Physics · Rotation

Angular momentum of a rigid body

Q: How can I check a calculation involving rigid-body angular momentum?

Two 1.0 kg particles sit on a massless rigid rod at $\vec r_1=(0.30,0,0.40)\,\mathrm m$ and $\vec r_2=-\vec r_1$. O is the centre of mass. Instantaneously $\vec\omega=(0,0,6.0)\,\mathrm{rad\,s^{-1}}$. Velocities are $\vec v_1=(0,1.80,0)\,\mathrm{m\,s^{-1}}$ and $\vec v_2=-\vec v_1$. Calculate each cross product. Both particles contribute the same angular momentum; add them. Lz=1.08 kg·m²/s agrees with $I_{zz}\omega_z$. The full vector additionally has $L_x=-1.44\,\mathrm{kg\,m^2\,s^{-1}}$ and is not parallel to ω. Its magnitude is $\sqrt{1.44^2+1.08^2}=1.80\,\mathrm{kg\,m^2\,s^{-1}}$, not 1.08. Direct energy is $2\cdot\frac12\cdot1\cdot1.80^2\,\mathrm J=3.24\,\mathrm J$, also $\frac12I_{zz}\omega_z^2$. Keeping the z-axis fixed requires transverse bearing torques. The tensor form $\vec L=\mathsf I\vec\omega$ gives the same components; pure z-rotation specifically has Izz = 0.18 kg·m².

Q: What assumptions and limits apply to rigid-body angular momentum?

For general spatial rotation, $\vec L=\mathsf I\vec\omega$ uses a tensor. Even in principal-axis coordinates, $L_i=I_i\omega_i$; with several nonzero ω-components and unequal principal moments, L and ω are generally not parallel. One scalar I therefore cannot be applied to every component. A body-fixed coordinate system itself rotates. The inertial derivative is then the body-frame derivative of L plus $\vec\omega\times\vec L$. A vector apparently constant in that frame may change in the inertial frame. Simple rotor problems can use the fixed-axis component; gyroscopes, free tumbling and bearing loads require the full vector balance.

Q: When should I combine rigid-body angular momentum with another relation?

Figure skating and retractable weights illustrate $I_1\omega_1=I_2\omega_2$ when external axial torque is negligible. During retraction, the system is not a single rigid body. Use the relation only when the relevant axial angular momentum is genuinely represented by Iω and additional relative motion is accounted for appropriately. Conserved angular momentum does not imply conserved rotational energy: $E=\frac{L_z^2}{2I}$. Halving I at constant Lz doubles ωz and E. The additional energy comes from internal work, such as muscles or a mechanism. A system with no external torque can therefore change rotational energy without violating angular-momentum conservation.

For pure rigid rotation about a fixed axis, the axial component obeys $L_z=I_{zz}\omega_z$ . Along a principal axis, scalar I also describes the entire parallel angular-momentum vector.

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Formula

Calculate

L = I ω

L in $\mathrm{kg\,m^2\,s^{-1}}$ ; I in $\mathrm{kg\,m^2}$ ; $\omega$ in $\mathrm{rad\,s^{-1}}$ .

Variables & units

Variables & units
Symbol	Meaning	Unit
$L$	Angular momentum about the specified reference; Lz is the signed axial component. The scalar full-vector relation L⃗=Iω⃗ applies to principal-axis rotation.	kg·m²/s
$I$	Mass moment of inertia about the same axis	kg·m²
$ω$	Angular velocity about the same axis; ωz uses a specified positive axis direction	rad/s

LaTeX

L = I \omega

QUANTA STUDYquanta-study.de

xz-plane section: equal masses on a tilted rod through O. Angular velocity along positive z; angular momentum points up-left, perpendicular to the rod. L and ω are not parallel. — Non-principal-axis rotation: ω points along z, but L also has a negative x component. This two-particle body gives L = (−1.44, 0, 1.08) kg·m²/s. L and ω use separate arrow scales.

Angular momentum calculator

Find the angular momentum, moment of inertia or angular velocity of a rigid body turning about a fixed axis; the sign shows the direction of rotation.

Find

Moment of inertia Izz

Moment of inertia about exactly this axis, always positive, e.g. 0.24 kg·m².

