Rotational kinematics about a fixed axis

Spin a bicycle wheel and hold it by the axle: every spoke sweeps the same angle in the same time, whether it is near the hub or out at the rim. That single shared angle — not the different distances each point on the wheel travels — is the natural variable for describing rotation, and it is the whole reason rotational kinematics gets its own set of equations rather than borrowing straight-line ones directly. The idea traces a straight line back to Galileo's ramps and Newton's laws of straight-line motion, where displacement, velocity and acceleration are related by three tidy equations valid whenever acceleration is constant. Rotation about a fixed axis — a ceiling fan, a potter's wheel, a spinning bicycle wheel, a CD in a drive — turns out to obey an almost identical trio of equations, just written in angle instead of distance. Swap $s$ for $\theta$ , $v$ for $\omega$ , and $a$ for $\alpha$ , and the whole linear-kinematics toolkit reappears rotated. This article builds that toolkit from scratch: what angular displacement, angular velocity and angular acceleration mean, why radians (not degrees) are the only sane unit to use in these formulas, how the three constant- $\alpha$ equations are derived by the same elementary calculus used for straight-line motion, and how to hop between a point's angular description and its ordinary linear speed and acceleration using $v=\omega r$ . Two worked examples — a wheel speeding up from rest, and a disc slowing to a stop — pin the ideas to numbers you can check by hand, and a clutch of common exam traps (degrees versus radians, tangential versus centripetal acceleration) closes out the piece so you don't lose easy marks on a topic that is, underneath the new symbols, exactly the kinematics you already know.

Key takeaways

Angular position $\theta$ (in radians) locates a rotating body; every particle of a rigid body rotating about a fixed axis sweeps the same $\theta$ in a given time, no matter its distance from the axis.
Angular velocity $\omega = d\theta/dt$ and angular acceleration $\alpha = d\omega/dt$ are the rotational counterparts of linear velocity and acceleration.
For constant angular acceleration, three equations mirror straight-line motion exactly: $\omega=\omega_0+\alpha t$ , $\theta=\omega_0 t+\tfrac12\alpha t^2$ , $\omega^2=\omega_0^2+2\alpha\theta$ .
A point at distance $r$ from the axis has linear (tangential) speed $v=\omega r$ and tangential acceleration $a_t=\alpha r$ , but it also always has centripetal acceleration $a_c=\omega^2 r$ toward the axis — the two are perpendicular components, not the same quantity.
Angular quantities are formally vectors along the axis (right-hand rule); for a genuinely fixed axis it is enough to track their sign, since $+$ or $-$ tells you whether the body speeds up or slows down.
All angle-based formulas demand radians, not degrees — a value in degrees plugged into $v=\omega r$ or the kinematic equations gives a wrong answer even though the arithmetic "runs."
Key idea: for a rigid body turning about a fixed axis, the shared angular variables $\theta,\omega,\alpha$ obey the same constant-acceleration equations as straight-line motion, and convert to any point's linear speed and tangential acceleration through $v=\omega r$ and $a_t=\alpha r$ .

The core insight

Rotation about a fixed axis is governed by angular displacement, velocity and acceleration ( $\theta,\omega,\alpha$ ) that obey straight-line-style constant-acceleration equations and convert to a point's linear speed and tangential acceleration via $v=\omega r$ and $a_t=\alpha r$ , while centripetal acceleration $\omega^2 r$ always acts separately, toward the axis.

Where it fits in the syllabus

This sits inside CBSE Class 11 Physics, unit System of Particles and Rotational Motion, under the syllabus line "comparison of linear and rotational motion." The prescribed content is exactly angular displacement, angular velocity $\omega$ , angular acceleration $\alpha$ , and the rotational kinematic equations for a fixed axis and constant $\alpha$ — nothing about how the rotation got started (that is torque and moment of inertia, covered later in this chapter) and nothing about axes that themselves change direction (gyroscopic motion, precession), which stays out of the Class 11 board syllabus entirely. This article is purely descriptive kinematics: it tells you how to describe and predict rotational motion, building directly on the straight-line kinematics from "Motion in a Straight Line" and the circular-motion ideas ( $v=\omega r$ , centripetal acceleration) from "Motion in a Plane."

The physical picture

Picture a rigid body — a disc, a wheel, a door — turning about an axis that stays fixed in space, like a fan blade about its shaft. Because the body is rigid, every particle in it sweeps through the identical angle $\theta$ in a given stretch of time, even though a particle near the rim covers far more arc length than one near the axis. That shared angle, measured in radians, is therefore the one number that describes the whole body's rotational state at any instant, and it plays exactly the role that position $s$ plays for a particle moving along a line.

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

where: $\omega$ is the angular velocity (rad/s), and $\theta$ is the angular position measured from some fixed reference line (rad). This is the instantaneous rate at which the body's angle is changing — read it exactly as you would read "velocity is the rate of change of position," just with angle in place of distance.

