Kinematics of Rotational Motion

Angular displacement θ

Consider a rigid body rotating about a fixed axis — take it as the $z$-axis. NCERT §6.6 instructs us to pick any particle $P$ of the body lying in the plane of motion (the $x$–$y$ plane). Its angular displacement $\theta$, measured from a fixed reference direction, is the angular displacement of the whole body, because in pure rotation every particle turns through the same angle in the same time. We write $\theta_0$ for the angular displacement at $t=0$.

Angular displacement is measured in radians. The radian is defined through arc length: a particle at perpendicular distance $r$ from the axis that sweeps an arc of length $s$ has turned through $\theta = s/r$. One complete revolution corresponds to an arc equal to the full circumference $2\pi r$, so one revolution is $2\pi$ radians. This single fact — $1~\text{rev} = 2\pi~\text{rad}$ — is the conversion that NEET repeatedly tests.

Angular velocity ω = dθ/dt

Let the particle move from $P$ to $P'$ in time $\Delta t$, sweeping angle $\Delta\theta$. The average angular velocity over that interval is $\Delta\theta/\Delta t$. As $\Delta t \to 0$ this ratio approaches the instantaneous angular velocity, denoted $\omega$:

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

The same $\omega$ describes every particle of the body at a given instant — this is precisely what we mean by pure rotation: NCERT characterises it as "all parts of the body having the same angular velocity at any instant of time." We therefore call $\omega$ the angular velocity of the whole body, not of a single particle. Its SI unit is the radian per second ($\text{rad s}^{-1}$).

Strictly, $\omega$ is a vector. For rotation about a fixed axis it lies along the axis of rotation, in the direction a right-handed screw would advance if turned with the body. Because the axis is fixed, the direction of $\omega$ does not change with time — only its magnitude can. This is why, for fixed-axis problems, NCERT notes "there is no need to treat angular velocity as a vector" and we may work with the scalar $\omega = d\theta/dt$.

Angular acceleration α = dω/dt

By exact analogy with linear acceleration as the time rate of change of velocity, NCERT §6.6.1 defines angular acceleration $\alpha$ as the time rate of change of angular velocity:

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

Its SI unit is the radian per second squared ($\text{rad s}^{-2}$). For rotation about a fixed axis the direction of $\omega$ is fixed, so the direction of $\alpha$ is fixed too — it lies along the axis of rotation. When the body is speeding up, $\alpha$ points the same way as $\omega$; when the body is slowing down, $\alpha$ points opposite to $\omega$ (so its scalar value is taken negative in the equations). The vector definition then reduces to the scalar equation $\alpha = d\omega/dt$.

Linear–angular relations

Each particle's linear (translational) quantities are tied to the body's angular quantities through its perpendicular distance $r$ from the axis. From $s = r\theta$, differentiating once gives the speed and twice gives the rate of change of speed:

Two consequences matter for NEET. First, $v$, $a_t$ and $a_c$ all scale with $r$, so different particles of the same rigid body move at different linear speeds even though they share one $\omega$; particles on the axis ($r=0$) are stationary. Second, even a body spinning at constant $\omega$ has a non-zero centripetal acceleration $a_c=\omega^2 r$ on every off-axis particle, while its tangential acceleration is zero. Only when $\alpha \neq 0$ does $a_t = \alpha r$ appear.

The rotational kinematic equations

When the angular acceleration $\alpha$ is constant (uniform), the angular variables obey three equations that mirror the linear SUVAT set term for term. NCERT §6.10 states them as Eqs. (6.36)–(6.38):

$$\omega = \omega_0 + \alpha t \qquad\text{(6.36)}$$

$$\theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2 \qquad\text{(6.37)}$$

$$\omega^2 = \omega_0^{\,2} + 2\alpha(\theta - \theta_0) \qquad\text{(6.38)}$$

Here $\omega_0$ is the initial angular velocity and $\theta_0$ the initial angular displacement, both at $t=0$. The first equation can be obtained from first principles by integrating $\alpha = d\omega/dt = \text{constant}$ (NCERT Example 6.10); integrating $\omega = d\theta/dt$ then yields the second, and eliminating $t$ between them gives the third. The derivation is identical in form to the linear case — only the symbols change.

A caution that NEET exploits: these three equations hold only for constant $\alpha$, exactly as the linear SUVAT equations require constant $a$. If $\alpha$ varies with time, you must return to the definitions $\omega = d\theta/dt$ and $\alpha = d\omega/dt$ and integrate directly. The phrase "uniform angular acceleration" in a problem is the green light to use the boxed equations.

The linear ↔ angular analogy

NCERT Table 6.2 lays out the correspondence between translational motion and rotation about a fixed axis. Reading it as a substitution dictionary — $x\to\theta$, $v\to\omega$, $a\to\alpha$ — lets you convert any linear result into its rotational twin without re-deriving it. The grid below collects the kinematic rows together with the dynamic analogues you will meet next.

Worked examples

A standard solving strategy

Rotational kinematics problems reduce to a four-step routine that mirrors how you attack linear SUVAT problems.

1. Convert all angular speeds to $\text{rad s}^{-1}$. Strip out rpm and rev/s before any substitution.
2. List the five quantities $\omega_0,\ \omega,\ \alpha,\ t,\ \theta$. Mark the three you know and the one you want.
3. Pick the equation that omits the unwanted variable. No time given or asked $\Rightarrow$ use $\omega^2 = \omega_0^2 + 2\alpha\theta$. Need $\theta$ from time $\Rightarrow$ use $\theta = \omega_0 t + \tfrac12\alpha t^2$.
4. Watch the sign of $\alpha$ (negative for deceleration) and convert $\theta$ back to revolutions only at the end, dividing by $2\pi$.