Angular velocity ωz

Angular velocity, e.g. 450 rpm; positive if the body turns anticlockwise seen from the tip of the axis.

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

Working · rearrange for Lz, substitute, evaluate

1. Rearrange for Lz
$L_z=I_{zz}\omega_z$
2. Substitute values (in SI units)
- ωz = 450 rpm = 47.1239 rad/s
$L_z=0.24\cdot 47.1239$
3. Result
Lz = 11.31 kg·m²/s

Result: Angular momentum about the axis Lz

Lz = 11.31 kg·m²/s

With Izz = 0.24 kg·m² and ωz = 450 rpm, Lz = 11.31 kg·m²/s.

Quick tips

Formula triangle: L on top, I and ω below. Cover the quantity you need and the rest is the calculation, e.g. ω = L ÷ I.
ω from the rotation rate: ω = 2π · n. 450 revolutions per minute are 7.5 per second, so ω ≈ 47 rad/s. By hand, never insert revolutions per minute or Hz directly for ω.

Assumptions and limits

Rigid body rotating purely about the selected fixed inertial z axis through a fixed reference point, with Izz > 0. Positive orientation follows the right-hand rule. Lz = Izzωz is the signed axial component; other vector components are not determined. The full angular momentum is parallel to ω only for a principal axis. No moving axis, centre-of-mass translation or changing mass distribution.

Derivation & background

About a fixed inertial reference point, $\vec L=\sum_i\vec r_i\times m_i\vec v_i$ . Pure fixed-z rotation gives $L_z=I_{zz}\omega_z$ , including nonprincipal axes with possible transverse components. Generally $\vec L=\mathsf I\vec\omega$ ; along a principal axis its scalar moment suffices. Net external torque satisfies $\sum\vec\tau_{\mathrm{ext}}=\frac{d\vec L}{dt}$ . Only zero total torque conserves the entire vector; zero axial torque conserves the corresponding fixed component.

Exam blueprint

Validity range

$L_z=I_{zz}\omega_z$ is the axial component for pure rigid rotation about fixed z even without a principal axis. The full vector obeys $\vec L=\mathsf I\vec\omega$ ; L=Iω with scalar I and parallel vectors requires principal-axis rotation. Reference point and velocity frame must match.

Derivation steps

The z-component of particle cross products gives $I_{zz}$ ωz; principal-axis alignment is required for the entire parallel vector.

1Choose a fixed inertial reference point on z and pure rigid rotation about that axis.
2From $\vec v=\vec\omega\times\vec r$ , obtain $v_x=-\omega_z y$ and $v_y=\omega_z x$ .
3The z component of $\vec r\times m\vec v$ is $m(xv_y-yv_x)$ .
4Substitution gives one particle’s $L_{z,i}=m_i(x_i^2+y_i^2)\omega_z$ .
5Summing gives $L_z=\sum_i m_ir_{\perp i}^2\omega_z=I_{zz}\omega_z$ .
6Generally $\vec L=\mathsf I\vec\omega$ . Along a principal axis, this reduces to parallel vectors using that principal moment.

Rearrangements

Angular speed from angular momentum

\omega_z=\frac{L_z}{I_{zz}}

For the signed axial component with $I_{zz}>0$ . This does not determine transverse components of the full vector.

Task variant

A rotor has $I=0.60\,\mathrm{kg\,m^2}$ and L=3.0 $\mathrm{kg\,m^2\,s^{-1}}$ . Find $\omega$ .

$\omega=\frac{L}{I}=\frac{3.0\,\mathrm{kg\,m^2\,s^{-1}}}{0.60\,\mathrm{kg\,m^2}}=5.0\,\mathrm{rad\,s^{-1}}$ .

Common mistakes

Confusing angular momentum with torque or using I for another axis.

L has units $\mathrm{kg\,m^2\,s^{-1}}$ ; torque has units $\mathrm{N\,m}$ and satisfies $\tau$ = $\frac{d\vec L}{dt}$ .

Exam context

Changing spin rate when arms are pulled in, rotors and conservation with negligible external torque.

Formula cluster

Angular momentum and torque

Angular-momentum balance explains changes in rotation.