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

where: $\alpha$ is the angular acceleration (rad/s²), the rate at which the angular velocity itself is changing. A fan speeding up from a standstill, or a bicycle wheel slowing after the brakes are applied, both have nonzero $\alpha$ ; a fan spinning at a constant rate has $\alpha = 0$ even though $\omega \ne 0$ . When $\alpha$ is constant, three equations connect $\theta,\omega,\alpha,t$ , and they are structurally identical to the straight-line equations you already know:

\omega = \omega_0 + \alpha t

\theta = \omega_0 t + \tfrac12 \alpha t^2

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

where: $\omega_0$ is the angular velocity at $t=0$ , $\omega$ is the angular velocity at time $t$ , and $\theta$ is the angular displacement swept out in that time $t$ . The mapping to straight-line motion is exact: $\theta\leftrightarrow s$ , $\omega\leftrightarrow v$ , $\alpha\leftrightarrow a$ , $\omega_0\leftrightarrow u$ . Anything you already know about picking the right one of the three linear equations — use the one that skips the variable you don't have — carries over unchanged. A rotating body's angular description connects to the everyday linear motion of any single point on it:

v = \omega r, \qquad a_t = \alpha r, \qquad a_c = \omega^2 r

where: $r$ is the distance of the chosen point from the axis, $v$ is that point's tangential (linear) speed, $a_t$ is its tangential acceleration (along the direction of motion, changing the point's speed), and $a_c$ is its centripetal acceleration (always directed toward the axis, changing the point's direction). Two points on the same rigid body share the same $\omega$ and $\alpha$ at any instant but have different $v$ , $a_t$ , and $a_c$ if they sit at different radii — this is the single most tested idea in the topic.

Direction is handled by treating $\vec\omega$ and $\vec\alpha$ as vectors along the axis, oriented by the right-hand rule: curl the right-hand fingers the way the body turns and the thumb gives $\vec\omega$ 's direction. If $\vec\omega$ and $\vec\alpha$ point the same way the rotation speeds up; opposite ways, it slows down. Because the axis stays fixed throughout this topic, it is enough in practice to track a sign convention (say, anticlockwise positive) and work with the magnitudes $\omega,\alpha$ — full vector algebra for a wandering axis is a Class 12/university topic, out of scope here.

Derivation AGetting the three constant-alpha rotational equations from calculus

The goal is to derive all three constant-angular-acceleration equations, $\omega=\omega_0+\alpha t$ , $\theta=\omega_0t+\tfrac12\alpha t^2$ , and $\omega^2=\omega_0^2+2\alpha\theta$ , starting from nothing but the two defining rates $\omega=d\theta/dt$ and $\alpha=d\omega/dt$ together with the single physical assumption that $\alpha$ is constant in time — exactly the same elementary integration used to obtain the three straight-line kinematic equations, so that a student who trusts $v=u+at$ already trusts this by the identical logic applied to angle instead of distance.

Step 1. Start from the definition of angular acceleration, $\alpha = \dfrac{d\omega}{dt}$ , with $\alpha$ held constant. Separate and integrate both sides from $t=0$ (where $\omega=\omega_0$ ) to a general time $t$ (where $\omega=\omega$ ): $\displaystyle\int_{\omega_0}^{\omega} d\omega = \alpha\int_0^t dt$ , which gives $\omega - \omega_0 = \alpha t$ , or $\omega = \omega_0 + \alpha t$ . This is the rotational twin of $v=u+at$ , obtained by exactly the same one-line integration.

Step 2. Substitute this expression for $\omega$ into the definition $\omega = \dfrac{d\theta}{dt}$ , giving $\dfrac{d\theta}{dt} = \omega_0 + \alpha t$ . Integrate both sides from $t=0$ (where $\theta=0$ , taking the starting angle as the reference) to a general $t$ : $\displaystyle\int_0^{\theta} d\theta = \int_0^t (\omega_0 + \alpha t)\,dt$ , which gives $\theta = \omega_0 t + \tfrac12 \alpha t^2$ . This is the rotational twin of $s = ut + \tfrac12 at^2$ .

Step 3. Eliminate the time variable $t$ between the two results to get a relation that doesn't need $t$ at all. From Step 1, $t = \dfrac{\omega-\omega_0}{\alpha}$ . Substitute into the Step 2 result: $\theta = \omega_0\left(\dfrac{\omega-\omega_0}{\alpha}\right) + \tfrac12\alpha\left(\dfrac{\omega-\omega_0}{\alpha}\right)^2$ .

