axis 0.25 m away. Find IA.

Show hint

Add $Md^2$ , not Md.

Learning goal: Add the quadratic offset contribution to centre-of-mass inertia.

Show solution and calculation

Approach: The reference is the parallel centre-of-mass axis, so $I_A=I_{\mathrm{CM}}+Md^2$ .
$Md^2=1.6\cdot0.25^2\,\mathrm{kg\,m^2}=0.10\,\mathrm{kg\,m^2}$ .
Thus $I_A=I_{\mathrm{CM}}+Md^2=0.28\,\mathrm{kg\,m^2}$ .
It exceeds $I_{\mathrm{CM}}$ , as required for a centre-of-mass reference axis.

I_A=0.18+1.6(0.25)^2=0.28\,\mathrm{kg\,m^2}

Calculate · Application

$I_A=0.50\,\mathrm{kg\,m^2}$ , $M=2.0\,\mathrm{kg}$ and $d=0.40\,\mathrm m$ to a parallel centre-of-mass axis. Find $I_{\mathrm{CM}}$ .

Show hint

Subtract the addition when reversing the shift.

Learning goal: Recover centre-of-mass inertia from an offset axis.

Show solution and calculation

Approach: Returning to the parallel centre-of-mass axis requires subtracting the shift contribution.
$I_{\mathrm{CM}}=I_A-Md^2=(0.50-2.0\cdot0.40^2)\,\mathrm{kg\,m^2}=0.18\,\mathrm{kg\,m^2}$ .

Check: $(0.18+0.32)\,\mathrm{kg\,m^2}=0.50\,\mathrm{kg\,m^2}$ . A negative result would signal inconsistent data or an unsuitable axis assumption.

I_{\mathrm{CM}}=0.50-2.0(0.40)^2=0.18\,\mathrm{kg\,m^2}

Error analysis · Application

For $M=2.0\,\mathrm{kg}$ and $I_{\mathrm{CM}}=0.10\,\mathrm{kg\,m^2}$ , parallel axes A and B lie 0.20 m and 0.50 m from the same centre-of-mass axis on the same side. Find IA and IB.

Show hint

Shift each axis separately from the centre of mass.

Learning goal: Compare two noncentral parallel axes through their common centre-of-mass axis.

Show solution and calculation

Approach: Both axes share the same known centre reference; calculate $I_A=I_{\mathrm{CM}}+Md_A^2$ and $I_B=I_{\mathrm{CM}}+Md_B^2$ separately.
$I_A=0.18\,\mathrm{kg\,m^2}$ and $I_B=0.60\,\mathrm{kg\,m^2}$ .
Their mutual separation is 0.30 m; $I_A+M(0.30\,\mathrm m)^2=0.36\,\mathrm{kg\,m^2}$ would be wrong.
Correctly, $I_B-I_A=M(0.50^2-0.20^2)\,\mathrm{m^2}=0.42\,\mathrm{kg\,m^2}$ .

I_B-I_A=M(d_B^2-d_A^2)=0.42\,\mathrm{kg\,m^2}

Interpret · Application

A rigid 2.0 kg body has $I_{\mathrm{CM}}=0.12\,\mathrm{kg\,m^2}$ about a specified centre-of-mass axis. A graph plots I for parallel axes against signed displacement d to the right or left. Find I at $d=-0.30\,\mathrm m$ , 0 and +0.30 m. What shape and minimum must the curve have?

Show hint

Axis offset enters quadratically. Axes stay parallel; only their position changes.

Learning goal: Infer symmetry of an inertia–offset graph from the parallel-axis theorem.

Show solution and calculation

Approach: use the stated centre-of-mass axis as the reference and add the quadratic offset contribution.
Both offsets ±0.30 m give $I=0.30\,\mathrm{kg\,m^2}$ ; the centre axis gives 0.12 kg·m².
The curve is an upward parabola with its minimum at $d=0$ .

Check: the added term is never negative; displacement cannot lower inertia below the centre value for this family of parallel axes.

\begin{aligned}I(d)&=I_{\mathrm{CM}}+Md^2\\I(\pm0.30\,\mathrm m)&=0.12+2.0(0.30)^2=0.30\,\mathrm{kg\,m^2}\\I(0)&=0.12\,\mathrm{kg\,m^2}\end{aligned}

[2][1]

Error analysis · Application

A massless rigid rod carries two point masses of 0.50 kg each at $x=\pm0.40\,\mathrm m$ , with centre of mass at the origin. Axis A is the x-axis along the rod; axis B is the z-axis through the same origin. A solution uses $d=0$ in the parallel-axis theorem and states $I_B=I_A$ . Check using the point masses’ perpendicular distances to each axis.

Show hint

The axes intersect but are not parallel. Use the definition of mass moment of inertia separately for each axis.

Learning goal: Avoid replacing axis orientation by separation alone.

Show solution and calculation

Approach: perpendicular distances are zero for both point masses about the x-axis, but 0.40 m about the z-axis.
Thus $I_A=0$ and $I_B=0.16\,\mathrm{kg\,m^2}$ .
The parallel-axis theorem does not apply to this axis pair.

