Physics · Rotational motion

Newton’s second law for rotation

Q: What does Newton’s second law for rotation describe physically?

$\sum\tau=I\alpha$ relates net external torque component to signed angular acceleration of a rigid body about a fixed inertial axis. Include every contribution: drive, load and bearing friction. Motor torque alone therefore is not automatically net torque. With balanced torques, $\alpha=0$ can mean uniform rotation rather than rest. Larger I produces less acceleration for the same net torque. This scalar statement concerns the axial component and needs no principal axis. It does not assert that the full angular momentum vector is parallel to angular velocity about every axis; the inertia tensor governs that general vector relation.

Q: When should I combine Newton’s second law for rotation with another relation?

With constant net torque and I, $\alpha=\frac{\sum\tau}I$ leads to $\omega=\omega_0+\alpha t$ and $\Delta\theta=\omega_0t+\frac12\alpha t^2$. Work and power for the same fixed axis follow from $W=\int\tau\,d\theta$ and $P=\tau\omega$. A falling mass driving a drum adds $mg-T=ma$ and cord constraint $a=R\alpha$; rotational torque is $TR$ and tension is not automatically mg. For rolling, combine centre-of-mass translation with rotation about the centre of mass and the contact constraint. Every additional relation needs the same system, appropriate reference points and consistent signs. Energy provides an independent check but does not replace the requested tension or friction force.

Net external torque component about a fixed inertial axis equals I times signed angular acceleration when I is constant.

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Formula

Calculate

\sum τ = I α

$\tau$ in $\mathrm{N\,m}$ ; I in $\mathrm{kg\,m^2}$ ; $\alpha$ in $\mathrm{rad\,s^{-2}}$ .

Variables & units

Variables & units
Symbol	Meaning	Unit
$\sum τ$	Signed sum of all external torque components about the fixed inertial axis	N·m
$I$	Constant moment of inertia of the rigid body about that same axis	kg·m²
$α$	Signed angular acceleration component about the same axis	rad/s²

LaTeX

\sum\tau=I\alpha

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Signed net-torque versus angular-acceleration line through zero: from −6 rad/s², −1.920 N·m to +6 rad/s², +1.920 N·m. Rotor example is marked upper right; zero torque means no angular acceleration, not necessarily rest. — For the same rotor about a fixed axis with constant I = 0.320 kg·m², τnet = Iα. Sum motor and load torques first: +2.40 − 0.480 = +1.920 N·m gives α = +6 rad/s².

Torque and angular acceleration calculator

Find the torque, moment of inertia or angular acceleration of a rigid body about a fixed axis, using τ = I · α.

Find

Net torque τ

Net torque with its sign, so drive minus load and friction, e.g. 2 N·m.

Axial moment of inertia I

Moment of inertia about the same axis, e.g. 0.5 kg·m² for a heavy disc.

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

Working · rearrange for α, substitute, evaluate

1. Rearrange for α
$\alpha_z=\frac{\tau_{\mathrm{net},z}}{I_z}$
2. Substitute values (in SI units)
$\frac{2}{0.5}$
3. Result
α = 4 rad/s²

Result: Signed angular acceleration α

α = 4 rad/s²

With τ = 2 N·m and I = 0.5 kg·m², α = 4 rad/s².

Quick tips

Formula triangle: τ on top, I and α below. Cover the quantity you need and the rest is the calculation, e.g. α = τ ÷ I.
Like F = m · a for rotation: torque instead of force, moment of inertia instead of mass, angular acceleration instead of acceleration.

Assumptions and limits

Rigid body with constant positive inertia about an axis fixed in an inertial frame. Net torque includes drive, load and bearing friction; all signed components refer to the same axis and direction. No general free 3D rotation or coupled drum solution.

Derivation & background

For the fixed axis, $L_z=I_z\omega_z$ and $\sum\tau_{z,\mathrm{ext}}=\frac{dL_z}{dt}$ . Constant inertia therefore gives $\sum\tau_{z,\mathrm{ext}}=I_z\alpha_z$ . This scalar balance requires no principal axis; full angular momentum need not be parallel to angular velocity. Include drive, load and friction. Changing inertia adds $\dot I\omega_z$ if the product expression still holds. Other internal angular motion requires full angular momentum.

Exam blueprint

Validity range

Rigid body with constant moment of inertia about an axis fixed in an inertial frame; all quantities are signed components about that same axis. Changing mass distribution or other axis motion requires the appropriate general angular momentum balance.

Derivation steps

Net external torque changes axial angular momentum; constant I turns this into $I\alpha$ .

1Choose the body, fixed inertial z axis and positive rotation direction.
2Calculate $I_z=\sum_i m_i r_{i,\perp}^2$ from perpendicular distances to that same axis.
3Rigid rotation gives every mass the same $\omega_z$ , hence $L_z=I_z\omega_z$ .
4The net external axial torque satisfies $\sum\tau_{z,\mathrm{ext}}=\frac{dL_z}{dt}$ .
5Constant axial distances keep $I_z$ constant, so differentiation gives $I_z\frac{d\omega_z}{dt}$ .
6With $\alpha_z=\frac{d\omega_z}{dt}$ , obtain $\sum\tau_{z,\mathrm{ext}}=I_z\alpha_z$ . Include drive, load and friction.
7This scalar axial balance requires no principal axis. The full vector relation generally needs the inertia tensor.

Rearrangements

Angular acceleration

\alpha_z=\frac{\sum\tau_{z,\mathrm{ext}}}{I_z}

The net torque sign determines the direction of angular acceleration.

Task variant

A rigid rotor has $I=0.20\,\mathrm{kg\,m^2}$ about its fixed inertial axis. Net external axial torque is $+0.60\,\mathrm{N\,m}$ . Find angular acceleration.