I = Σ m_ir_i²

Moment of inertia of a particle system

M = F·r·sin(θ)

Torque

ω = 2πf; v = ω·r

Angular Velocity and Angular Frequency

p_vor = p_nach

Conservation of Momentum

Worked example

A rotor has $I=0.80\,\mathrm{kg\,m^2}$ about a fixed principal axis and $\omega_z=+5.0\,\mathrm{rad\,s^{-1}}$ . This gives $L_z=I\omega_z=+4.0\,\mathrm{kg\,m^2\,s^{-1}}$ . Check: conserved axial momentum with halved inertia would double angular velocity.

Applications

Gyroscopes, rotors, figure skating and rotation with changing mass distribution.

How do you calculate with Angular momentum of a rigid body?

Here is how to work through a typical Angular momentum of a rigid body (L = Iω) task step by step:

1
Task
A rotor has $I=0.60\,\mathrm{kg\,m^2}$ and L=3.0 $\mathrm{kg\,m^2\,s^{-1}}$ . Find $\omega$ .
Solution path
$\omega=\frac{L}{I}=\frac{3.0\,\mathrm{kg\,m^2\,s^{-1}}}{0.60\,\mathrm{kg\,m^2}}=5.0\,\mathrm{rad\,s^{-1}}$ .

Understand and apply the equation

Angular momentum describes rotational motion relative to a specified point or axis. L=Iω is a useful shorthand, but it may describe an axial component or, for a principal axis, the entire vector. This distinction determines which torques must be included.

Keeping or changing rotational motion

Angular momentum is the quantity changed by external torque. About a fixed axis, begin with a signed component distinguishing the two rotation directions. Larger inertia or angular velocity gives a larger axial component. The spatial vector description is introduced afterwards because not every body rotates about a principal axis.[1]

Prior knowledge: rotation direction, moment of inertia and the right-hand rule. A principal axis is a special axis along which angular momentum and angular velocity are parallel.

Reference point, axis and rotation direction

\mathbf L_O=\sum_i\mathbf r_i\times m_i\mathbf v_i

For a fixed point O in an inertial frame, position vectors and velocities are measured in the same frame.[1]

The vector does not necessarily point along the geometric rotation axis. Its axial component $L_z$ is the projection onto a specified positive z-direction. Pure rigid rotation about a fixed z-axis gives $L_z=I_{zz}\omega_z$ even if z is not a principal axis. Additional transverse components may then exist. For the entire vector equation with one scalar I, rotation must align with a principal axis of the inertia tensor.[2]

\begin{aligned}L_z&=I_{zz}\omega_z;\\\mathbf L&=I_{\mathrm{principal}}\boldsymbol\omega\end{aligned}

I is nonnegative; the sign of $L_z$ follows the sign of $\omega_z$ . The second vector relation applies to principal-axis rotation.[2]

Angular momentum retains the sign

QUANTA STUDYquanta-study.de

I_{zz}=0.35\,\mathrm{kg\,m^2}

as in exercise 1. At

\omega_z=-12\,\mathrm{rad\,s^{-1}}

,

L_z=-4.2\,\mathrm{kg\,m^2\,s^{-1}}

; at +12

\mathrm{rad\,s^{-1}}

it is +4.2

\mathrm{kg\,m^2\,s^{-1}}

. This shows only the fixed-axis component, not possible transverse components.[1]

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From particles to an axial component

For pure rotation about z, substitute $\vec v=\vec\omega\times\vec r$ . A particle at $(x,y,z)$ has $v_x=-\omega_z y$ and $v_y=\omega_z x$ . The z-component of the cross product is then $xm v_y-ym v_x=m(x^2+y^2)\omega_z$ . Sum over all particles.[1][2]

L_z=\sum_i m_i(x_i^2+y_i^2)\omega_z=I_{zz}\omega_z

The shared axial angular velocity factors out. This does not mean that individual particle angular momenta point exclusively along z.[2]

\mathbf L_O=\int\mathbf r\times(\boldsymbol\omega\times\mathbf r)\,dm=\mathsf I_O\boldsymbol\omega

The integral gives the full inertia tensor. The same form may alternatively be used relative to the centre of mass with relative velocities.[2]

