Equations of Rotational Motion

In short

For constant angular acceleration, three equations describe all rotational motion: \omega = \omega_0 + \alpha t, \theta = \omega_0 t + \frac{1}{2}\alpha t^2, and \omega^2 = \omega_0^2 + 2\alpha\theta. These are the exact rotational twins of the linear kinematic equations v = u + at, s = ut + \frac{1}{2}at^2, and v^2 = u^2 + 2as — with angle \theta replacing displacement s, angular velocity \omega replacing velocity v, and angular acceleration \alpha replacing acceleration a. The number of complete revolutions is N = \theta / 2\pi.

Switch off a ceiling fan. The blades do not stop instantly — they coast, slower and slower, each revolution taking a little longer than the last, until they finally come to rest. If you counted the revolutions from the moment you hit the switch, you could predict when the fan would stop and how many turns it would make. The physics behind that prediction is exactly the algebra you already know from straight-line kinematics — just applied to rotation.

A car braking on a highway uses v = u + at and s = ut + \frac{1}{2}at^2. The ceiling fan does the same thing, but in a circle. Replace distance with angle, speed with angular speed, and acceleration with angular acceleration, and every kinematic equation you learned for straight-line motion works again, letter for letter. That is not a coincidence — it is the same calculus applied to a different geometry.

The linear–rotational dictionary

Every quantity in straight-line kinematics has a rotational counterpart:

Linear (straight-line)	Rotational (about a fixed axis)	SI unit
Displacement s	Angular displacement \theta	rad
Initial velocity u	Initial angular velocity \omega_0	rad/s
Final velocity v	Final angular velocity \omega	rad/s
Acceleration a	Angular acceleration \alpha	rad/s²
Time t	Time t	s

The structure is identical. The linear equations are:

v = u + at \qquad s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as

The rotational equations are:

\omega = \omega_0 + \alpha t \qquad \theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 \qquad \omega^2 = \omega_0^2 + 2\alpha\theta

If you can solve one set, you can solve the other. The algebra is the same because the definitions are the same: angular velocity is to angular displacement what velocity is to displacement, and angular acceleration is to angular velocity what acceleration is to velocity. The calculus — integrating acceleration to get velocity, integrating velocity to get displacement — produces the same algebraic forms regardless of whether the motion is along a line or around an axis.

A point P on the rim of a spinning wheel. The angular displacement $\theta$ is the angle swept from a reference direction. Angular velocity $\omega = d\theta/dt$ tells you how fast the wheel turns. Angular acceleration $\alpha = d\omega/dt$ tells you how fast $\omega$ changes — the same hierarchy as $s$, $v$, $a$ in linear motion.

Deriving the three equations

Take a rigid body rotating about a fixed axis with constant angular acceleration \alpha. At time t = 0 its angular velocity is \omega_0 and the angular displacement is zero. You want to find \omega(t) and \theta(t).

Assumption: The angular acceleration \alpha is constant throughout the motion. This is the rotational equivalent of the constant-acceleration assumption in linear kinematics. It is a model — a ceiling fan's air resistance changes with speed, a motor's torque varies with RPM — but it is the right starting point, and it is accurate enough for a wide range of real problems: a grinding wheel spun up by a steady motor, a flywheel slowed by constant friction, a merry-go-round pushed at a steady rate.

Equation 1: \omega = \omega_0 + \alpha t

Angular acceleration is defined as the rate of change of angular velocity:

\alpha = \frac{d\omega}{dt}

Why: this is the definition of angular acceleration — exactly as a = dv/dt defines linear acceleration.

Since \alpha is constant, rearrange and integrate:

d\omega = \alpha\,dt

\int_{\omega_0}^{\omega} d\omega' = \int_0^t \alpha\,dt'

\omega - \omega_0 = \alpha t

\boxed{\omega = \omega_0 + \alpha t} \tag{1}

Why: with constant \alpha, angular velocity changes at a steady rate. After time t, it has gained (or lost, if \alpha is negative) exactly \alpha t of angular velocity. Compare with v = u + at — the structure is identical.

This equation answers: what is the angular velocity at time t? If \alpha and \omega_0 have opposite signs, the body is decelerating — it spins slower and slower until it stops (when \omega = 0) or reverses.

Equation 2: \theta = \omega_0 t + \frac{1}{2}\alpha t^2

Angular velocity is the rate of change of angular displacement:

\omega = \frac{d\theta}{dt}

Substitute \omega = \omega_0 + \alpha t from Equation (1):

\frac{d\theta}{dt} = \omega_0 + \alpha t

Why: replacing \omega with its expression in terms of t turns this into a single equation in \theta and t, ready to integrate.