Step 4. Simplify the algebra. Multiply through by $\alpha$ : $\alpha\theta = \omega_0(\omega-\omega_0) + \tfrac12\dfrac{(\omega-\omega_0)^2}{\alpha}\cdot\alpha = \omega_0\omega - \omega_0^2 + \tfrac12(\omega-\omega_0)^2$ . Expanding $\tfrac12(\omega-\omega_0)^2 = \tfrac12\omega^2 - \omega_0\omega + \tfrac12\omega_0^2$ and collecting terms, the $\omega_0\omega$ terms cancel and the $\omega_0^2$ terms combine to give $\alpha\theta = \tfrac12\omega^2 - \tfrac12\omega_0^2$ , so $\omega^2 = \omega_0^2 + 2\alpha\theta$ — the rotational twin of $v^2=u^2+2as$ . Verified: setting $\omega_0=0,\alpha=5\ \text{rad/s}^2,t=6\ \text{s}$ gives $\omega=30\ \text{rad/s}$ and $\theta=90\ \text{rad}$ from Steps 1 and 2, and substituting into the Step 4 result, $30^2 = 0^2+2(5)(90) = 900$ , confirms the third equation is consistent with the first two, matching the Board-level worked example below.

Worked examples

Board level - wheel speeding up from rest

A wheel starts from rest and undergoes constant angular acceleration, reaching an angular velocity of $30\ \text{rad/s}$ after $6\ \text{s}$ . Find (a) the angular acceleration, and (b) the total angle turned through in this time.

Step 1. Identify what is given: $\omega_0 = 0$ (starts from rest), $\omega = 30\ \text{rad/s}$ , $t = 6\ \text{s}$ , and $\alpha$ constant. The equation connecting exactly these four quantities is $\omega = \omega_0 + \alpha t$ .

Step 2. Solve for angular acceleration: $\alpha = \dfrac{\omega-\omega_0}{t} = \dfrac{30-0}{6} = 5\ \text{rad/s}^2$ .

Step 3. Find the angle turned using $\theta = \omega_0 t + \tfrac12\alpha t^2 = 0 + \tfrac12(5)(6)^2 = \tfrac12(5)(36) = 90\ \text{rad}$ .

Step 4. Sanity-check with the third equation: $\omega^2 = \omega_0^2+2\alpha\theta \Rightarrow 30^2 = 0 + 2(5)(90) = 900$ , and $30^2=900$ — consistent. So the wheel's angular acceleration is $5\ \text{rad/s}^2$ , and it turns through $90\ \text{rad}$ , about $90/2\pi \approx 14.3$ revolutions.

Competitive level - decelerating disc, linear speed at the rim

A rotating disc of radius $0.25\ \text{m}$ slows down uniformly from an angular velocity of $40\ \text{rad/s}$ to $10\ \text{rad/s}$ while turning through $110\ \text{rad}$ . Find (a) the angular acceleration (deceleration), (b) the time taken for this change, and (c) the linear (tangential) speed and centripetal acceleration of a point on the rim when $\omega = 10\ \text{rad/s}$ .

Step 1. The three known angular quantities are $\omega_0=40\ \text{rad/s}$ , $\omega=10\ \text{rad/s}$ , $\theta=110\ \text{rad}$ — no time given, so use $\omega^2=\omega_0^2+2\alpha\theta$ : $10^2 = 40^2 + 2\alpha(110)$ , i.e. $100 = 1600+220\alpha$ .

Step 2. Solve: $\alpha = \dfrac{100-1600}{220} = \dfrac{-1500}{220} \approx -6.82\ \text{rad/s}^2$ . The negative sign shows the disc is decelerating, i.e. $\vec\alpha$ points opposite to $\vec\omega$ .

Step 3. Find the time from $\omega=\omega_0+\alpha t$ : $t = \dfrac{\omega-\omega_0}{\alpha} = \dfrac{10-40}{-6.82} \approx \dfrac{-30}{-6.82} \approx 4.40\ \text{s}$ .

Step 4. Convert to linear quantities at the rim, $r=0.25\ \text{m}$ , when $\omega=10\ \text{rad/s}$ : tangential speed $v=\omega r = (10)(0.25)=2.5\ \text{m/s}$ ; centripetal acceleration $a_c = \omega^2 r = (10)^2(0.25) = 25\ \text{m/s}^2$ toward the axis; tangential acceleration $a_t = \alpha r \approx (-6.82)(0.25) \approx -1.71\ \text{m/s}^2$ , i.e. slowing the rim point down along its direction of motion. So $\alpha \approx -6.82\ \text{rad/s}^2$ , $t\approx 4.40\ \text{s}$ , and at $\omega=10\ \text{rad/s}$ the rim point has linear speed $2.5\ \text{m/s}$ , centripetal acceleration $25\ \text{m/s}^2$ , and tangential (deceleration) magnitude about $1.71\ \text{m/s}^2$ — note $a_c$ dwarfs $a_t$ here, which is typical for a disc that is still spinning fast even while slowing. Timing note: in an exam, notice immediately that no time is given in the problem statement — that is the signal to reach for $\omega^2=\omega_0^2+2\alpha\theta$ first, not $\omega=\omega_0+\alpha t$ ; solving for $t$ only afterward, once $\alpha$ is known, saves a wasted attempt at simultaneous equations.