Moment of inertia of a system of particles gives the factor between angular velocity and the axial angular momentum component here. Use distances to the selected axis, not arbitrary distances from the origin.[5][1]

Units and useful rearrangements

\begin{aligned}[L]&=\mathrm{kg\,m^2/s}=\mathrm{J\,s}\\\omega_z&=\frac{L_z}{I_{zz}}\ (I_{zz}>0)\\I_{zz}&=\frac{L_z}{\omega_z}\ (\omega_z\ne0)\end{aligned}

Angular momentum is not torque: $\mathrm{N\,m}$ contains one additional inverse second compared with $\mathrm{kg\,m^2\,s^{-1}}$ .

State $\omega_z$ in $\mathrm{rad\,s^{-1}}$ . The radian is dimensionless in SI but identifies angular quantities. For n revolutions per minute, $|\omega_z|=\frac{2\pi n_{\mathrm{rpm}}}{60\,\mathrm{s\,min^{-1}}}$ ; rotation direction must be specified separately. Dividing $L_z$ by a frequency in Hz rather than by $\omega_z$ would omit the factor 2π.

With translation and rotation together, angular momentum about an external point also contains centre-of-mass orbital angular momentum. Using relative quantities about the centre of mass gives $\vec L_O=\vec R_{\mathrm{CM}}\times M\vec v_{\mathrm{CM}}+\vec L_{\mathrm{CM}}$ . Therefore $I_{\mathrm{CM}}\omega$ alone does not automatically describe total angular momentum about an arbitrary observation point.[2][1]

The unit kg·m²/s can also be written as N·m·s: a torque acting over a time interval changes angular momentum. With $I_{zz}>0$ , $L_z=0$ implies $\omega_z=0$ . Conversely, $L_z=0$ and $\omega_z=0$ do not determine inertia; any positive $I_{zz}$ fits. To solve for $I_{zz}$ when $\omega_z\ne0$ , $L_z$ and $\omega_z$ must have the same sign.[1]

Signs and a tilted angular momentum

Rotor with a specified positive direction

Given: A rotor turns about its fixed principal z-axis. $I_{zz}=0.24\,\mathrm{kg\,m^2}$ and n=450 rpm. Viewed from +z, it turns clockwise.
Find: Find $\omega_z$ , $L_z$ and rotational energy.

Approach: For a fixed principal axis, set the sign by the right-hand rule; angular momentum and energy use the same angular velocity.
$\begin{aligned}\omega_z&=-2\pi\frac{n_{\mathrm{rpm}}}{60\,\mathrm{s/min}}\\L_z&=I_{zz}\omega_z\\E_{\mathrm{rot}}&=\frac12 I_{zz}\omega_z^2\end{aligned}$
Right-hand rule: clockwise when viewed from +z means negative $\omega_z$ .
$\omega_z=-2\pi\frac{450}{60}=-15\pi\,\mathrm{rad/s}$
Multiply by inertia about that same axis; energy instead uses the square.
$\begin{aligned}L_z&=-3.6\pi\,\mathrm{kg\,m^2/s}\\{} E&=\frac12(0.24)(15\pi)^2=27\pi^2\,\mathrm J\end{aligned}$

Result: $\omega_z\approx-47.12\,\mathrm{rad\,s^{-1}}$ , $L_z\approx-11.31\,\mathrm{kg\,m^2\,s^{-1}}$ and $E\approx266.5\,\mathrm{J}$ . Negative rotation does not give negative kinetic energy.

Check: Check: $E=\frac{L_z^2}{2I_{zz}}=266.5\,\mathrm J$ . The rotation rate is 7.5 revolutions/s; a revolution takes 0.1333 s, agreeing with $\frac{2\pi}{|\omega_z|}$ .

A rigid body not rotating about a principal axis

Given: Two point masses of 1.0 kg each lie on a massless rigid rod at $\vec r_1=(0.30,0,0.40)\,\mathrm m$ and $\vec r_2=-\vec r_1$ . O is the centre of mass. At this instant $\vec\omega=(0,0,6.0)\,\mathrm{rad\,s^{-1}}$ .
Find: Find the full L and its z-component.