Integrate both sides from t = 0 (where \theta = 0) to time t:

\int_0^{\theta} d\theta' = \int_0^t (\omega_0 + \alpha t')\,dt'

\theta = \omega_0 t + \frac{1}{2}\alpha t^2

\boxed{\theta = \omega_0 t + \frac{1}{2}\alpha t^2} \tag{2}

Why: the first term (\omega_0 t) is the angle the body would sweep at its initial angular velocity alone. The second term (\frac{1}{2}\alpha t^2) is the extra angle gained (or lost) due to the acceleration. Together they give the total angular displacement. Compare with s = ut + \frac{1}{2}at^2.

This equation answers: how much angle has the body swept by time t? For a body starting from rest (\omega_0 = 0), it simplifies to \theta = \frac{1}{2}\alpha t^2 — angular displacement grows as the square of time, exactly like distance does for a car accelerating from rest.

Equation 3: \omega^2 = \omega_0^2 + 2\alpha\theta

This equation eliminates time. From Equation (1), solve for t:

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

Why: isolating t lets you substitute into Equation (2) to get a relationship between \omega and \theta that does not involve time — useful when the problem gives you angles but not time.

Substitute into Equation (2):

\theta = \omega_0 \cdot \frac{\omega - \omega_0}{\alpha} + \frac{1}{2}\alpha \left(\frac{\omega - \omega_0}{\alpha}\right)^2

\theta = \frac{\omega_0(\omega - \omega_0)}{\alpha} + \frac{(\omega - \omega_0)^2}{2\alpha}

Why: expand both terms so you can combine them over the common denominator 2\alpha.

Multiply both sides by 2\alpha:

2\alpha\theta = 2\omega_0(\omega - \omega_0) + (\omega - \omega_0)^2

2\alpha\theta = 2\omega_0\omega - 2\omega_0^2 + \omega^2 - 2\omega_0\omega + \omega_0^2

2\alpha\theta = \omega^2 - \omega_0^2

\boxed{\omega^2 = \omega_0^2 + 2\alpha\theta} \tag{3}

Why: the 2\omega_0\omega terms cancel, leaving a clean relationship between final angular velocity and angular displacement. This is the rotational twin of v^2 = u^2 + 2as. Reach for it when the problem gives you revolutions or angle instead of time.

When to use which equation

Known quantities	Unknown	Use
\omega_0, \alpha, t	\omega	Equation (1)
\omega_0, \alpha, t	\theta	Equation (2)
\omega_0, \alpha, \theta	\omega	Equation (3)
\omega_0, \omega, t	\alpha	Equation (1)
\omega_0, \omega, t	\theta	Equation (2) or use \theta = \frac{1}{2}(\omega_0 + \omega)t

The last entry — \theta = \frac{1}{2}(\omega_0 + \omega)t — is the rotational version of s = \frac{1}{2}(u + v)t. It comes from recognising that for constant acceleration, the average angular velocity is \frac{1}{2}(\omega_0 + \omega).

Watching it happen: a wheel from rest

The simulation below shows a point on the rim of a wheel that starts from rest and accelerates at \alpha = \pi/2 rad/s². Watch the point: it starts slowly and speeds up with every revolution. In 4 seconds, it completes exactly two full revolutions — because \theta = \frac{1}{2} \times \frac{\pi}{2} \times 16 = 4\pi rad.

A point on the rim of a wheel accelerating from rest ($\alpha = \pi/2$ rad/s²). The trail is bunched early (the point moves slowly) and spread later (it moves fast) — visual proof that $\omega$ is increasing. The wheel completes exactly two revolutions in 4 seconds. Click replay to watch again.

Explore the ω–t graph

For constant angular acceleration starting from rest, the \omega–t graph is a straight line through the origin with slope \alpha. The angular displacement \theta equals the area under this line — and since the line grows with time, \theta grows quadratically.

Drag the point along the line to see how \omega and \theta change with time.

Drag the red point along the $\omega$–$t$ line ($\alpha = 2$ rad/s², starting from rest). Angular velocity $\omega$ grows linearly, but angular displacement $\theta$ grows quadratically — the wheel sweeps more angle in each successive second. At $t = 3$ s: $\omega = 6$ rad/s, $\theta = 9$ rad $\approx 1.4$ revolutions.

Revolutions and angular displacement

One complete revolution is 2\pi radians. If a body rotates through a total angular displacement \theta (in radians), the number of revolutions is:

N = \frac{\theta}{2\pi}

Going the other way: if a body makes N revolutions, its angular displacement is \theta = 2\pi N.

This conversion appears in almost every rotational kinematics problem. The equations need \theta in radians, but the problem statement usually gives revolutions ("the fan makes 40 revolutions before stopping") or RPM ("the wheel spins at 300 RPM"). Convert first, apply the equations, then convert back.

