Physics · Rotational motion

Angular velocity under constant angular acceleration

Q: What does Angular velocity under constant angular acceleration describe physically?

The relation $\omega=\omega_0+\alpha t$ describes a signed angular velocity component about a fixed axis. Choose positive rotation direction first. Constant α is the slope of the line in an ω-t diagram; $\omega_0$ is its initial value. Negative α does not automatically mean braking: with negative ω, speed magnitude can increase further. A zero crossing marks instantaneous rest and, if the model continues, reversal. Area under the line also gives signed angular displacement, rather than total angular travel during reversal. The equation specifies motion but cannot determine driving torque or energy requirements without inertia and a torque balance.

Q: How do I check units and scale for Angular velocity under constant angular acceleration?

ω and $\omega_0$ are in rad/s, α in rad/s² and t in s. Product αt therefore has the required rad/s unit. Convert rpm using ω=2πn/60; 600 rpm is 20π rad/s, not 600 rad/s. Radian is dimensionless in SI but useful here as an angular indicator. Also check t=0: the result must be $\omega_0$. At $\alpha=0$, ω must not change. Doubling time doubles only the change αt, not generally the full final value including initial velocity. Larger negative values can represent a larger speed magnitude, so inspect the sign as well as the numerical size.

Q: When should I combine Angular velocity under constant angular acceleration with another relation?

For angular displacement integrate angular velocity: $\Delta\theta=\omega_0t+\frac12\alpha t^2$ when α is constant. For total angular travel during reversal integrate |ω| instead. At fixed radius r, $v=r|\omega|$ and accelerations at=rα and ar=rω² give linear quantities. Constant α therefore does not make ar constant. To explain the cause of motion, add $\sum\tau=I\alpha$ for a rigid body about a fixed axis. Energy follows from K=$I\omega^2$/2 with appropriate I. Matching axis, times and direction convention connects these calculations. Energy depends on the square rather than rotation direction, while tangential acceleration retains the chosen sign.

Constant signed angular acceleration about a fixed axis makes signed angular velocity change linearly with time.

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Formula

Calculate

ω = ω_{0} + α t

$\omega,\omega_0$ in rad/s; $\alpha$ in rad/s²; t in seconds. Convert rotation rate to angular velocity first.

Variables & units

Variables & units
Symbol	Meaning	Unit
$ω$	Signed angular velocity component about the fixed axis at time t	rad/s
$ω₀$	Signed initial component at t=0	rad/s
$α$	Signed angular acceleration about the same axis, constant over the interval	rad/s²
$t$	Elapsed time since the initial state	s

LaTeX

\omega=\omega_0+\alpha t

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Angular velocity against time: line from 12 rad/s at 0 s through zero at 4 s to −3 rad/s at 5 s. Positive triangular area 24 rad, negative area −1.5 rad, net angle 22.5 rad. — Constant α = −3 rad/s² passes through rest at t = 4 s. Over 0 to 5 s, signed area is 22.5 rad; total angular travel is 25.5 rad.

Angular velocity calculator

Find angular velocity, angular acceleration, time and angle turned for constant angular acceleration about a fixed axis.

Find

Initial angular velocity ω₀

Angular velocity at the start with its sign, e.g. 10 rad/s; 0 for a start from rest.

Signed angular acceleration α

Constant angular acceleration with its sign, e.g. 2 rad/s²; against the direction of turning means braking.

Elapsed time interval Δt

Duration of the steady acceleration, e.g. 5 s.

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

Result: Final angular velocity ω

ωf = 20 rad/s

With ω₀ = 10 rad/s, α = 2 rad/s² and Δt = 5 s, ωf = 20 rad/s.

Working · rearrange for ωf, substitute, evaluate

1. Rearrange for ωf
$\omega_{f,z}=\omega_{0,z}+\alpha_z\Delta t$
2. Substitute values (in SI units)
$10+2\cdot 5$
3. Result
ωf = 20 rad/s

Quick tips

Just like straight-line motion: ω matches v, α matches a and the angle matches distance. ω = ω₀ + α · t is the partner of v = v₀ + a · t.
Shortcut for the angle turned: average angular velocity times time, (ω₀ + ω)/2 · t. From 10 to 20 rad/s in 5 s gives 15 rad/s · 5 s = 75 rad.

Assumptions and limits

Fixed axis and constant signed acceleration throughout a nonnegative elapsed interval. Reversal is calculated only if acceleration actually continues; a passive brake does not automatically continue through rest. Angular displacement is signed; angular travel adds magnitudes.

Derivation & background

Integrating $\alpha=\frac{d\omega}{dt}$ at constant value gives $\omega=\omega_0+\alpha t$ . Both components use the same fixed axis and positive rotation direction; $|\omega|$ is magnitude. Negative acceleration may slow rotation or increase speed depending on direction. A passive brake must not be modelled as a continuing constant torque beyond rest without justification.

Exam blueprint

Validity range

Fixed axis with chosen positive direction; $\omega$ and $\alpha$ are signed components. $\alpha$ must be constant over the entire interval. After stopping, the line continues only if the assumed acceleration is actually maintained.

Derivation steps

Integrating constant signed angular acceleration gives linear angular velocity.

1Choose a fixed axis and positive rotation direction; $\omega$ and $\alpha$ are signed components.
2Define $\alpha=\frac{d\omega}{dt}$ and initial value $\omega(0)=\omega_0$ .
3Integrate $\omega(t)-\omega_0=\int_0^t\alpha(t^{\prime})\,dt^{\prime}$ .
4Only an $\alpha$ constant throughout the interval may leave the integral, yielding $\alpha t$ .
5Add the initial value: $\omega(t)=\omega_0+\alpha t$ . The added term has the same $\mathrm{rad\,s^{-1}}$ unit as $\omega$ .
6For $\alpha\ne0$ , check the zero $t_s=-\frac{\omega_0}{\alpha}$ . A later reversal requires positive $t_s$ and continuation of the model.
7With $\alpha=0$ , angular velocity stays constant. Varying acceleration requires the integral rather than endpoint acceleration times time.