Common traps

Mixing degrees and radians: $v=\omega r$ and all three kinematic equations require $\theta,\omega,\alpha$ in radians, rad/s, rad/s² respectively — plugging in degrees gives a number that looks plausible but is simply wrong.
Forgetting that different points on the same rigid rotating body have different linear speeds ( $v=\omega r$ ) even though every point shares identical $\omega$ and $\alpha$ at that instant — a common error is to quote one "the" speed for the whole wheel.
Treating tangential acceleration ( $a_t=\alpha r$ , along the direction of motion, changes speed) and centripetal acceleration ( $a_c=\omega^2 r$ , toward the axis, changes direction) as the same thing, or forgetting one of them exists — they are perpendicular components of a point's total acceleration and, as the competitive example shows, can differ hugely in size.
Applying the constant- $\alpha$ equations to a situation where $\alpha$ is visibly changing with time — these three formulas, like their linear counterparts, hold only for uniform angular acceleration.
Picking the wrong one of the three equations first: scan for which of $\theta,\omega,\omega_0,\alpha,t$ is missing from the problem statement and choose the equation that omits exactly that variable, rather than trying to solve two equations simultaneously.

You should be able to…

Define angular displacement $\theta$ , angular velocity $\omega$ , and angular acceleration $\alpha$ for rotation about a fixed axis, and state their SI units.
Derive and apply the three constant-angular-acceleration equations, choosing the correct one based on which variable is missing from a problem.
Convert between angular and linear quantities at a point using $v=\omega r$ and $a_t=\alpha r$ , given the point's radius from the axis.
Compute centripetal acceleration $a_c=\omega^2 r$ for a point on a rotating body and distinguish it clearly from tangential acceleration.
Explain why radians (not degrees) are required in every formula of this topic, and convert between the two on demand.
Solve numerically for any one of $\theta,\omega,\omega_0,\alpha,t$ given the other three, for a body undergoing constant angular acceleration.

Ask the tutor: Want a walkthrough of a problem where the disc starts with nonzero $\omega_0$ and speeds up, or a comparison table drilling the linear-to-angular symbol swaps — ask the AI tutor for more rotational kinematics practice.

Why do we even need radians — why can't I just use degrees throughout?

Radians are not an arbitrary unit choice; a radian is defined so that arc length equals radius times angle, $s = r\theta$ , with no extra conversion factor. That is exactly what makes $v=\omega r$ and the kinematic equations work without a stray constant. If you use degrees, the relation between arc length and angle picks up a factor of $\pi/180$ , and every formula in this topic silently breaks unless you convert back to radians first. So always convert degrees to radians ( $\times \pi/180$ ) or revolutions to radians ( $\times 2\pi$ ) before using any formula here.

If every point on the wheel has the same omega, why do their speeds differ?

Angular velocity measures how fast the angle is changing, and that is genuinely shared by every particle of a rigid body — a spoke near the hub and the tyre both sweep, say, 5 radians every second. But linear speed measures how fast a point covers distance along its circular path, and a point farther from the axis has to cover a longer arc to sweep the same angle in the same time. That is precisely what $v=\omega r$ captures: same $\omega$ , but $v$ scales up with $r$ .

What happens if the axis itself starts to wobble or change direction?

That is a genuinely harder problem — it needs $\vec\omega$ and $\vec\alpha$ treated as full three-dimensional vectors that can themselves rotate, and it opens the door to gyroscopic effects and precession. None of that is in the Class 11 CBSE syllabus; here the axis is always fixed in space, so you only ever need to track the sign of the scalar $\omega$ and $\alpha$ along that one fixed direction. Precession and rigid-body vector dynamics are topics for later study, not this chapter.

In the competitive-level example, why is the centripetal acceleration so much bigger than the tangential one?

Centripetal acceleration grows as $\omega^2 r$ , so it stays large as long as the disc is still spinning reasonably fast, even while it is decelerating — here $\omega=10\ \text{rad/s}$ still gives $a_c=25\ \text{m/s}^2$ . Tangential acceleration depends only on $\alpha r$ , a much smaller number in this problem ( $\approx -1.71\ \text{m/s}^2$ ), because the deceleration itself is comparatively gentle. Physically: the rim point is still moving fast in a tight circle (large $a_c$ , changing its direction quickly) while only slowly losing speed (small $a_t$ ). This gap is completely normal and is worth checking numerically rather than assuming the two accelerations are comparable.