A useful shortcut for RPM: to convert from revolutions per minute to rad/s, multiply by 2\pi/60:

\omega \text{ (rad/s)} = \text{RPM} \times \frac{2\pi}{60}

Worked examples

Example 1: Ceiling fan slowing down

A ceiling fan spinning at 300 RPM is switched off. It decelerates uniformly and comes to rest after 25 seconds. Find (a) the angular deceleration and (b) the total number of revolutions before it stops.

The fan decelerates uniformly from 300 RPM to rest in 25 seconds.

Step 1. Convert RPM to rad/s.

\omega_0 = 300 \times \frac{2\pi}{60} = 10\pi \text{ rad/s} \approx 31.4 \text{ rad/s}

Why: the equations use rad/s, not RPM. Multiply RPM by 2\pi/60 to convert.

Step 2. Find \alpha using Equation (1). The fan stops, so \omega = 0.

0 = 10\pi + \alpha \times 25

\alpha = -\frac{10\pi}{25} = -\frac{2\pi}{5} \text{ rad/s}^2 \approx -1.26 \text{ rad/s}^2

Why: the negative sign confirms the fan is decelerating — the angular acceleration opposes the angular velocity, slowing the fan down.

Step 3. Find \theta using Equation (2).

\theta = 10\pi \times 25 + \frac{1}{2}\left(-\frac{2\pi}{5}\right) \times 625

\theta = 250\pi - 125\pi = 125\pi \text{ rad}

Why: the first term (250\pi rad) is the angle the fan would sweep in 25 s if it kept spinning at 10\pi rad/s without decelerating. The second term (-125\pi rad) subtracts the angle it did not sweep because it was slowing down. The difference is the actual angle swept.

Step 4. Convert to revolutions.

N = \frac{125\pi}{2\pi} = 62.5 \text{ revolutions}

Why: divide the total angle in radians by 2\pi to get the number of complete turns.

Result: The angular deceleration is 2\pi/5 \approx 1.26 rad/s². The fan makes 62.5 revolutions before stopping.

What this shows: A ceiling fan at 300 RPM takes 62.5 turns to stop — over a minute of coasting. You could also reach the same \theta using Equation (3): 0 = (10\pi)^2 + 2(-2\pi/5)\theta, giving \theta = 125\pi rad. Both routes agree, as they must.

Example 2: Potter's wheel starting up

A potter in Khurja kicks a pottery wheel from rest. The wheel accelerates uniformly at 3 rad/s² for 4 seconds. Find (a) the angular velocity after 4 seconds, (b) the number of revolutions completed, and (c) the angular velocity after exactly 2 revolutions.

The potter's wheel accelerates from rest to 12 rad/s in 4 seconds.

Step 1. Find \omega at t = 4 s using Equation (1).

\omega = 0 + 3 \times 4 = 12 \text{ rad/s}

Why: starting from rest (\omega_0 = 0), the angular velocity after 4 s is simply \alpha t. Converting: 12 \times 60/(2\pi) \approx 115 RPM.

Step 2. Find \theta at t = 4 s using Equation (2).

\theta = 0 + \frac{1}{2} \times 3 \times 16 = 24 \text{ rad}

N = \frac{24}{2\pi} = \frac{12}{\pi} \approx 3.82 \text{ revolutions}

Why: with \omega_0 = 0, the displacement is just \frac{1}{2}\alpha t^2. In 4 seconds of uniform acceleration, the wheel completes nearly 4 full turns.

Step 3. Find \omega after exactly 2 revolutions using Equation (3).

Two revolutions means \theta = 2 \times 2\pi = 4\pi rad.

\omega^2 = 0 + 2 \times 3 \times 4\pi = 24\pi

\omega = \sqrt{24\pi} \approx \sqrt{75.4} \approx 8.68 \text{ rad/s}

Why: Equation (3) is the right tool because the problem gives angle, not time. After 2 revolutions the wheel is at about 8.7 rad/s — less than the final 12 rad/s at t = 4 s, because 2 revolutions occur before the full 4 seconds are up.

Result: (a) \omega = 12 rad/s after 4 s. (b) 3.82 revolutions in that time. (c) After exactly 2 revolutions, \omega = \sqrt{24\pi} \approx 8.68 rad/s.

What this shows: Equation (3) is the tool to reach for when the problem gives you revolutions instead of time. The potter's wheel reaches 2 revolutions before the 4-second mark, so its angular velocity at that point (8.68 rad/s) is less than the final 12 rad/s — it is still accelerating.