Rearrangements

Angular acceleration

\alpha=\frac{\omega-\omega_0}t

Only for constant angular acceleration and t>0.

Task variant

A rotor starts at $\omega_0=2.0\,\mathrm{rad\,s^{-1}}$ . Its angular acceleration stays at $\alpha=+0.50\,\mathrm{rad\,s^{-2}}$ for four seconds. Find the final value.

Approach: $\omega=\omega_0+\alpha t$ .
Substitution gives $\omega=4.0\,\mathrm{rad\,s^{-1}}$ .

Check: the 2.0 $\mathrm{rad\,s^{-1}}$ increase over four seconds recovers the given acceleration.

Common mistakes

Always interpreting negative acceleration as braking, or signed angular change as total angular travel.

Check the sign of $\omega$ as well. Add area magnitudes across reversals; a passive brake does not automatically continue the line beyond rest.

Exam context

Find angular speed and revolutions in uniformly accelerated rotation.

Formula cluster

Rotational motion

Related relations: winkelgeschwindigkeit-kreisfrequenz, drehmoment, rotationsenergie-starrer-koerper.

ω = 2πf; v = ω·r

Angular Velocity and Angular Frequency

M = F·r·sin(θ)

Torque

Erot = ½Iω²

Rotational kinetic energy of a rigid body

Worked example

A rotor begins at $\omega_0=2.0\,\mathrm{rad\,s^{-1}}$ and maintains $\alpha=+0.50\,\mathrm{rad\,s^{-2}}$ . After $t=4.0\,\mathrm s$ , $\omega=\omega_0+\alpha t=4.0\,\mathrm{rad\,s^{-1}}$ . The 2.0 rad/s change over four seconds matches the specified slope.

Applications

Find angular speed and revolutions in uniformly accelerated rotation.

How do you calculate with Angular velocity under constant angular acceleration?

Here is how to work through a typical Angular velocity under constant angular acceleration (ω = ω₀ + αt) task step by step:

1
Task
A rotor starts at $\omega_0=2.0\,\mathrm{rad\,s^{-1}}$ . Its angular acceleration stays at $\alpha=+0.50\,\mathrm{rad\,s^{-2}}$ for four seconds. Find the final value.
Solution path
1. Approach: $\omega=\omega_0+\alpha t$ .
2. Substitution gives $\omega=4.0\,\mathrm{rad\,s^{-1}}$ .
Check: the 2.0 $\mathrm{rad\,s^{-1}}$ increase over four seconds recovers the given acceleration.

Understand and apply the equation

A rotor can speed up, slow down and reverse direction. Under constant angular acceleration, signed angular velocity is a straight line in time. Its slope, zero crossing and area answer different questions: acceleration, reversal and angular displacement.

A straight line describes spin-up

How fast is a rotor turning after a few seconds? If angular velocity changes by the same amount each second, angular acceleration is constant and its time graph is a straight line. Signs indicate direction; area gives angular displacement. With a reversal, net displacement differs from total angular travel.[1]

Prior knowledge: line slopes and triangle or trapezoid areas. One revolution corresponds to an angle of two pi radians.

A signed quantity about a fixed axis

\omega_z(t)=\omega_{z,0}+\alpha_z(t-t_0)

The upper short form sets t0=0 and suppresses the axis index. Choose the positive rotation direction first.[1]

The sign of $\omega_z$ describes rotation direction; $|\omega_z|$ is nonnegative angular speed. Negative $\alpha_z$ therefore does not always mean slower rotation: with already negative $\omega_z$ , negative $\alpha_z$ increases its magnitude. The formula by itself says nothing about driving torque or required energy.[2]

Reversal separates displacement and travel

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Original example:

\omega(t)=12\,\mathrm{rad\,s^{-1}}-(3\,\mathrm{rad\,s^{-2}})t

. Reversal occurs at 4 s. Areas of +24 rad and −1.5 rad give net displacement 22.5 rad; total travel is 25.5 rad. A drive maintains negative acceleration after reversal.[1]

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Integrate the rate of change

\begin{aligned}\alpha_z&=\frac{d\omega_z}{dt}\\\omega_z(t)-\omega_{z,0}&=\int_{t_0}^{t}\alpha_z(t^{\prime})\,dt^{\prime}\end{aligned}

Angular acceleration is the time derivative of the signed component.[1]

If $\alpha_z$ is constant throughout the considered interval, take it outside the integral. The remaining time integral gives $t-t_0$ . Adding the initial value gives straight-line $\omega_z(t)$ . With variable $\alpha_z(t)$ , retain the integral; final acceleration multiplied by time generally is not the change in $\omega$ .[2]

\begin{aligned}\alpha_z&=\frac{\omega_z-\omega_{z,0}}{t-t_0}\\t-t_0&=\frac{\omega_z-\omega_{z,0}}{\alpha_z}\end{aligned}

The first rearrangement needs a nonzero time interval; the second additionally requires $\alpha_z$ ≠0. At $\alpha_z=0$ , $\omega_z$ remains constant.