Check: sharing a centre of mass does not specify orientation. Zero inertia here belongs to ideal point masses on a massless line, not a rod with finite cross-section.

\begin{aligned}I_A&=\sum_i m_i r_{\perp,A,i}^2=0\\I_B&=2(0.50\,\mathrm{kg})(0.40\,\mathrm m)^2=0.16\,\mathrm{kg\,m^2}\\\hat{\vec e}_A\cdot\hat{\vec e}_B&=0\end{aligned}

[1][3]

What the theorem does not replace

Key points

Sources for this explanation

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October 2, 2026Linked mass distribution and rotational energy and added conditions of the mass inverse; preserved geometry, examples, exercises and figure.
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October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Parallel-axis theorem for mass moment of inertia

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Parallel axis theorem calculator

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Derivation & background

Exam blueprint

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Derivation steps

Rearrangements

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Common mistakes

Exam context

Rotational motion

Worked example

Applications

How do you calculate with Parallel-axis theorem for mass moment of inertia?

Two parallel axes, one centre of mass

The centre axis gives the minimum

Explore further: Integral proof in centre-of-mass coordinates

Rearrange and check the geometry

From a rod to a composite rotor

A uniform thin rod

Two small disks, two axes

Explore further: Composite objects and removed material

Use the same axis in energy and dynamics

Five exercises with solutions

An object with ICM=0.18 kg m2I_{\mathrm{CM}}=0.18\,\mathrm{kg\,m^2}ICM​=0.18kgm2 and M=1.6 kgM=1.6\,\mathrm{kg}M=1.6kg has a parallel axis 0.25 m away. Find IA.

IA=0.50 kg m2I_A=0.50\,\mathrm{kg\,m^2}IA​=0.50kgm2, M=2.0 kgM=2.0\,\mathrm{kg}M=2.0kg and d=0.40 md=0.40\,\mathrm md=0.40m to a parallel centre-of-mass axis. Find ICMI_{\mathrm{CM}}ICM​.

For M=2.0 kgM=2.0\,\mathrm{kg}M=2.0kg and ICM=0.10 kg m2I_{\mathrm{CM}}=0.10\,\mathrm{kg\,m^2}ICM​=0.10kgm2, parallel axes A and B lie 0.20 m and 0.50 m from the same centre-of-mass axis on the same side. Find IA and IB.

What the theorem does not replace

Key points

Sources for this explanation

Frequently asked questions about Parallel-axis theorem for mass moment of inertia

Subject references and further reading

More formulas: Mechanics

Retain Parallel-axis theorem for mass moment of inertia for exams

Parallel-axis theorem for mass moment of inertia

Variables & units

Parallel axis theorem calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Rotational motion

Worked example

Applications

How do you calculate with Parallel-axis theorem for mass moment of inertia?

Two parallel axes, one centre of mass

The centre axis gives the minimum

Explore further: Integral proof in centre-of-mass coordinates

Rearrange and check the geometry

From a rod to a composite rotor

A uniform thin rod

Two small disks, two axes

Explore further: Composite objects and removed material

Use the same axis in energy and dynamics

Five exercises with solutions

An object with ICM=0.18 kg m2I_{\mathrm{CM}}=0.18\,\mathrm{kg\,m^2}ICM​=0.18kgm2 and M=1.6 kgM=1.6\,\mathrm{kg}M=1.6kg has a parallel axis 0.25 m away. Find IA.

IA=0.50 kg m2I_A=0.50\,\mathrm{kg\,m^2}IA​=0.50kgm2, M=2.0 kgM=2.0\,\mathrm{kg}M=2.0kg and d=0.40 md=0.40\,\mathrm md=0.40m to a parallel centre-of-mass axis. Find ICMI_{\mathrm{CM}}ICM​.

For M=2.0 kgM=2.0\,\mathrm{kg}M=2.0kg and ICM=0.10 kg m2I_{\mathrm{CM}}=0.10\,\mathrm{kg\,m^2}ICM​=0.10kgm2, parallel axes A and B lie 0.20 m and 0.50 m from the same centre-of-mass axis on the same side. Find IA and IB.

What the theorem does not replace

Key points

Sources for this explanation

Frequently asked questions about Parallel-axis theorem for mass moment of inertia

Subject references and further reading

More formulas: Mechanics

Retain Parallel-axis theorem for mass moment of inertia for exams

An object with $I_{\mathrm{CM}}=0.18\,\mathrm{kg\,m^2}$ and $M=1.6\,\mathrm{kg}$ has a parallel axis 0.25 m away. Find IA.

$I_A=0.50\,\mathrm{kg\,m^2}$ , $M=2.0\,\mathrm{kg}$ and $d=0.40\,\mathrm m$ to a parallel centre-of-mass axis. Find $I_{\mathrm{CM}}$ .

For $M=2.0\,\mathrm{kg}$ and $I_{\mathrm{CM}}=0.10\,\mathrm{kg\,m^2}$ , parallel axes A and B lie 0.20 m and 0.50 m from the same centre-of-mass axis on the same side. Find IA and IB.

An object with $I_{\mathrm{CM}}=0.18\,\mathrm{kg\,m^2}$ and $M=1.6\,\mathrm{kg}$ has a parallel axis 0.25 m away. Find IA.

$I_A=0.50\,\mathrm{kg\,m^2}$ , $M=2.0\,\mathrm{kg}$ and $d=0.40\,\mathrm m$ to a parallel centre-of-mass axis. Find $I_{\mathrm{CM}}$ .

For $M=2.0\,\mathrm{kg}$ and $I_{\mathrm{CM}}=0.10\,\mathrm{kg\,m^2}$ , parallel axes A and B lie 0.20 m and 0.50 m from the same centre-of-mass axis on the same side. Find IA and IB.