Common confusions

"The equations only work for full revolutions." No — \theta can be any real number. The equations work for half a revolution, a quarter turn, or 17.3 radians. You convert between radians and revolutions only because the problem statement uses revolutions; the physics is always in radians.
"Angular acceleration must be in RPM per second." The equations require \alpha in rad/s². Using RPM without converting is a common source of wrong answers. If a fan decelerates from 300 RPM to 0 in 25 seconds, \alpha is not 300/25 = 12 "RPM/s" — it is (10\pi)/25 = 2\pi/5 rad/s². Always convert to radians first.
"Negative \alpha always means slowing down." Only if \omega is positive. If a body spins clockwise (\omega < 0) and \alpha is also negative, the body is speeding up in the clockwise direction. What matters is whether \alpha and \omega have the same sign (speeding up) or opposite signs (slowing down).
"These equations describe circular motion." They describe rotation — any rigid body spinning about a fixed axis. A disc, a turbine blade, a door swinging on its hinges, a cricket ball spinning in the air — all use these equations. Uniform circular motion (constant speed on a circle) is a special case where \alpha = 0 and \omega is constant.
"Equation (3) gives you information that Equations (1) and (2) don't." It does not. Equation (3) is derived by eliminating t from (1) and (2). It is a convenience — you could always solve (1) for t and substitute into (2) — but it saves time when the problem gives angle instead of time.
"You need to know the radius to use these equations." The three kinematic equations involve only angular quantities (\theta, \omega, \alpha, t). The radius R of the rotating body does not appear. If the problem asks for the linear speed of a point on the rim, you use v = R\omega after solving the rotational problem — but the rotational kinematics themselves are independent of R.

If you came here to use the three rotational kinematic equations and solve problems, you have what you need. What follows is for readers who want to see what happens when angular acceleration is not constant, and how these equations connect to the dynamics of rotation.

When angular acceleration is not constant

The three boxed equations assume \alpha is constant. Real rotating systems rarely maintain perfectly constant angular acceleration — air resistance on a fan depends on speed, friction in a bearing changes with temperature, and a motor's torque varies with RPM.

When \alpha varies with time, the definitions are still valid:

\omega(t) = \omega_0 + \int_0^t \alpha(t')\,dt'

\theta(t) = \int_0^t \omega(t')\,dt'

These are the general forms. The constant-\alpha equations are what you get when you pull \alpha out of the integral.

For example, suppose a fan's angular deceleration increases linearly with time: \alpha(t) = -bt for some constant b > 0. Then:

\omega(t) = \omega_0 + \int_0^t (-bt')\,dt' = \omega_0 - \frac{1}{2}bt^2

\theta(t) = \int_0^t \left(\omega_0 - \frac{1}{2}bt'^2\right)dt' = \omega_0 t - \frac{1}{6}bt^3

The structure is the same — integrate the angular acceleration to get velocity, integrate velocity to get displacement — but the results no longer follow the clean quadratic forms. JEE Advanced occasionally tests this: "given \alpha as a function of t or \theta, find \omega and \theta." The approach is always the same: go back to the definitions and integrate.

When \alpha depends on \theta (not t), use the chain rule:

\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta} \cdot \frac{d\theta}{dt} = \omega\frac{d\omega}{d\theta}

This converts the problem into an ODE in \omega and \theta:

\alpha(\theta) = \omega\frac{d\omega}{d\theta} \implies \omega\,d\omega = \alpha(\theta)\,d\theta

Integrate both sides to find \omega as a function of \theta. This is the rotational analogue of the v\,dv = a(s)\,ds trick from linear kinematics.

Connection to the rotational work-energy theorem

Equation (3), \omega^2 = \omega_0^2 + 2\alpha\theta, has a deeper meaning. Multiply both sides by \frac{1}{2}I (where I is the moment of inertia):

\frac{1}{2}I\omega^2 - \frac{1}{2}I\omega_0^2 = I\alpha\theta = \tau\theta

The left side is the change in rotational kinetic energy. The right side is the work done by the constant torque \tau = I\alpha through angle \theta. This is the rotational work-energy theorem — the rotational twin of \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = Fs. Equation (3) is not just a kinematic identity; it encodes energy conservation for rotational motion. The connection is explored further in the articles on torque and moment of inertia.

Rolling without slipping

A rolling wheel both rotates and translates. The kinematic equations derived here describe only the rotation about the axis. For the full picture — the centre of mass moving along a surface while the wheel turns — you need the rolling constraint v_{\text{cm}} = R\omega (for rolling without slipping), which links linear and rotational kinematics into a single coupled system. That topic is covered in Rotation with Translation.

Where this leads next

Angular Displacement, Velocity, and Acceleration — the definitions behind the symbols \theta, \omega, and \alpha.
Rotation with Translation — what happens when a body rotates and its centre of mass moves simultaneously.
Moment of Inertia — Definition and Physical Meaning — the rotational equivalent of mass, and the bridge from kinematics to dynamics.
Torque — the rotational equivalent of force, and the cause of angular acceleration.
Kinematics Equations for Constant Acceleration — the linear originals that these rotational equations mirror.