Distinguish radians and revolutions

\begin{aligned}\omega&=2\pi f=\frac{2\pi n_{\mathrm{rpm}}}{60}\\1\,\mathrm{rev}&=2\pi\,\mathrm{rad}\end{aligned}

n in revolutions per minute is rotational frequency, not directly $\mathrm{rad\,s^{-1}}$ . For signed rotational frequency use the same direction convention.[2]

$\alpha_z$ in $\mathrm{rad\,s^{-2}}$ multiplied by time in s gives $\mathrm{rad\,s^{-1}}$ . Radian is dimensionless in SI but preserves the angular meaning. Convert degree values using $\frac\pi{180}$ before using relations such as $v=r|\omega|$ . Two measured $\omega$ values initially give an average acceleration. Constancy throughout the interval is an additional model assumption.[2]

Angular velocity and angular frequency clarify the conversion from revolutions per time to radians per time. Then apply the chosen axis sign convention; a positive rotation-rate magnitude does not replace a signed component.[1]

Area gives displacement, not automatically travel

\Delta\theta=\int_{t_0}^{t}\omega_z(t^{\prime})\,dt^{\prime}=\omega_{z,0}\Delta t+\frac12\alpha_z\Delta t^2=\frac{\omega_{z,0}+\omega_z}{2}\Delta t

For constant $\alpha$ , signed area under the line is a trapezoidal balance.[1]

\Theta_{\mathrm{travel}}=\int_{t_0}^{t}|\omega_z(t^{\prime})|\,dt^{\prime}

Total angular travel adds both directions positively. It equals | $\Delta\theta$ | only when there is no reversal.[2]

The relation $\omega^2=\omega_0^2+2\alpha\Delta\theta$ eliminates unknown time but squaring loses the sign of $\omega$ . A reversal therefore needs additional timing or rotation-direction information. $\theta$ may count full turns; an angle given only $\bmod 2\pi$ contains less information.[1]

Evaluate two different time histories

Controlled rotor spin-up

Given: A rotor accelerates at constant $\alpha$ from +300 to +900 rpm in 8.00 s; fixed axis and the same positive direction.
Find: Find $\alpha$ , angular displacement and number of turns.

Approach: Convert rotational rates to angular velocities and use constant acceleration. $\omega=2\pi\frac{n}{60},\quad\alpha=\frac{\omega_f-\omega_0}{\Delta t}$ . First convert both frequencies to $\mathrm{rad\,s^{-1}}$ .
$\begin{aligned}\omega_0&=10\pi\,\mathrm{rad/s}\\\omega_f&=30\pi\,\mathrm{rad/s}\end{aligned}$
Divide change by time interval.
$\alpha=\frac{30\pi-10\pi}{8}=2.5\pi\approx7.853982\,\mathrm{rad/s^2}$
Mean angular velocity determines area.
$\begin{aligned}\Delta\theta&=\tfrac12(10\pi+30\pi)8=160\pi\,\mathrm{rad}\\{} N&=\frac{\Delta\theta}{2\pi}=80\end{aligned}$

Result: $\alpha\approx7.854\,\mathrm{rad\,s^{-2}}$ and $\Delta\theta\approx502.655\,\mathrm{rad}$ , or 80 complete turns.

Check: Independent formula $\omega_0t+\frac{\alpha t^2}{2}$ also gives 160π. Mean frequency 600 rpm over $\frac8{60}\,\mathrm{min}$ gives 80 turns. Using only ωf·t would incorrectly give 120.

Actively pass through zero and reverse

Given: $\omega_0=+12.0\,\mathrm{rad\,s^{-1}}$ ; a drive maintains $\alpha=-3.00\,\mathrm{rad\,s^{-2}}$ for 5.00 s. The model continues after the instant of rest.
Find: Find final value, zero crossing, signed displacement and total angular travel.

Approach: The signed line may continue through zero into the opposite direction. $\omega(t)=\omega_0+\alpha t,\quad t_s=-\frac{\omega_0}{\alpha}$ . Find final value and zero crossing from the line.
$\begin{aligned}\omega_f&=12-3(5)=-3\,\mathrm{rad/s}\\{} t_s&=\frac{12}{3}=4\,\mathrm s\end{aligned}$
Calculate areas before and after the zero crossing separately.
$\begin{aligned}\Delta\theta_+&=\tfrac12(12)(4)=24\,\mathrm{rad}\\\Delta\theta_-&=-\tfrac12(3)(1)=-1.5\,\mathrm{rad}\end{aligned}$
Subtract signed contributions and add travel magnitudes.
$\begin{aligned}\Delta\theta&=22.5\,\mathrm{rad}\\\Theta_{\mathrm{travel}}&=25.5\,\mathrm{rad}\end{aligned}$

Result: After 5 s, the rotor turns at 3 $\mathrm{rad\,s^{-1}}$ in the reverse direction. Net displacement 22.5 rad differs from total travel 25.5 rad.

Check: Independently, $12\cdot5-\frac{3\cdot25}{2}=22.5\,\mathrm{rad}$ . The check $\omega_f^2=144-6\cdot22.5=9\,\mathrm{rad^2\,s^{-2}}$ confirms magnitude but cannot select direction. Passive dry braking would not automatically maintain the same negative acceleration after rest.

Separate tangential and radial acceleration

\begin{aligned}v&=r|\omega_z|\\a_t&=r\alpha_z\\a_r&=r\omega_z^2\end{aligned}

For a point at fixed perpendicular axis distance r, at is signed tangentially; $a_r$ is the inward radial magnitude.[2]

Constant $\alpha$ means constant tangential acceleration at fixed r, not constant radial acceleration. During spin-up, $\omega^2$ and therefore $a_r$ rise. $\sum\tau_z=I\alpha_z$ can supply required net dynamics when the system is rigid with constant axis-specific I. Kinematics alone determines neither friction losses nor motor power.[3]

Newton’s second law for rotation obtains angular acceleration from net external torque when constant inertia is known. Treat that result as constant in the time calculation only when the underlying torque balance supports it throughout the interval.[3]

Five exercises with solutions

Calculate · Foundation

$\omega_0=-5\,\mathrm{rad\,s^{-1}}$ , $\alpha=+2\,\mathrm{rad\,s^{-2}}$ and $t=4\,\mathrm s$ . Find $\omega$ and reversal time.