Approach: At this instant about the centre of mass O, obtain velocities from rigid rotation and add individual angular-momentum vectors.
$\begin{aligned}\mathbf v_i&=\boldsymbol\omega\times\mathbf r_i\\\mathbf L_O&=\sum_i\mathbf r_i\times m_i\mathbf v_i\\L_z&=I_{zz}\omega_z\end{aligned}$
Velocities are $\vec v_1=(0,1.80,0)\,\mathrm{m\,s^{-1}}$ and $\vec v_2=-\vec v_1$ . Calculate each cross product separately.
$\mathbf r_1\times m\mathbf v_1=(-0.72,0,0.54)\,\mathrm{kg\,m^2/s}$
Both particles contribute the same angular momentum; add them.
$\begin{aligned}\mathbf L&=(-1.44,0,1.08)\,\mathrm{kg\,m^2/s}\\{} I_{zz}&=2(1.0)(0.30)^2=0.18\,\mathrm{kg\,m^2}\end{aligned}$

Result: Axial momentum $L_z=1.08\,\mathrm{kg\,m^2\,s^{-1}}$ agrees with $I_{zz}\omega_z$ . The full vector also has $L_x=-1.44\,\mathrm{kg\,m^2\,s^{-1}}$ , so it is not parallel to angular velocity.

Check: Its magnitude is $\sqrt{1.44^2+1.08^2}=1.80\,\mathrm{kg\,m^2\,s^{-1}}$ , not 1.08. Direct energy is $2\cdot\frac12\cdot1\cdot1.80^2\,\mathrm J=3.24\,\mathrm J$ , also $\frac12 I_{zz}\omega_z^2$ . Keeping the z-axis fixed requires transverse bearing torques.

Balance torque and conservation correctly

\sum\boldsymbol\tau_{O,\mathrm{ext}}=\left(\frac{d\mathbf L_O}{dt}\right)_{\mathrm{inertial}}

This form refers to a fixed inertial reference point O; the corresponding centre-of-mass balance can be used for the centre of mass.[1][2]

If the entire external torque vanishes, the full vector is conserved. If only $\tau_z$ about a fixed inertial z-axis vanishes, $L_z$ is conserved; other components need not stay constant. Bearings may transmit transverse torques even when drive torque about z is zero. “No motor drive” and “no total external torque” are therefore different statements.[2][4]

\tau_z=\frac{d(I_{zz}\omega_z)}{dt}=I_{zz}\alpha_z+\dot I_{zz}\omega_z

For a rigid body with unchanged mass distribution about the fixed axis, Īzz=0. Retractable weights require a suitable description of the changing system.[4]

Explore further: When mass distribution changes

Figure skating and retractable weights illustrate $I_1\omega_1=I_2\omega_2$ when external axial torque is negligible. During retraction, the system is not a single rigid body. Use the relation only when the relevant axial angular momentum is genuinely represented by Iω and additional relative motion is accounted for appropriately.[4]

Conserved angular momentum does not imply conserved rotational energy: $E=\frac{L_z^2}{2I}$ . Halving I at constant $L_z$ doubles $\omega_z$ and E. The additional energy comes from internal work, such as muscles or a mechanism. A system with no external torque can therefore change rotational energy without violating angular-momentum conservation.[4]

Five exercises with solutions

Calculate · Foundation

$I_{zz}=0.35\,\mathrm{kg\,m^2}$ and $\omega_z=-12\,\mathrm{rad\,s^{-1}}$ about a fixed axis. Find $L_z$ and E.

Show hint

$L_z$ is signed; energy uses $\omega_z^2$ .

Learning goal: Separate signed momentum from positive rotational energy.

Show solution and calculation

Approach: $L_z=I_{zz}\omega_z$ retains the sign, whereas $E=\frac12I_{zz}\omega_z^2$ uses the square.
This gives $L_z=-4.2\,\mathrm{kg\,m^2\,s^{-1}}$ and $E=25.2\,\mathrm J$ .