Approach: $\alpha=\frac{\tau_{\mathrm{net}}}I$ .
This gives $\alpha=+3.0\,\mathrm{rad\,s^{-2}}$ .

Check: $I\alpha=0.60\,\mathrm{N\,m}$ . The positive sign identifies acceleration direction rather than automatically increasing angular speed.

Common mistakes

Using motor torque alone instead of every external axial torque.

Add drive, load and friction with signs. Use the same axis for $I$ and torque; positive torque can slow a negatively rotating body.

Exam context

Analyze motors and drums on fixed bearings; for rolling, combine centre-of-mass force and torque balances with the rolling constraint.

Formula cluster

Rotational motion

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

M = F·r·sin(θ)

Torque

L = Iω

Angular momentum of a rigid body

I = Σ m_ir_i²

Moment of inertia of a particle system

Worked example

With $I=0.20\,\mathrm{kg\,m^2}$ and $\tau_{\mathrm{net}}=+0.60\,\mathrm{N\,m}$ , $\alpha=\frac{\tau_{\mathrm{net}}}I=+3.0\,\mathrm{rad\,s^{-2}}$ . Check: inertia times acceleration recovers net torque. Its sign alone does not determine whether angular speed increases.

Applications

Analyze motors and drums on fixed bearings; for rolling, combine centre-of-mass force and torque balances with the rolling constraint.

How do you calculate with Newton’s second law for rotation?

Here is how to work through a typical Newton’s second law for rotation (Στ = Iα) task step by step:

1
Task
A rigid rotor has $I=0.20\,\mathrm{kg\,m^2}$ about its fixed inertial axis. Net external axial torque is $+0.60\,\mathrm{N\,m}$ . Find angular acceleration.
Solution path
1. Approach: $\alpha=\frac{\tau_{\mathrm{net}}}I$ .
2. This gives $\alpha=+3.0\,\mathrm{rad\,s^{-2}}$ .
Check: $I\alpha=0.60\,\mathrm{N\,m}$ . The positive sign identifies acceleration direction rather than automatically increasing angular speed.

Understand and apply the equation

A motor must balance external torques and also change rotational motion. The rotational equation links net torque about a fixed axis to angular acceleration. The chosen axis, sign convention and moment of inertia about that very axis are essential.

What changes rotational motion?

A heavy door is easier to turn by its outer handle than close to the hinge. Force alone therefore does not determine its turning effect: the lever arm matters too. That turning effect is torque. For a rotor, compare the sum of external torques with its inertia about the chosen axis. This explains why the same motor accelerates different rotors at different rates.[1]

Prior knowledge: force balance and angular velocity. Moment of inertia describes how mass is distributed about an axis, rather than mass alone.

Net axial torque produces acceleration

\sum\tau_{z,\mathrm{ext}}=I_z\alpha_z

For a rigid body about a fixed inertial axis, net external axial torque equals constant axial moment of inertia times signed angular acceleration.[1][2]

The left side sums all external contributions: drive, load, bearing friction and, where relevant, forces at supports or guides. Internal forces of the complete chosen body do not add external net torque. Positive motor torque may be fully balanced by negative load torque. Then $\alpha_z=0$ although both torques are nonzero; the body may rotate uniformly.[3]

View diagram: Net torque sets the slope

Net torque sets the slope

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The first example has

I=0.320\,\mathrm{kg\,m^2}

and net torque +1.920

\mathrm{N\,m}

. The line rises from 4 to 13

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

in 1.50 s. Its area of 12.75 rad gives 24.48 J work with the net torque, matching the energy change.[1]

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From angular momentum to acceleration

\begin{aligned}L_z&=\sum_i m_i r_{i,\perp}^2\omega_z=I_z\omega_z\\\sum\tau_{z,\mathrm{ext}}&=\frac{dL_z}{dt}\end{aligned}

Each particle shares angular velocity; its perpendicular distance to the fixed axis determines its axial angular momentum contribution.[2]

Rigidity and an unchanged body-fixed rotation axis keep all axial distances constant. Thus $I_z$ is time independent and differentiating $I_z\omega_z$ gives $I_z\alpha_z$ . This derivation also explains why the formula concerns the axial component. The full angular momentum vector need not be parallel to angular velocity about an arbitrary axis. The more general vector balance sets external torque equal to the time derivative of angular momentum.[2]

\begin{aligned}\alpha_z&=\frac{\sum\tau_z}{I_z}\\I_z&=\frac{\sum\tau_z}{\alpha_z}\end{aligned}

The first division requires $I_z$ >0. The second requires $\alpha_z$ ≠0 and a valid model; zero torque and zero acceleration do not determine I.

Angular momentum of a rigid body explains why this equation is one axial component of a more general balance. With unchanged inertia, differentiating signed angular velocity is sufficient here; a changing distribution or free spatial rotation requires checking the full balance.[2][1]

Explore further: Distinguish axial balance from a vector law

The scalar z balance for a constrained fixed axis needs no principal-axis condition. Bearings can supply transverse torques to maintain that direction. The additional statement that the full vector is $\vec L=I\vec\omega$ with a single scalar I needs a suitable principal axis. A moment-of-inertia tensor describes the general vector relation.[2]

For a rolling body the contact location moves. It therefore is not automatically a permanently fixed inertial rotation axis. For planar rolling, a reliable balance combines centre-of-mass translation $\sum\vec F=M\vec a$ with rotation about the centre of mass $\sum\tau_{\mathrm{CM}}=I_{\mathrm{CM}}\alpha$ . The rolling condition then links a and $\alpha$ . Energy can alternatively be expressed using the instantaneous contact axis without automatically justifying every torque balance about that moving location.[5][3]