Show hint

Keep signs; separately solve $\omega=0$ .

Learning goal: Find signed final rotation and reversal time.

Show solution and calculation

Approach: Use $\omega=\omega_0+\alpha t$ , with $t_s=-\frac{\omega_0}{\alpha}$ for its zero.
The results are $\omega(4\,\mathrm s)=+3\,\mathrm{rad\,s^{-1}}$ and $t_s=2.5\,\mathrm s$ .

Check: $-5+2\cdot2.5=0$ . Magnitude decreases until the zero and then grows in the positive direction.

\begin{aligned}\omega&=3\,\mathrm{rad/s}\\t_s&=2.5\,\mathrm s\end{aligned}

Calculate · Application

A rotor at +600 rpm stops after 12 s under constant angular acceleration. Find $\alpha$ and $\Delta\theta$ up to stopping.

Show hint

600 rpm equals 20π $\mathrm{rad\,s^{-1}}$ .

Learning goal: Convert rotational speed for braking kinematics.

Show solution and calculation

Approach: Convert rotation rate to $\omega_0$ , then use $\alpha=-\frac{\omega_0}{t_s}$ and $\Delta\theta=\frac12\omega_0t_s$ .
This gives $\omega_0=20\pi\,\mathrm{rad\,s^{-1}}$ , $\alpha=-\frac{5\pi}3\,\mathrm{rad\,s^{-2}}$ and $\Delta\theta=120\pi\,\mathrm{rad}$ , or 60 turns.

Check: average rotation rate during uniform braking is 300 revolutions per minute; over 0.2 minutes this also gives 60 turns.

\begin{aligned}\alpha&=-\frac{5\pi}{3}\,\mathrm{rad/s^2}\\\Delta\theta&=120\pi\,\mathrm{rad}\end{aligned}

Error analysis · Application

Now $\alpha_z(t)=(2\,\mathrm{rad\,s^{-3}})t$ and $\omega_0=1\,\mathrm{rad\,s^{-1}}$ . Find angular velocity after three seconds and explain why substituting final acceleration into the constant-acceleration formula is wrong.

Show hint

Acceleration is explicitly time dependent.

Learning goal: Integrate time-dependent angular acceleration.

Show solution and calculation

Approach: Acceleration varies, so use $\omega(t)=\omega_0+\int_0^t\alpha(t^{\prime})\,dt^{\prime}$ .
Integration gives $\omega(t)=1\,\mathrm{rad\,s^{-1}}+(1\,\mathrm{rad\,s^{-3}})t^2$ , hence 10 $\mathrm{rad\,s^{-1}}$ after three seconds.
The incorrect endpoint method gives 19 $\mathrm{rad\,s^{-1}}$ .

Check: differentiating the correct result recovers the specified acceleration.

\omega_z(3\,\mathrm s)=10\,\mathrm{rad/s}

Interpret · Application

An $\omega$ -versus-t line runs from −4 $\mathrm{rad\,s^{-1}}$ at t=0 to +4 $\mathrm{rad\,s^{-1}}$ at $t=4\,\mathrm s$ , with constant angular acceleration. Find net angle, angular travel and reversal time. Explain why zero net angle does not mean no motion.

Show hint

Distinguish negative and positive triangular areas.

Learning goal: Separate signed areas across reversal.

Show solution and calculation

Approach: Zero crossing occurs at two seconds.
Each triangular area has magnitude 4 rad.
Signed areas cancel, while total angular travel is 8 rad.
Rotation is first negative, then positive; only final orientation returns to its initial value.
Zero net angle is a balance, not evidence of rest.

\begin{aligned}\alpha&=\frac{4-(-4)}4=2\,\mathrm{rad\,s^{-2}}\\{}t_s&=2\,\mathrm s\\{}\Delta\theta&=-4+4=0\,\mathrm{rad}\\{}s_\theta&=4+4=8\,\mathrm{rad}\end{aligned}

[1]

Error analysis · Application

$\omega_0=+6\,\mathrm{rad\,s^{-1}}$ and constant $\alpha=-2\,\mathrm{rad\,s^{-2}}$ until $t=5\,\mathrm{s}$ . Someone uses the squared rotation equation and reports +4 $\mathrm{rad\,s^{-1}}$ because the square root is positive. Find the correct final value and explain the lost branch.

Show hint

The linear time equation preserves sign.

Learning goal: Recognize lost direction information in a squared relation.

Show solution and calculation

Approach: The time equation gives −4 $\mathrm{rad\,s^{-1}}$ after five seconds; reversal occurs after three seconds.
The squared endpoint is 16 but does not choose between +4 and −4.
Taking the positive root gives magnitude rather than signed angular velocity.
Additional timing or direction information is required and provided here.

\begin{aligned}\omega(5\,\mathrm s)&=6-2\cdot5=-4\,\mathrm{rad\,s^{-1}}\\{}t_s&=\frac6{2}\,\mathrm s=3\,\mathrm s\\{}\omega^2&=16\,\mathrm{rad^2\,s^{-2}}\end{aligned}

[1]

A constant component is a checkable assumption

The scalar straight line assumes unchanged axis direction and constant $\alpha_z$ over the entire interval. With a changing axis, account for vector kinematics; angular velocity and acceleration magnitudes cannot simply be added as scalars. For time-varying $\alpha$ or a controller with multiple operating phases, calculate piecewise or integrate the actual history.[2]

The existing upper calculation $\omega_0=2\,\mathrm{rad\,s^{-1}}$ , $\alpha=0.50\,\mathrm{rad\,s^{-2}}$ and $t=4\,\mathrm s$ remains correct: $\omega=4\,\mathrm{rad\,s^{-1}}$ . It is a kinematic example, not a general claim about a constant-input-power motor. Constant torque with constant I can yield constant $\alpha$ ; constant mechanical power $P=\tau_z\omega_z$ generally produces a different history.[3]