Check: $\frac{L_z^2}{2I_{zz}}=25.2\,\mathrm J$ . Negative axial momentum does not imply negative inertia.

\begin{aligned}L_z&=-4.2\,\mathrm{kg\,m^2/s}\\{} E&=25.2\,\mathrm J\end{aligned}

Transfer · Application

A system starts at $I_1$ =0.60 $\mathrm{kg\,m^2}$ and $\omega_{1z}=4.0\,\mathrm{rad\,s^{-1}}$ . In an appropriate axial model, I falls to 0.40 $\mathrm{kg\,m^2}$ with no external $\tau_z$ . Find $\omega_{2z}$ and the rotational-energy change.

Show hint

Conserve $L_z$ , not E.

Learning goal: Apply angular momentum conservation during redistribution.

Show solution and calculation

Approach: With zero external axial torque, $I_1\omega_{1z}=I_2\omega_{2z}$ .
Compare energies separately using $E=\frac12I\omega_z^2$ .
Conserved axial momentum is 2.4 $\mathrm{kg\,m^2\,s^{-1}}$ , giving $\omega_{2z}=6.0\,\mathrm{rad\,s^{-1}}$ .
Energy rises from 4.8 J to 7.2 J, a 2.4 J increase supplied by internal work.

Check: $0.40\cdot6.0=0.60\cdot4.0$ . Redistribution itself is not a rigid phase.

I_1\omega_{1z}=I_2\omega_{2z}=2.4\,\mathrm{kg\,m^2/s}

Calculate · Application

A rigid rotor with $I_{zz}=0.20\,\mathrm{kg\,m^2}$ receives constant net torque $\tau_z=-0.30\,\mathrm{N\,m}$ for 2.0 s. Initially $\omega_z=5.0\,\mathrm{rad\,s^{-1}}$ . Find its final value and direction.

Show hint

Use $\Delta L_z=\tau_z\Delta t$ and keep the axis reference fixed.

Learning goal: Convert torque impulse to final angular velocity.

Show solution and calculation

Approach: Use torque impulse $\Delta L_z=\tau_z\Delta t$ , then $\omega_{z,f}=\frac{L_{z,0}+\Delta L_z}{I_{zz}}$ .
Initial momentum is 1.0 $\mathrm{kg\,m^2\,s^{-1}}$ and impulse −0.60 $\mathrm{kg\,m^2\,s^{-1}}$ .
This gives 0.40 $\mathrm{kg\,m^2\,s^{-1}}$ and $\omega_{z,f}=2.0\,\mathrm{rad\,s^{-1}}$ .
Rotation remains positive but slows.

Check: $\alpha_z=-1.5\,\mathrm{rad\,s^{-2}}$ gives the same end value through $5-1.5\cdot2=2$ .

\omega_{z,\mathrm f}=\frac{I\omega_{z,0}+\tau_z\Delta t}{I}=2.0\,\mathrm{rad/s}

Interpret · Challenge

A rigid body has principal moments 1, 2 and 3 $\mathrm{kg\,m^2}$ about its centre of mass. At one instant, principal-axis components are $\omega$ =(1,1,0) $\mathrm{rad\,s^{-1}}$ . Find L and test whether L is parallel to $\omega$ .

Show hint

Multiply each component by its own principal moment.

Learning goal: Use the inertia tensor rather than one scalar moment.

Show solution and calculation

Approach: $L_i=I_i\omega_i$ gives L=(1,2,0) $\mathrm{kg\,m^2\,s^{-1}}$ .
Ratios $\frac{L_x}{\omega_x}$ and $\frac{L_y}{\omega_y}$ differ, so the vectors are not parallel.

Check: $\vec\omega\times\vec L$ has a nonzero z component. Principal coordinates diagonalize the tensor but do not make all moments equal. A common scalar relation $\vec L=I\vec\omega$ is therefore inappropriate here.

\mathbf L=\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\!\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\2\\0\end{pmatrix}\,\mathrm{kg\,m^2\,s^{-1}}

[2]

Error analysis · Challenge

A bearing-constrained rigid rotor has body-frame components $\omega$ =(0,0,2) $\mathrm{rad\,s^{-1}}$ and L=(1,0,2) $\mathrm{kg\,m^2\,s^{-1}}$ , constant in that rotating frame. A solution infers zero external torque. Test it using the article’s derivative rule. The reference point for angular momentum and external torque is the centre of mass, fixed in the inertial frame.