Specify lever arm, direction and units

\begin{aligned}\tau_z&=(\mathbf r\times\mathbf F)_z\\{}[I_z\alpha_z]&=\mathrm{kg\,m^2\,s^{-2}}=\mathrm{N\,m}\end{aligned}

r points from the chosen axis to the point of application. The right-hand rule sets the sign, not force magnitude alone.[1]

A tangential force at radius R produces magnitude $RF$ . A radial force has no lever arm or torque about that axis. $\mathrm{N\,m}$ and joule have the same SI dimension, but torque is not energy. Work requires torque acting through angular displacement. Angular acceleration in $\mathrm{rad\,s^{-2}}$ retains angular meaning even though radian is dimensionless in SI.[1]

Drive and coupled descending mass

Motor with opposing load

Given: A rigid rotor about a fixed axis has $I=0.320\,\mathrm{kg\,m^2}$ . Motor torque is +2.40 $\mathrm{N\,m}$ and load torque −0.480 $\mathrm{N\,m}$ . Initially $\omega_0=+4.00\,\mathrm{rad\,s^{-1}}$ ; both torques stay constant for 1.50 s.
Find: Find $\alpha$ , final angular velocity and energy change.

Approach: Add all external axial torques with their signs. $\alpha=\frac{\tau_{\mathrm{Motor}}+\tau_{\mathrm{Last}}}{I}$ . First sum signed torques.
$\begin{aligned}\tau_{\mathrm{net}}&=2.40-0.480=1.920\,\mathrm{N\,m}\\{} \alpha&=\frac{1.920}{0.320}=6.00\,\mathrm{rad/s^2}\end{aligned}$
With constant I and torque, $\alpha$ is constant.
$\begin{aligned}\omega_f&=4+6(1.50)=13.0\,\mathrm{rad/s}\\\Delta\theta&=4(1.50)+\tfrac12(6)(1.50)^2=12.75\,\mathrm{rad}\end{aligned}$
Calculate rotational energy change from initial and final values.
$\Delta K=\tfrac12(0.320)(13^2-4^2)=24.48\,\mathrm J$

Result: The results are $\alpha=+6.00\,\mathrm{rad\,s^{-2}}$ , $\omega_f=+13.0\,\mathrm{rad\,s^{-1}}$ and $\Delta K=24.48\,\mathrm J$ . The load opposes positive rotation throughout the interval.

Check: Independently, net work is $W_{\mathrm{net}}=\tau_{\mathrm{net}}\Delta\theta=1.920\cdot12.75\,\mathrm J=24.48\,\mathrm J$ . Motor work of 30.60 J and negative load work of −6.12 J give the same sum.

Descending mass drives a drum

Given: A mass $m=1.80\,\mathrm{kg}$ hangs from a massless taut cord. The cord unwinds without slipping from a drum with $R=0.150\,\mathrm{m}$ and $I=0.120\,\mathrm{kg\,m^2}$ . Fixed frictionless axis, constant R, release from rest, drop $h=0.650\,\mathrm m$ , $g=9.81\,\mathrm{m\,s^{-2}}$ .
Find: Find a, cord tension and speed after the descent.

Approach: Combine translation of the mass, drum rotation and rope coupling. $mg-T=ma,\quad TR=I\alpha,\quad a=R\alpha$ . Choose downward translation and unwinding rotation positive.
$\begin{aligned}mg-T&=ma\\TR&=I\alpha\\a&=R\alpha\end{aligned}$
Eliminate cord tension and include rotational inertia.
$\begin{aligned}a&=\frac{mg}{m+\frac{I}{R^2}}=2.475421\,\mathrm{m/s^2}\\{} \alpha&=16.502804\,\mathrm{rad/s^2}\end{aligned}$
Tension is below mg because the mass accelerates.
$\begin{aligned}T&=m(g-a)=13.202243\,\mathrm N\\TR&=1.980336\,\mathrm{N\,m}\end{aligned}$
Use the descent distance with constant acceleration.
$\begin{aligned}v&=\sqrt{2ah}=1.793892\,\mathrm{m/s}\\{} \omega&=\frac{v}{R}=11.959277\,\mathrm{rad/s}\end{aligned}$

Result: $a\approx2.475\,\mathrm{m\,s^{-2}}$ , T≈13.202 N and $v\approx1.794\,\mathrm{m\,s^{-1}}$ . The drum receives part of the released gravitational energy.

Check: Independent energy check using unrounded values: $mgh=11.4777\,\mathrm J$ , $\frac12mv^2=2.896242\,\mathrm J$ and $\frac12I\omega^2=8.581458\,\mathrm J$ . Their sum recovers 11.4777 J. The incorrect assumption $T=mg$ would give $a=0$ , contradicting the coupled dynamics.

Torque and power complement the time balance

\begin{aligned}W_{\mathrm{net}}&=\int\tau_{z,\mathrm{net}}\,d\theta=\Delta\!\left(\frac12 I_z\omega_z^2\right)\\P_{\mathrm{net}}&=\tau_{z,\mathrm{net}}\omega_z\end{aligned}

For the same fixed axis and constant I, torque work equals change in rotational kinetic energy.[3]

Instantaneous torque can be used in $\sum\tau_z=I_z\alpha_z$ even when it varies with time. Only constant net torque with constant I justifies linear $\omega$ (t). Under constant mechanical P with $\omega\ne0$ , torque instead is $\tau=\frac P\omega$ , so acceleration depends on rotational state. Do not divide by $\omega$ at rest. An actual drive needs a startup characteristic instead of an infinite torque demand.[1][3]

Mechanical power describes work per time. For the same fixed axis, instantaneous $P=\tau\omega$ ; constant torque therefore does not mean constant power when angular velocity changes. The sign shows whether a torque is instantaneously adding energy to the body or removing it.[1]

Check five axial balances

Calculate · Foundation

A rotor has $I=0.450\,\mathrm{kg\,m^2}$ and $\omega_0=+6.00\,\mathrm{rad\,s^{-1}}$ . Constant braking torque −0.900 $\mathrm{N\,m}$ acts until the first stop. Find $\alpha$ and duration.