What to remember

The slope of the angular-velocity line is constant angular acceleration. Its zero crossing marks possible reversal, while its signed area gives net displacement. Add area magnitudes for total travel. Convert rotation rates before substitution and physically justify any assumed acceleration continuing after rest.[1]

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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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Frequently asked questions about Angular velocity under constant angular acceleration

What does Angular velocity under constant angular acceleration describe physically?+

The relation $\omega=\omega_0+\alpha t$ describes a signed angular velocity component about a fixed axis. Choose positive rotation direction first. Constant α is the slope of the line in an ω-t diagram; $\omega_0$ is its initial value. Negative α does not automatically mean braking: with negative ω, speed magnitude can increase further. A zero crossing marks instantaneous rest and, if the model continues, reversal. Area under the line also gives signed angular displacement, rather than total angular travel during reversal. The equation specifies motion but cannot determine driving torque or energy requirements without inertia and a torque balance.

How can I check a calculation involving Angular velocity under constant angular acceleration?+

First check that initial velocity, acceleration and time use the same axis and direction convention. With $\omega_0$ =+12 rad/s, α=−3 rad/s² and t=5 s, the result is ω=−3 rad/s. The zero crossing occurs at 4 s; the drive must maintain negative acceleration afterwards. Independently, areas of 24 rad before and −1.5 rad after the crossing give total signed displacement 22.5 rad. Total angular travel instead is 25.5 rad. The check $\omega^2=\omega_0^2+2\alpha\Delta\theta$ yields $\omega^2=9\,\mathrm{rad^2\,s^{-2}}$ but cannot choose the sign. That requires the time history or known rotation direction.

What assumptions and limits apply to Angular velocity under constant angular acceleration?+

Axis direction remains fixed and signed angular acceleration must be constant throughout the interval. A measured constant average alone does not establish that history. For time-varying α integrate $\omega-\omega_0=\int\alpha\,dt$ ; for multiple operating phases calculate piecewise with appropriate initial values. A passive brake does not automatically become a reverse drive after stopping. Continuing the straight-line model requires actual acceleration to continue. With a changing rotation axis, scalar magnitudes are insufficient and vector kinematics is needed. The equation itself does not assume constant torque, although constant net torque with constant I can justify the constant-acceleration model.

How do I check units and scale for Angular velocity under constant angular acceleration?+

ω and $\omega_0$ are in rad/s, α in rad/s² and t in s. Product αt therefore has the required rad/s unit. Convert rpm using ω=2πn/60; 600 rpm is 20π rad/s, not 600 rad/s. Radian is dimensionless in SI but useful here as an angular indicator. Also check t=0: the result must be $\omega_0$ . At $\alpha=0$ , ω must not change. Doubling time doubles only the change αt, not generally the full final value including initial velocity. Larger negative values can represent a larger speed magnitude, so inspect the sign as well as the numerical size.

When should I combine Angular velocity under constant angular acceleration with another relation?+

For angular displacement integrate angular velocity: $\Delta\theta=\omega_0t+\frac12\alpha t^2$ when α is constant. For total angular travel during reversal integrate |ω| instead. At fixed radius r, $v=r|\omega|$ and accelerations at=rα and ar=rω² give linear quantities. Constant α therefore does not make ar constant. To explain the cause of motion, add $\sum\tau=I\alpha$ for a rigid body about a fixed axis. Energy follows from K= $I\omega^2$ /2 with appropriate I. Matching axis, times and direction convention connects these calculations. Energy depends on the square rather than rotation direction, while tangential acceleration retains the chosen sign.

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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, §10.2: Rotation with Constant Angular Acceleration

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

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

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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 velocity under constant angular acceleration

Constant signed angular acceleration about a fixed axis makes signed angular velocity change linearly with time.

BasicExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

ω

Signed angular velocity component about the fixed axis at time t

rad/s

ω₀

Signed initial component at t=0

rad/s

α

Signed angular acceleration about the same axis, constant over the interval

rad/s²

t

Elapsed time since the initial state

Derivation & background

Exam blueprint

Validity range

Derivation steps

Integrating constant signed angular acceleration gives linear angular velocity.

1Choose a fixed axis and positive rotation direction; $\omega$ and $\alpha$ are signed components.
2Define $\alpha=\frac{d\omega}{dt}$ and initial value $\omega(0)=\omega_0$ .
3Integrate $\omega(t)-\omega_0=\int_0^t\alpha(t^{\prime})\,dt^{\prime}$ .
4Only an $\alpha$ constant throughout the interval may leave the integral, yielding $\alpha t$ .
5Add the initial value: $\omega(t)=\omega_0+\alpha t$ . The added term has the same $\mathrm{rad\,s^{-1}}$ unit as $\omega$ .
6For $\alpha\ne0$ , check the zero $t_s=-\frac{\omega_0}{\alpha}$ . A later reversal requires positive $t_s$ and continuation of the model.
7With $\alpha=0$ , angular velocity stays constant. Varying acceleration requires the integral rather than endpoint acceleration times time.

How do you calculate with Angular velocity under constant angular acceleration?

Here is how to work through a typical Angular velocity under constant angular acceleration (ω = ω₀ + αt) task step by step:

Task

A rotor starts at $\omega_0=2.0\,\mathrm{rad\,s^{-1}}$ . Its angular acceleration stays at $\alpha=+0.50\,\mathrm{rad\,s^{-2}}$ for four seconds. Find the final value.

Solution path

Approach: $\omega=\omega_0+\alpha t$ .
Substitution gives $\omega=4.0\,\mathrm{rad\,s^{-1}}$ .