Show hint

The inertial derivative additionally includes $\vec\omega\times\vec L$ .

Learning goal: Distinguish body-frame constancy from inertial conservation.

Show solution and calculation

Approach: the body-frame derivative vanishes, but $\vec\omega\times\vec L=(0,2,0)\,\mathrm{N\,m}$ .
That bearing torque is required.

Check: the z component of L is constant and z torque is zero; this does not eliminate transverse vector change. A rotor-fixed transverse momentum component rotates in inertial space. The data describe constrained nonprincipal-axis rotation, not free torque-free motion.

\boldsymbol\tau=\left(\frac{d\mathbf L}{dt}\right)_{\mathrm{body}}+\boldsymbol\omega\times\mathbf L=\begin{pmatrix}0\\2\\0\end{pmatrix}\,\mathrm{N\,m}

[3][2]

What a scalar formula omits

For general spatial rotation, $\vec L=\mathsf I\vec\omega$ uses a tensor. Even in principal-axis coordinates, $L_i=I_i\omega_i$ ; with several nonzero $\omega$ -components and unequal principal moments, L and $\omega$ are generally not parallel. One scalar I therefore cannot be applied to every component.[2]

A body-fixed coordinate system itself rotates. The inertial derivative is then the body-frame derivative of L plus $\vec\omega\times\vec L$ . A vector apparently constant in that frame may change in the inertial frame. Simple rotor problems can use the fixed-axis component; gyroscopes, free tumbling and bearing loads require the full vector balance.[3]

What to remember

Reference point and positive axis are part of angular momentum. The fixed-axis component is inertia times signed angular velocity. The full vector may include transverse components. Conservation requires the appropriate external torque to vanish; unchanged angular momentum does not guarantee unchanged rotational energy.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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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.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Linked axial inertia and torque in context and added N·m·s units and zero cases of the inverse; preserved the vector example, exercises and diagram.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an individual introduction and conclusion, checked existing calculations, clarified notation and model references, and added an original learning figure.

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Quanta. Angular momentum of a rigid body. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/drehimpuls-starrer-koerper

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Frequently asked questions about Angular momentum of a rigid body

What does rigid-body angular momentum describe physically?+

The vector does not necessarily point along the geometric rotation axis. Its axial component Lz is the projection onto a specified positive z-direction. Pure rigid rotation about a fixed z-axis gives $L_z=I_{zz}\omega_z$ even if z is not a principal axis. Additional transverse components may then exist. For the entire vector equation with one scalar I, rotation must align with a principal axis of the inertia tensor. I is nonnegative; the sign of Lz follows the sign of ωz. The shared axial angular velocity factors out. This does not mean that individual particle angular momenta point exclusively along z.

How can I check a calculation involving rigid-body angular momentum?+

Two 1.0 kg particles sit on a massless rigid rod at $\vec r_1=(0.30,0,0.40)\,\mathrm m$ and $\vec r_2=-\vec r_1$ . O is the centre of mass. Instantaneously $\vec\omega=(0,0,6.0)\,\mathrm{rad\,s^{-1}}$ . Velocities are $\vec v_1=(0,1.80,0)\,\mathrm{m\,s^{-1}}$ and $\vec v_2=-\vec v_1$ . Calculate each cross product. Both particles contribute the same angular momentum; add them. Lz=1.08 kg·m²/s agrees with $I_{zz}\omega_z$ . The full vector additionally has $L_x=-1.44\,\mathrm{kg\,m^2\,s^{-1}}$ and is not parallel to ω. Its magnitude is $\sqrt{1.44^2+1.08^2}=1.80\,\mathrm{kg\,m^2\,s^{-1}}$ , not 1.08. Direct energy is $2\cdot\frac12\cdot1\cdot1.80^2\,\mathrm J=3.24\,\mathrm J$ , also $\frac12I_{zz}\omega_z^2$ . Keeping the z-axis fixed requires transverse bearing torques. The tensor form $\vec L=\mathsf I\vec\omega$ gives the same components; pure z-rotation specifically has Izz = 0.18 kg·m².