Show hint

Calculate only until rest; do not extend the braking model into reversal without justification.

Learning goal: Calculate braking angular acceleration and time to first stop.

Show solution and calculation

Approach: Use $\alpha=\frac{\tau_{\mathrm{net}}}I$ ; at the first stop, $0=\omega_0+\alpha t_s$ .
This gives $\alpha=-2.00\,\mathrm{rad\,s^{-2}}$ and $t_s=3.00\,\mathrm s$ .

Check: $6.00-2.00\cdot3.00=0$ confirms rest. Subsequent motion depends on whether the brake holds or a drive continues acting.

\begin{aligned}\alpha&=-2.00\,\mathrm{rad/s^2}\\t_s&=3.00\,\mathrm s\end{aligned}

Rearrange · Application

A rigid rotor on fixed bearings accelerates at +3.00 $\mathrm{rad\,s^{-2}}$ . Motor torque +1.60 $\mathrm{N\,m}$ and load torque −0.400 $\mathrm{N\,m}$ are the only axial contributions. Find I.

Show hint

Sum torques first, then divide by $\alpha$ .

Learning goal: Find inertia from resultant torque and angular acceleration.

Show solution and calculation

Approach: Add signed torques and rearrange $I=\frac{\tau_{\mathrm{motor}}+\tau_{\mathrm{load}}}{\alpha}$ .
Net torque is $+1.20\,\mathrm{N\,m}$ , giving $I=0.400\,\mathrm{kg\,m^2}$ .

Check: $I\alpha=0.400\cdot3.00\,\mathrm{N\,m}=1.20\,\mathrm{N\,m}$ . Using motor torque alone would overestimate inertia.

I=0.400\,\mathrm{kg\,m^2}

Transfer · Challenge

A nonrigid rotor changes only radial mass distances; all masses share $\omega_z=+10.0\,\mathrm{rad\,s^{-1}}$ with no additional relative angular motion. At this instant $I=0.200\,\mathrm{kg\,m^2}$ , $\frac{dI}{dt}=+0.0300\,\mathrm{kg\,m^2\,s^{-1}}$ and $\tau_{z,\mathrm{ext}}$ =0. Find $\alpha_z$ .

Show hint

Differentiate $L_z$ =Iωz using the product rule.

Learning goal: Apply the angular-momentum product rule when inertia changes.

Show solution and calculation

Approach: This explicitly nonrigid system satisfies $0=\frac{d(I\omega_z)}{dt}=I\alpha_z+\dot I\omega_z$ .
Hence $\alpha_z=-\frac{\dot I\omega_z}I=-1.50\,\mathrm{rad\,s^{-2}}$ .

Check: the two angular-momentum derivative terms are $-0.300\,\mathrm{N\,m}$ and $+0.300\,\mathrm{N\,m}$ . Zero external torque keeps angular momentum constant rather than angular velocity. This is a counterexample to the rigid-body short form.

\alpha_z=-\frac{\dot I\omega_z}{I}=-1.50\,\mathrm{rad/s^2}

Interpret · Foundation

A rigid rotor rotates about a fixed axis with constant $I=0.30\,\mathrm{kg\,m^2}$ and initial $\omega=+8\,\mathrm{rad\,s^{-1}}$ . Motor and load simultaneously supply +1.2 $\mathrm{N\,m}$ and −1.2 $\mathrm{N\,m}$ , with no other torques. How does $\omega$ evolve during 5 s? A graph shows a horizontal $\omega$ -versus-time line above zero. Is this consistent with Newton’s rotational law?

Show hint

The law uses the sum of torques. Zero angular acceleration preserves the existing angular velocity.

Learning goal: Connect zero resultant torque to constant angular velocity rather than rest.

Show solution and calculation

Approach: sum the directed torques and divide by constant inertia.
Angular acceleration is zero, so $\omega$ stays at +8 $\mathrm{rad\,s^{-1}}$ for the five seconds.
The graph is correct.

Check: motor and load each transfer 9.6 W with opposite signs; net power and rotational-energy change vanish. Zero resultant torque does not automatically mean rest.

\begin{aligned}\tau_{\mathrm{net}}&=1.2-1.2=0\,\mathrm{N\,m}\\\alpha&=\frac{\tau_{\mathrm{net}}}{I}=0\\\omega(5\,\mathrm s)&=\omega_0+\alpha t=+8\,\mathrm{rad/s}\\P_{\mathrm{motor}}&=-P_{\mathrm{load}}=9.6\,\mathrm W\end{aligned}

[1]

Error analysis · Application

A rigid rotor with $I=0.20\,\mathrm{kg\,m^2}$ has $\omega_0=-6\,\mathrm{rad\,s^{-1}}$ . A constant positive resultant torque +0.40 $\mathrm{N\,m}$ acts about the same fixed axis for 1.0 s. A solution says: “Positive torque makes the rotor faster.” Calculate $\omega$ after one second and assess angular-speed magnitude; find the rotational-energy change.

Show hint

Positive angular acceleration increases signed angular velocity. With negative initial velocity, its magnitude can decrease.

Learning goal: Interpret torque sign relative to existing rotation direction.

Show solution and calculation

Approach: obtain angular acceleration from torque and add its contribution to the negative initial velocity.
Acceleration is +2 $\mathrm{rad\,s^{-2}}$ , giving $\omega=-4\,\mathrm{rad\,s^{-1}}$ after one second.
Magnitude falls from 6 to 4 $\mathrm{rad\,s^{-1}}$ , so the rotor slows.
Rotational energy decreases by 2.0 J.