Check: the 2.0 $\mathrm{rad\,s^{-1}}$ increase over four seconds recovers the given acceleration.

A straight line describes spin-up

Prior knowledge: line slopes and triangle or trapezoid areas. One revolution corresponds to an angle of two pi radians.

A signed quantity about a fixed axis

\omega_z(t)=\omega_{z,0}+\alpha_z(t-t_0)

The upper short form sets t0=0 and suppresses the axis index. Choose the positive rotation direction first.[1]

Reversal separates displacement and travel

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Original example:

\omega(t)=12\,\mathrm{rad\,s^{-1}}-(3\,\mathrm{rad\,s^{-2}})t

. Reversal occurs at 4 s. Areas of +24 rad and −1.5 rad give net displacement 22.5 rad; total travel is 25.5 rad. A drive maintains negative acceleration after reversal.[1]

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Integrate the rate of change

\begin{aligned}\alpha_z&=\frac{d\omega_z}{dt}\\\omega_z(t)-\omega_{z,0}&=\int_{t_0}^{t}\alpha_z(t^{\prime})\,dt^{\prime}\end{aligned}

Angular acceleration is the time derivative of the signed component.[1]

\begin{aligned}\alpha_z&=\frac{\omega_z-\omega_{z,0}}{t-t_0}\\t-t_0&=\frac{\omega_z-\omega_{z,0}}{\alpha_z}\end{aligned}

The first rearrangement needs a nonzero time interval; the second additionally requires $\alpha_z$ ≠0. At $\alpha_z=0$ , $\omega_z$ remains constant.

Distinguish radians and revolutions

\begin{aligned}\omega&=2\pi f=\frac{2\pi n_{\mathrm{rpm}}}{60}\\1\,\mathrm{rev}&=2\pi\,\mathrm{rad}\end{aligned}

n in revolutions per minute is rotational frequency, not directly $\mathrm{rad\,s^{-1}}$ . For signed rotational frequency use the same direction convention.[2]

Area gives displacement, not automatically travel

\Delta\theta=\int_{t_0}^{t}\omega_z(t^{\prime})\,dt^{\prime}=\omega_{z,0}\Delta t+\frac12\alpha_z\Delta t^2=\frac{\omega_{z,0}+\omega_z}{2}\Delta t

For constant $\alpha$ , signed area under the line is a trapezoidal balance.[1]

\Theta_{\mathrm{travel}}=\int_{t_0}^{t}|\omega_z(t^{\prime})|\,dt^{\prime}

Total angular travel adds both directions positively. It equals | $\Delta\theta$ | only when there is no reversal.[2]

Evaluate two different time histories

Controlled rotor spin-up

Given: A rotor accelerates at constant $\alpha$ from +300 to +900 rpm in 8.00 s; fixed axis and the same positive direction.
Find: Find $\alpha$ , angular displacement and number of turns.

Approach: Convert rotational rates to angular velocities and use constant acceleration. $\omega=2\pi\frac{n}{60},\quad\alpha=\frac{\omega_f-\omega_0}{\Delta t}$ . First convert both frequencies to $\mathrm{rad\,s^{-1}}$ .
$\begin{aligned}\omega_0&=10\pi\,\mathrm{rad/s}\\\omega_f&=30\pi\,\mathrm{rad/s}\end{aligned}$
Divide change by time interval.
$\alpha=\frac{30\pi-10\pi}{8}=2.5\pi\approx7.853982\,\mathrm{rad/s^2}$
Mean angular velocity determines area.
$\begin{aligned}\Delta\theta&=\tfrac12(10\pi+30\pi)8=160\pi\,\mathrm{rad}\\{} N&=\frac{\Delta\theta}{2\pi}=80\end{aligned}$

Result: $\alpha\approx7.854\,\mathrm{rad\,s^{-2}}$ and $\Delta\theta\approx502.655\,\mathrm{rad}$ , or 80 complete turns.

Check: Independent formula $\omega_0t+\frac{\alpha t^2}{2}$ also gives 160π. Mean frequency 600 rpm over $\frac8{60}\,\mathrm{min}$ gives 80 turns. Using only ωf·t would incorrectly give 120.

Actively pass through zero and reverse

Given: $\omega_0=+12.0\,\mathrm{rad\,s^{-1}}$ ; a drive maintains $\alpha=-3.00\,\mathrm{rad\,s^{-2}}$ for 5.00 s. The model continues after the instant of rest.
Find: Find final value, zero crossing, signed displacement and total angular travel.

Approach: The signed line may continue through zero into the opposite direction. $\omega(t)=\omega_0+\alpha t,\quad t_s=-\frac{\omega_0}{\alpha}$ . Find final value and zero crossing from the line.
$\begin{aligned}\omega_f&=12-3(5)=-3\,\mathrm{rad/s}\\{} t_s&=\frac{12}{3}=4\,\mathrm s\end{aligned}$
Calculate areas before and after the zero crossing separately.
$\begin{aligned}\Delta\theta_+&=\tfrac12(12)(4)=24\,\mathrm{rad}\\\Delta\theta_-&=-\tfrac12(3)(1)=-1.5\,\mathrm{rad}\end{aligned}$
Subtract signed contributions and add travel magnitudes.
$\begin{aligned}\Delta\theta&=22.5\,\mathrm{rad}\\\Theta_{\mathrm{travel}}&=25.5\,\mathrm{rad}\end{aligned}$

Result: After 5 s, the rotor turns at 3 $\mathrm{rad\,s^{-1}}$ in the reverse direction. Net displacement 22.5 rad differs from total travel 25.5 rad.