What assumptions and limits apply to rigid-body angular momentum?+

For general spatial rotation, $\vec L=\mathsf I\vec\omega$ uses a tensor. Even in principal-axis coordinates, $L_i=I_i\omega_i$ ; with several nonzero ω-components and unequal principal moments, L and ω are generally not parallel. One scalar I therefore cannot be applied to every component. A body-fixed coordinate system itself rotates. The inertial derivative is then the body-frame derivative of L plus $\vec\omega\times\vec L$ . A vector apparently constant in that frame may change in the inertial frame. Simple rotor problems can use the fixed-axis component; gyroscopes, free tumbling and bearing loads require the full vector balance.

How do I check units and scale for rigid-body angular momentum?+

Angular momentum is not torque: N·m contains one additional inverse second compared with kg·m²/s. State ωz in rad/s. The radian is dimensionless in SI but identifies angular quantities. For n revolutions per minute, $|\omega_z|=\frac{2\pi n_{\mathrm{rpm}}}{60\,\mathrm{s\,min^{-1}}}$ ; rotation direction must be specified separately. Dividing Lz by a frequency in Hz rather than by ωz would omit the factor 2π. With translation and rotation together, angular momentum about an external point also contains centre-of-mass orbital angular momentum. Using relative quantities about the centre of mass gives $\vec L_O=\vec R_{\mathrm{CM}}\times M\vec v_{\mathrm{CM}}+\vec L_{\mathrm{CM}}$ . Therefore $I_{\mathrm{CM}}\omega$ alone does not automatically describe total angular momentum about an arbitrary observation point.

When should I combine rigid-body angular momentum with another relation?+

Figure skating and retractable weights illustrate $I_1\omega_1=I_2\omega_2$ when external axial torque is negligible. During retraction, the system is not a single rigid body. Use the relation only when the relevant axial angular momentum is genuinely represented by Iω and additional relative motion is accounted for appropriately. Conserved angular momentum does not imply conserved rotational energy: $E=\frac{L_z^2}{2I}$ . Halving I at constant Lz doubles ωz and E. The additional energy comes from internal work, such as muscles or a mechanism. A system with no external torque can therefore change rotational energy without violating angular-momentum conservation.

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

Kinetic Energy

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

Potential Energy (Gravitational)

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Angular momentum of a rigid body

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Izz=0.35 kg m2I_{zz}=0.35\,\mathrm{kg\,m^2}Izz​=0.35kgm2 and ωz=−12 rad s−1\omega_z=-12\,\mathrm{rad\,s^{-1}}ωz​=−12rads−1 about a fixed axis. Find LzL_zLz​ and E.

A rigid rotor with Izz=0.20 kg m2I_{zz}=0.20\,\mathrm{kg\,m^2}Izz​=0.20kgm2 receives constant net torque τz=−0.30 N m\tau_z=-0.30\,\mathrm{N\,m}τz​=−0.30Nm for 2.0 s. Initially ωz=5.0 rad s−1\omega_z=5.0\,\mathrm{rad\,s^{-1}}ωz​=5.0rads−1. Find its final value and direction.

A rigid body has principal moments 1, 2 and 3 kg m2\mathrm{kg\,m^2}kgm2 about its centre of mass. At one instant, principal-axis components are ω\omegaω=(1,1,0) rad s−1\mathrm{rad\,s^{-1}}rads−1. Find L and test whether L is parallel to ω\omegaω.

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Retain Angular momentum of a rigid body for exams

$I_{zz}=0.35\,\mathrm{kg\,m^2}$ and $\omega_z=-12\,\mathrm{rad\,s^{-1}}$ about a fixed axis. Find $L_z$ and E.

A rigid rotor with $I_{zz}=0.20\,\mathrm{kg\,m^2}$ receives constant net torque $\tau_z=-0.30\,\mathrm{N\,m}$ for 2.0 s. Initially $\omega_z=5.0\,\mathrm{rad\,s^{-1}}$ . Find its final value and direction.

A rigid body has principal moments 1, 2 and 3 $\mathrm{kg\,m^2}$ about its centre of mass. At one instant, principal-axis components are $\omega$ =(1,1,0) $\mathrm{rad\,s^{-1}}$ . Find L and test whether L is parallel to $\omega$ .