Check: torque and angular velocity have opposite signs throughout the interval, so power is negative.

\begin{aligned}\alpha&=\frac{0.40\,\mathrm{N\,m}}{0.20\,\mathrm{kg\,m^2}}=+2\,\mathrm{rad/s^2}\\\omega_f&=-6+2\cdot1=-4\,\mathrm{rad/s}\\\Delta E_{\mathrm{rot}}&=\frac12I(\omega_f^2-\omega_0^2)\\&=\frac12(0.20)(16-36)\,\mathrm J=-2.0\,\mathrm J\end{aligned}

[1][2]

Choose the model before rearranging

The fixed axis must be at rest relative to an inertial frame, and the body must not change its moment of inertia about that axis during the calculation. Free three-dimensional rotation, deformable bodies, internal rotational degrees of freedom or an arbitrarily moving reference axis require the general angular momentum balance with appropriate reference treatment. The upper example $I=0.20\,\mathrm{kg\,m^2}$ and $\tau_{\mathrm{net}}=0.60\,\mathrm{N\,m}$ remains correct: $\alpha=3.0\,\mathrm{rad\,s^{-2}}$ .[2][3]

If $L_z=I(t)\omega_z$ still holds, variable I adds the term $\dot I\omega_z$ . Additional relative angular motion can invalidate even this product form unless included in total angular momentum. The most useful check is therefore whether torque, inertia and acceleration refer to the same physical system and axis, rather than numerical substitution alone.[2]

What to remember

Choose the body, axis and positive rotation direction. Sum all external axial torques and use inertia about that same axis. Balanced torques preserve angular velocity when inertia is constant, including during rotation. Coupled cords or rolling additionally require translation and contact constraints.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Linked relevant learning transitions; preserved existing derivations, worked examples, exercises and figures in full.
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Quanta. Newton’s second law for rotation. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/newton-rotation-starrer-koerper

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Frequently asked questions about Newton’s second law for rotation

What does Newton’s second law for rotation describe physically?+

$\sum\tau=I\alpha$ relates net external torque component to signed angular acceleration of a rigid body about a fixed inertial axis. Include every contribution: drive, load and bearing friction. Motor torque alone therefore is not automatically net torque. With balanced torques, $\alpha=0$ can mean uniform rotation rather than rest. Larger I produces less acceleration for the same net torque. This scalar statement concerns the axial component and needs no principal axis. It does not assert that the full angular momentum vector is parallel to angular velocity about every axis; the inertia tensor governs that general vector relation.

How can I check a calculation involving Newton’s second law for rotation?+

Choose body, axis and positive rotation direction first. With I=0.320 kg·m², motor torque +2.40 N·m and load torque −0.480 N·m, $\sum\tau=+1.920\,\mathrm{N\,m}$ and α=+6.00 rad/s². Starting at +4.00 rad/s, the rotor reaches +13.0 rad/s after 1.50 s. An independent energy check uses Δθ=12.75 rad: net work 1.920·12.75=24.48 J agrees with $\frac12I(\omega_f^2-\omega_0^2)$ . Check that the load really opposes positive rotation throughout the interval. Comparing only motor torque with inertia would omit the load and overestimate acceleration, even though the resulting units still look correct.

What assumptions and limits apply to Newton’s second law for rotation?+

The short form assumes a rigid body and constant I about an axis fixed in an inertial frame. Net torque may vary with time; the equation then holds instantaneously without guaranteeing constant α. For arbitrary free spatial rotation use vector balance $\vec\tau=\frac{d\vec L}{dt}$ . If mass distribution changes while $L_z=I(t)\omega_z$ still holds, the product rule adds $\dot I\omega_z$ . Further internal angular motion must be included in full angular momentum. A moving rolling contact location is not automatically a fixed inertial axis. In planar rolling, separate centre-of-mass force and torque balances combined with the rolling constraint provide a reliable approach.

How do I check units and scale for Newton’s second law for rotation?+

Torque is in N·m, I in kg·m² and α in rad/s². Radian is dimensionless in SI, so $I\alpha$ has dimension kg·m²/s²=N·m. Sharing the joule dimension does not turn torque into energy; work also requires angular displacement. Check signed torque addition and that I belongs to the chosen axis. Doubling I halves α at the same net torque. Zero net torque must give $\alpha=0$ with constant I. Rearrangement $I=\frac{\sum\tau}{\alpha}$ requires $\alpha\ne0$ . Uniform rotation with balanced loads therefore cannot determine I through that quotient, regardless of how accurately speed is measured.

When should I combine Newton’s second law for rotation with another relation?+

With constant net torque and I, $\alpha=\frac{\sum\tau}I$ leads to $\omega=\omega_0+\alpha t$ and $\Delta\theta=\omega_0t+\frac12\alpha t^2$ . Work and power for the same fixed axis follow from $W=\int\tau\,d\theta$ and $P=\tau\omega$ . A falling mass driving a drum adds $mg-T=ma$ and cord constraint $a=R\alpha$ ; rotational torque is $TR$ and tension is not automatically mg. For rolling, combine centre-of-mass translation with rotation about the centre of mass and the contact constraint. Every additional relation needs the same system, appropriate reference points and consistent signs. Energy provides an independent check but does not replace the requested tension or friction force.

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Newton’s second law for rotation

Net external torque component about a fixed inertial axis equals I times signed angular acceleration when I is constant.

BasicExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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Symbol

Meaning

Unit

\sum τ

Signed sum of all external torque components about the fixed inertial axis

N·m

I

Constant moment of inertia of the rigid body about that same axis

kg·m²

α

Signed angular acceleration component about the same axis

rad/s²

Derivation & background

Exam blueprint

Validity range

Derivation steps

Net external torque changes axial angular momentum; constant I turns this into $I\alpha$ .