Separate tangential and radial acceleration

\begin{aligned}v&=r|\omega_z|\\a_t&=r\alpha_z\\a_r&=r\omega_z^2\end{aligned}

For a point at fixed perpendicular axis distance r, at is signed tangentially; $a_r$ is the inward radial magnitude.[2]

Five exercises with solutions

Calculate · Foundation

$\omega_0=-5\,\mathrm{rad\,s^{-1}}$ , $\alpha=+2\,\mathrm{rad\,s^{-2}}$ and $t=4\,\mathrm s$ . Find $\omega$ and reversal time.

Show hint

Keep signs; separately solve $\omega=0$ .

Learning goal: Find signed final rotation and reversal time.

Show solution and calculation

Approach: Use $\omega=\omega_0+\alpha t$ , with $t_s=-\frac{\omega_0}{\alpha}$ for its zero.
The results are $\omega(4\,\mathrm s)=+3\,\mathrm{rad\,s^{-1}}$ and $t_s=2.5\,\mathrm s$ .

Check: $-5+2\cdot2.5=0$ . Magnitude decreases until the zero and then grows in the positive direction.

\begin{aligned}\omega&=3\,\mathrm{rad/s}\\t_s&=2.5\,\mathrm s\end{aligned}

Calculate · Application

A rotor at +600 rpm stops after 12 s under constant angular acceleration. Find $\alpha$ and $\Delta\theta$ up to stopping.

Show hint

600 rpm equals 20π $\mathrm{rad\,s^{-1}}$ .

Learning goal: Convert rotational speed for braking kinematics.

Show solution and calculation

Approach: Convert rotation rate to $\omega_0$ , then use $\alpha=-\frac{\omega_0}{t_s}$ and $\Delta\theta=\frac12\omega_0t_s$ .
This gives $\omega_0=20\pi\,\mathrm{rad\,s^{-1}}$ , $\alpha=-\frac{5\pi}3\,\mathrm{rad\,s^{-2}}$ and $\Delta\theta=120\pi\,\mathrm{rad}$ , or 60 turns.

Check: average rotation rate during uniform braking is 300 revolutions per minute; over 0.2 minutes this also gives 60 turns.

\begin{aligned}\alpha&=-\frac{5\pi}{3}\,\mathrm{rad/s^2}\\\Delta\theta&=120\pi\,\mathrm{rad}\end{aligned}

Error analysis · Application

Now $\alpha_z(t)=(2\,\mathrm{rad\,s^{-3}})t$ and $\omega_0=1\,\mathrm{rad\,s^{-1}}$ . Find angular velocity after three seconds and explain why substituting final acceleration into the constant-acceleration formula is wrong.

Show hint

Acceleration is explicitly time dependent.

Learning goal: Integrate time-dependent angular acceleration.

Show solution and calculation

Approach: Acceleration varies, so use $\omega(t)=\omega_0+\int_0^t\alpha(t^{\prime})\,dt^{\prime}$ .
Integration gives $\omega(t)=1\,\mathrm{rad\,s^{-1}}+(1\,\mathrm{rad\,s^{-3}})t^2$ , hence 10 $\mathrm{rad\,s^{-1}}$ after three seconds.
The incorrect endpoint method gives 19 $\mathrm{rad\,s^{-1}}$ .

Check: differentiating the correct result recovers the specified acceleration.

\omega_z(3\,\mathrm s)=10\,\mathrm{rad/s}

Interpret · Application

An $\omega$ -versus-t line runs from −4 $\mathrm{rad\,s^{-1}}$ at t=0 to +4 $\mathrm{rad\,s^{-1}}$ at $t=4\,\mathrm s$ , with constant angular acceleration. Find net angle, angular travel and reversal time. Explain why zero net angle does not mean no motion.

Show hint

Distinguish negative and positive triangular areas.

Learning goal: Separate signed areas across reversal.

Show solution and calculation

Approach: Zero crossing occurs at two seconds.
Each triangular area has magnitude 4 rad.
Signed areas cancel, while total angular travel is 8 rad.
Rotation is first negative, then positive; only final orientation returns to its initial value.
Zero net angle is a balance, not evidence of rest.

\begin{aligned}\alpha&=\frac{4-(-4)}4=2\,\mathrm{rad\,s^{-2}}\\{}t_s&=2\,\mathrm s\\{}\Delta\theta&=-4+4=0\,\mathrm{rad}\\{}s_\theta&=4+4=8\,\mathrm{rad}\end{aligned}

[1]

Error analysis · Application

$\omega_0=+6\,\mathrm{rad\,s^{-1}}$ and constant $\alpha=-2\,\mathrm{rad\,s^{-2}}$ until $t=5\,\mathrm{s}$ . Someone uses the squared rotation equation and reports +4 $\mathrm{rad\,s^{-1}}$ because the square root is positive. Find the correct final value and explain the lost branch.

Show hint

The linear time equation preserves sign.

Learning goal: Recognize lost direction information in a squared relation.

Show solution and calculation

Approach: The time equation gives −4 $\mathrm{rad\,s^{-1}}$ after five seconds; reversal occurs after three seconds.
The squared endpoint is 16 but does not choose between +4 and −4.
Taking the positive root gives magnitude rather than signed angular velocity.
Additional timing or direction information is required and provided here.

\begin{aligned}\omega(5\,\mathrm s)&=6-2\cdot5=-4\,\mathrm{rad\,s^{-1}}\\{}t_s&=\frac6{2}\,\mathrm s=3\,\mathrm s\\{}\omega^2&=16\,\mathrm{rad^2\,s^{-2}}\end{aligned}

[1]

A constant component is a checkable assumption

What to remember

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.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal. Clarified revolutions per minute in four text passages.
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 velocity under constant angular acceleration. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/winkelgeschwindigkeit-konstante-winkelbeschleunigung

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Angular velocity under constant angular acceleration

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Rotational motion

Worked example

Applications

How do you calculate with Angular velocity under constant angular acceleration?

A straight line describes spin-up

A signed quantity about a fixed axis

Reversal separates displacement and travel

Integrate the rate of change

Distinguish radians and revolutions

Area gives displacement, not automatically travel

Evaluate two different time histories

Controlled rotor spin-up

Actively pass through zero and reverse

Separate tangential and radial acceleration

Five exercises with solutions

ω0=−5 rad s−1\omega_0=-5\,\mathrm{rad\,s^{-1}}ω0​=−5rads−1, α=+2 rad s−2\alpha=+2\,\mathrm{rad\,s^{-2}}α=+2rads−2 and t=4 st=4\,\mathrm st=4s. Find ω\omegaω and reversal time.

A rotor at +600 rpm stops after 12 s under constant angular acceleration. Find α\alphaα and Δθ\Delta\thetaΔθ up to stopping.

Now αz(t)=(2 rad s−3)t\alpha_z(t)=(2\,\mathrm{rad\,s^{-3}})tαz​(t)=(2rads−3)t and ω0=1 rad s−1\omega_0=1\,\mathrm{rad\,s^{-1}}ω0​=1rads−1. Find angular velocity after three seconds and explain why substituting final acceleration into the constant-acceleration formula is wrong.

An ω\omegaω-versus-t line runs from −4 rad s−1\mathrm{rad\,s^{-1}}rads−1 at t=0 to +4 rad s−1\mathrm{rad\,s^{-1}}rads−1 at t=4 st=4\,\mathrm st=4s, with constant angular acceleration. Find net angle, angular travel and reversal time. Explain why zero net angle does not mean no motion.

A constant component is a checkable assumption

What to remember

Sources for this explanation

Frequently asked questions about Angular velocity under constant angular acceleration

Subject references and further reading

More formulas: Mechanics

Retain Angular velocity under constant angular acceleration for exams

Angular velocity under constant angular acceleration

Variables & units

Angular velocity 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 Angular velocity under constant angular acceleration?

A straight line describes spin-up

A signed quantity about a fixed axis

Reversal separates displacement and travel

Integrate the rate of change

Distinguish radians and revolutions

Area gives displacement, not automatically travel

Evaluate two different time histories

Controlled rotor spin-up

Actively pass through zero and reverse

Separate tangential and radial acceleration

Five exercises with solutions

ω0=−5 rad s−1\omega_0=-5\,\mathrm{rad\,s^{-1}}ω0​=−5rads−1, α=+2 rad s−2\alpha=+2\,\mathrm{rad\,s^{-2}}α=+2rads−2 and t=4 st=4\,\mathrm st=4s. Find ω\omegaω and reversal time.

A rotor at +600 rpm stops after 12 s under constant angular acceleration. Find α\alphaα and Δθ\Delta\thetaΔθ up to stopping.

Now αz(t)=(2 rad s−3)t\alpha_z(t)=(2\,\mathrm{rad\,s^{-3}})tαz​(t)=(2rads−3)t and ω0=1 rad s−1\omega_0=1\,\mathrm{rad\,s^{-1}}ω0​=1rads−1. Find angular velocity after three seconds and explain why substituting final acceleration into the constant-acceleration formula is wrong.

An ω\omegaω-versus-t line runs from −4 rad s−1\mathrm{rad\,s^{-1}}rads−1 at t=0 to +4 rad s−1\mathrm{rad\,s^{-1}}rads−1 at t=4 st=4\,\mathrm st=4s, with constant angular acceleration. Find net angle, angular travel and reversal time. Explain why zero net angle does not mean no motion.

A constant component is a checkable assumption

What to remember

Sources for this explanation

Frequently asked questions about Angular velocity under constant angular acceleration

Subject references and further reading

More formulas: Mechanics

Retain Angular velocity under constant angular acceleration for exams

$\omega_0=-5\,\mathrm{rad\,s^{-1}}$ , $\alpha=+2\,\mathrm{rad\,s^{-2}}$ and $t=4\,\mathrm s$ . Find $\omega$ and reversal time.

A rotor at +600 rpm stops after 12 s under constant angular acceleration. Find $\alpha$ and $\Delta\theta$ up to stopping.

Now $\alpha_z(t)=(2\,\mathrm{rad\,s^{-3}})t$ and $\omega_0=1\,\mathrm{rad\,s^{-1}}$ . Find angular velocity after three seconds and explain why substituting final acceleration into the constant-acceleration formula is wrong.

An $\omega$ -versus-t line runs from −4 $\mathrm{rad\,s^{-1}}$ at t=0 to +4 $\mathrm{rad\,s^{-1}}$ at $t=4\,\mathrm s$ , with constant angular acceleration. Find net angle, angular travel and reversal time. Explain why zero net angle does not mean no motion.

$\omega_0=-5\,\mathrm{rad\,s^{-1}}$ , $\alpha=+2\,\mathrm{rad\,s^{-2}}$ and $t=4\,\mathrm s$ . Find $\omega$ and reversal time.

A rotor at +600 rpm stops after 12 s under constant angular acceleration. Find $\alpha$ and $\Delta\theta$ up to stopping.

Now $\alpha_z(t)=(2\,\mathrm{rad\,s^{-3}})t$ and $\omega_0=1\,\mathrm{rad\,s^{-1}}$ . Find angular velocity after three seconds and explain why substituting final acceleration into the constant-acceleration formula is wrong.

An $\omega$ -versus-t line runs from −4 $\mathrm{rad\,s^{-1}}$ at t=0 to +4 $\mathrm{rad\,s^{-1}}$ at $t=4\,\mathrm s$ , with constant angular acceleration. Find net angle, angular travel and reversal time. Explain why zero net angle does not mean no motion.