1Choose the body, fixed inertial z axis and positive rotation direction.
2Calculate $I_z=\sum_i m_i r_{i,\perp}^2$ from perpendicular distances to that same axis.
3Rigid rotation gives every mass the same $\omega_z$ , hence $L_z=I_z\omega_z$ .
4The net external axial torque satisfies $\sum\tau_{z,\mathrm{ext}}=\frac{dL_z}{dt}$ .
5Constant axial distances keep $I_z$ constant, so differentiation gives $I_z\frac{d\omega_z}{dt}$ .
6With $\alpha_z=\frac{d\omega_z}{dt}$ , obtain $\sum\tau_{z,\mathrm{ext}}=I_z\alpha_z$ . Include drive, load and friction.
7This scalar axial balance requires no principal axis. The full vector relation generally needs the inertia tensor.

How do you calculate with Newton’s second law for rotation?

Here is how to work through a typical Newton’s second law for rotation (Στ = Iα) task step by step:

Task

A rigid rotor has $I=0.20\,\mathrm{kg\,m^2}$ about its fixed inertial axis. Net external axial torque is $+0.60\,\mathrm{N\,m}$ . Find angular acceleration.

Solution path

Approach: $\alpha=\frac{\tau_{\mathrm{net}}}I$ .
This gives $\alpha=+3.0\,\mathrm{rad\,s^{-2}}$ .

Check: $I\alpha=0.60\,\mathrm{N\,m}$ . The positive sign identifies acceleration direction rather than automatically increasing angular speed.

What changes rotational motion?

Prior knowledge: force balance and angular velocity. Moment of inertia describes how mass is distributed about an axis, rather than mass alone.

Net axial torque produces acceleration

\sum\tau_{z,\mathrm{ext}}=I_z\alpha_z

For a rigid body about a fixed inertial axis, net external axial torque equals constant axial moment of inertia times signed angular acceleration.[1][2]

View diagram: Net torque sets the slope

Net torque sets the slope

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The first example has

I=0.320\,\mathrm{kg\,m^2}

and net torque +1.920

\mathrm{N\,m}

. The line rises from 4 to 13

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

in 1.50 s. Its area of 12.75 rad gives 24.48 J work with the net torque, matching the energy change.[1]

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From angular momentum to acceleration

\begin{aligned}L_z&=\sum_i m_i r_{i,\perp}^2\omega_z=I_z\omega_z\\\sum\tau_{z,\mathrm{ext}}&=\frac{dL_z}{dt}\end{aligned}

Each particle shares angular velocity; its perpendicular distance to the fixed axis determines its axial angular momentum contribution.[2]

\begin{aligned}\alpha_z&=\frac{\sum\tau_z}{I_z}\\I_z&=\frac{\sum\tau_z}{\alpha_z}\end{aligned}

The first division requires $I_z$ >0. The second requires $\alpha_z$ ≠0 and a valid model; zero torque and zero acceleration do not determine I.

Explore further: Distinguish axial balance from a vector law

Specify lever arm, direction and units

\begin{aligned}\tau_z&=(\mathbf r\times\mathbf F)_z\\{}[I_z\alpha_z]&=\mathrm{kg\,m^2\,s^{-2}}=\mathrm{N\,m}\end{aligned}

r points from the chosen axis to the point of application. The right-hand rule sets the sign, not force magnitude alone.[1]

Drive and coupled descending mass

Motor with opposing load

Given: A rigid rotor about a fixed axis has $I=0.320\,\mathrm{kg\,m^2}$ . Motor torque is +2.40 $\mathrm{N\,m}$ and load torque −0.480 $\mathrm{N\,m}$ . Initially $\omega_0=+4.00\,\mathrm{rad\,s^{-1}}$ ; both torques stay constant for 1.50 s.
Find: Find $\alpha$ , final angular velocity and energy change.

Approach: Add all external axial torques with their signs. $\alpha=\frac{\tau_{\mathrm{Motor}}+\tau_{\mathrm{Last}}}{I}$ . First sum signed torques.
$\begin{aligned}\tau_{\mathrm{net}}&=2.40-0.480=1.920\,\mathrm{N\,m}\\{} \alpha&=\frac{1.920}{0.320}=6.00\,\mathrm{rad/s^2}\end{aligned}$
With constant I and torque, $\alpha$ is constant.
$\begin{aligned}\omega_f&=4+6(1.50)=13.0\,\mathrm{rad/s}\\\Delta\theta&=4(1.50)+\tfrac12(6)(1.50)^2=12.75\,\mathrm{rad}\end{aligned}$
Calculate rotational energy change from initial and final values.
$\Delta K=\tfrac12(0.320)(13^2-4^2)=24.48\,\mathrm J$

Result: The results are $\alpha=+6.00\,\mathrm{rad\,s^{-2}}$ , $\omega_f=+13.0\,\mathrm{rad\,s^{-1}}$ and $\Delta K=24.48\,\mathrm J$ . The load opposes positive rotation throughout the interval.

Descending mass drives a drum

Given: A mass $m=1.80\,\mathrm{kg}$ hangs from a massless taut cord. The cord unwinds without slipping from a drum with $R=0.150\,\mathrm{m}$ and $I=0.120\,\mathrm{kg\,m^2}$ . Fixed frictionless axis, constant R, release from rest, drop $h=0.650\,\mathrm m$ , $g=9.81\,\mathrm{m\,s^{-2}}$ .
Find: Find a, cord tension and speed after the descent.

Approach: Combine translation of the mass, drum rotation and rope coupling. $mg-T=ma,\quad TR=I\alpha,\quad a=R\alpha$ . Choose downward translation and unwinding rotation positive.
$\begin{aligned}mg-T&=ma\\TR&=I\alpha\\a&=R\alpha\end{aligned}$
Eliminate cord tension and include rotational inertia.
$\begin{aligned}a&=\frac{mg}{m+\frac{I}{R^2}}=2.475421\,\mathrm{m/s^2}\\{} \alpha&=16.502804\,\mathrm{rad/s^2}\end{aligned}$
Tension is below mg because the mass accelerates.
$\begin{aligned}T&=m(g-a)=13.202243\,\mathrm N\\TR&=1.980336\,\mathrm{N\,m}\end{aligned}$
Use the descent distance with constant acceleration.
$\begin{aligned}v&=\sqrt{2ah}=1.793892\,\mathrm{m/s}\\{} \omega&=\frac{v}{R}=11.959277\,\mathrm{rad/s}\end{aligned}$

Result: $a\approx2.475\,\mathrm{m\,s^{-2}}$ , T≈13.202 N and $v\approx1.794\,\mathrm{m\,s^{-1}}$ . The drum receives part of the released gravitational energy.

Torque and power complement the time balance

\begin{aligned}W_{\mathrm{net}}&=\int\tau_{z,\mathrm{net}}\,d\theta=\Delta\!\left(\frac12 I_z\omega_z^2\right)\\P_{\mathrm{net}}&=\tau_{z,\mathrm{net}}\omega_z\end{aligned}

For the same fixed axis and constant I, torque work equals change in rotational kinetic energy.[3]

Check five axial balances

Calculate · Foundation

A rotor has $I=0.450\,\mathrm{kg\,m^2}$ and $\omega_0=+6.00\,\mathrm{rad\,s^{-1}}$ . Constant braking torque −0.900 $\mathrm{N\,m}$ acts until the first stop. Find $\alpha$ and duration.

Show hint

Calculate only until rest; do not extend the braking model into reversal without justification.

Learning goal: Calculate braking angular acceleration and time to first stop.

Show solution and calculation

Approach: Use $\alpha=\frac{\tau_{\mathrm{net}}}I$ ; at the first stop, $0=\omega_0+\alpha t_s$ .
This gives $\alpha=-2.00\,\mathrm{rad\,s^{-2}}$ and $t_s=3.00\,\mathrm s$ .

Check: $6.00-2.00\cdot3.00=0$ confirms rest. Subsequent motion depends on whether the brake holds or a drive continues acting.

\begin{aligned}\alpha&=-2.00\,\mathrm{rad/s^2}\\t_s&=3.00\,\mathrm s\end{aligned}

Rearrange · Application

A rigid rotor on fixed bearings accelerates at +3.00 $\mathrm{rad\,s^{-2}}$ . Motor torque +1.60 $\mathrm{N\,m}$ and load torque −0.400 $\mathrm{N\,m}$ are the only axial contributions. Find I.

Show hint

Sum torques first, then divide by $\alpha$ .

Learning goal: Find inertia from resultant torque and angular acceleration.

Show solution and calculation

Approach: Add signed torques and rearrange $I=\frac{\tau_{\mathrm{motor}}+\tau_{\mathrm{load}}}{\alpha}$ .
Net torque is $+1.20\,\mathrm{N\,m}$ , giving $I=0.400\,\mathrm{kg\,m^2}$ .

Check: $I\alpha=0.400\cdot3.00\,\mathrm{N\,m}=1.20\,\mathrm{N\,m}$ . Using motor torque alone would overestimate inertia.

I=0.400\,\mathrm{kg\,m^2}

Transfer · Challenge

A nonrigid rotor changes only radial mass distances; all masses share $\omega_z=+10.0\,\mathrm{rad\,s^{-1}}$ with no additional relative angular motion. At this instant $I=0.200\,\mathrm{kg\,m^2}$ , $\frac{dI}{dt}=+0.0300\,\mathrm{kg\,m^2\,s^{-1}}$ and $\tau_{z,\mathrm{ext}}$ =0. Find $\alpha_z$ .

Show hint

Differentiate $L_z$ =Iωz using the product rule.

Learning goal: Apply the angular-momentum product rule when inertia changes.

Show solution and calculation

Approach: This explicitly nonrigid system satisfies $0=\frac{d(I\omega_z)}{dt}=I\alpha_z+\dot I\omega_z$ .
Hence $\alpha_z=-\frac{\dot I\omega_z}I=-1.50\,\mathrm{rad\,s^{-2}}$ .

\alpha_z=-\frac{\dot I\omega_z}{I}=-1.50\,\mathrm{rad/s^2}

Interpret · Foundation

A rigid rotor rotates about a fixed axis with constant $I=0.30\,\mathrm{kg\,m^2}$ and initial $\omega=+8\,\mathrm{rad\,s^{-1}}$ . Motor and load simultaneously supply +1.2 $\mathrm{N\,m}$ and −1.2 $\mathrm{N\,m}$ , with no other torques. How does $\omega$ evolve during 5 s? A graph shows a horizontal $\omega$ -versus-time line above zero. Is this consistent with Newton’s rotational law?

Show hint

The law uses the sum of torques. Zero angular acceleration preserves the existing angular velocity.

Learning goal: Connect zero resultant torque to constant angular velocity rather than rest.

Show solution and calculation

Approach: sum the directed torques and divide by constant inertia.
Angular acceleration is zero, so $\omega$ stays at +8 $\mathrm{rad\,s^{-1}}$ for the five seconds.
The graph is correct.

Check: motor and load each transfer 9.6 W with opposite signs; net power and rotational-energy change vanish. Zero resultant torque does not automatically mean rest.

\begin{aligned}\tau_{\mathrm{net}}&=1.2-1.2=0\,\mathrm{N\,m}\\\alpha&=\frac{\tau_{\mathrm{net}}}{I}=0\\\omega(5\,\mathrm s)&=\omega_0+\alpha t=+8\,\mathrm{rad/s}\\P_{\mathrm{motor}}&=-P_{\mathrm{load}}=9.6\,\mathrm W\end{aligned}

[1]

Error analysis · Application

A rigid rotor with $I=0.20\,\mathrm{kg\,m^2}$ has $\omega_0=-6\,\mathrm{rad\,s^{-1}}$ . A constant positive resultant torque +0.40 $\mathrm{N\,m}$ acts about the same fixed axis for 1.0 s. A solution says: “Positive torque makes the rotor faster.” Calculate $\omega$ after one second and assess angular-speed magnitude; find the rotational-energy change.

Show hint

Positive angular acceleration increases signed angular velocity. With negative initial velocity, its magnitude can decrease.

Learning goal: Interpret torque sign relative to existing rotation direction.

Show solution and calculation

Approach: obtain angular acceleration from torque and add its contribution to the negative initial velocity.
Acceleration is +2 $\mathrm{rad\,s^{-2}}$ , giving $\omega=-4\,\mathrm{rad\,s^{-1}}$ after one second.
Magnitude falls from 6 to 4 $\mathrm{rad\,s^{-1}}$ , so the rotor slows.
Rotational energy decreases by 2.0 J.

Check: torque and angular velocity have opposite signs throughout the interval, so power is negative.

\begin{aligned}\alpha&=\frac{0.40\,\mathrm{N\,m}}{0.20\,\mathrm{kg\,m^2}}=+2\,\mathrm{rad/s^2}\\\omega_f&=-6+2\cdot1=-4\,\mathrm{rad/s}\\\Delta E_{\mathrm{rot}}&=\frac12I(\omega_f^2-\omega_0^2)\\&=\frac12(0.20)(16-36)\,\mathrm J=-2.0\,\mathrm J\end{aligned}

[1][2]

Choose the model before rearranging

What to remember

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

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.

Editorial responsibility, sources and methods

Change history

October 2, 2026Linked relevant learning transitions; preserved existing derivations, worked examples, exercises and figures in full.
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.

Cite this reference page

Quanta. Newton’s second law for rotation. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/newton-rotation-starrer-koerper

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Frequently asked questions about Newton’s second law for rotation

What does Newton’s second law for rotation describe physically?+

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Newton’s second law for rotation

Variables & units

Torque and angular acceleration 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 Newton’s second law for rotation?

What changes rotational motion?

Net axial torque produces acceleration

Net torque sets the slope

From angular momentum to acceleration

Explore further: Distinguish axial balance from a vector law

Specify lever arm, direction and units

Drive and coupled descending mass

Motor with opposing load

Descending mass drives a drum

Torque and power complement the time balance

Check five axial balances

A rotor has I=0.450 kg m2I=0.450\,\mathrm{kg\,m^2}I=0.450kgm2 and ω0=+6.00 rad s−1\omega_0=+6.00\,\mathrm{rad\,s^{-1}}ω0​=+6.00rads−1. Constant braking torque −0.900 N m\mathrm{N\,m}Nm acts until the first stop. Find α\alphaα and duration.

A rigid rotor on fixed bearings accelerates at +3.00 rad s−2\mathrm{rad\,s^{-2}}rads−2. Motor torque +1.60 N m\mathrm{N\,m}Nm and load torque −0.400 N m\mathrm{N\,m}Nm are the only axial contributions. Find I.

Choose the model before rearranging

What to remember

Sources for this explanation

Frequently asked questions about Newton’s second law for rotation

Subject references and further reading

More formulas: Mechanics

Retain Newton’s second law for rotation for exams

Newton’s second law for rotation

Variables & units

Torque and angular acceleration 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 Newton’s second law for rotation?

What changes rotational motion?

Net axial torque produces acceleration

Net torque sets the slope

From angular momentum to acceleration

Explore further: Distinguish axial balance from a vector law

Specify lever arm, direction and units

Drive and coupled descending mass

Motor with opposing load

Descending mass drives a drum

Torque and power complement the time balance

Check five axial balances

A rotor has I=0.450 kg m2I=0.450\,\mathrm{kg\,m^2}I=0.450kgm2 and ω0=+6.00 rad s−1\omega_0=+6.00\,\mathrm{rad\,s^{-1}}ω0​=+6.00rads−1. Constant braking torque −0.900 N m\mathrm{N\,m}Nm acts until the first stop. Find α\alphaα and duration.

A rigid rotor on fixed bearings accelerates at +3.00 rad s−2\mathrm{rad\,s^{-2}}rads−2. Motor torque +1.60 N m\mathrm{N\,m}Nm and load torque −0.400 N m\mathrm{N\,m}Nm are the only axial contributions. Find I.

Choose the model before rearranging

What to remember

Sources for this explanation

Frequently asked questions about Newton’s second law for rotation

Subject references and further reading

More formulas: Mechanics

Retain Newton’s second law for rotation for exams

A rotor has $I=0.450\,\mathrm{kg\,m^2}$ and $\omega_0=+6.00\,\mathrm{rad\,s^{-1}}$ . Constant braking torque −0.900 $\mathrm{N\,m}$ acts until the first stop. Find $\alpha$ and duration.

A rigid rotor on fixed bearings accelerates at +3.00 $\mathrm{rad\,s^{-2}}$ . Motor torque +1.60 $\mathrm{N\,m}$ and load torque −0.400 $\mathrm{N\,m}$ are the only axial contributions. Find I.

A rotor has $I=0.450\,\mathrm{kg\,m^2}$ and $\omega_0=+6.00\,\mathrm{rad\,s^{-1}}$ . Constant braking torque −0.900 $\mathrm{N\,m}$ acts until the first stop. Find $\alpha$ and duration.

A rigid rotor on fixed bearings accelerates at +3.00 $\mathrm{rad\,s^{-2}}$ . Motor torque +1.60 $\mathrm{N\,m}$ and load torque −0.400 $\mathrm{N\,m}$ are the only axial contributions. Find I.