Angular acceleration

This topic belongs to the CBSE Class 11 chapter "System of Particles and Rotational Motion," specifically in the comparison-of-linear-and-rotational-motion thread that runs through the chapter alongside moment of inertia, torque, and angular momentum. The prescribed scope expects students to define angular acceleration, connect it to tangential acceleration through $a_t=r\alpha$, and use the constant-$\alpha$ rotational kinematics equations by direct analogy with straight-line motion — all of which is board-level content and none of which requires calculus beyond the simple derivative $\alpha=d\omega/dt$ already used for $\omega=d\theta/dt$. The equation $\tau=I\alpha$, where angular acceleration becomes the effect of an applied torque, is developed fully in the torque and moment-of-inertia topics that follow; here $\alpha$ is treated purely as a kinematic quantity, describing how rotation changes, before the chapter asks why it changes.

Angular velocity $\omega$ tells you how fast something is turning right now, in radians per second. Angular acceleration asks the next natural question: is that turning rate itself changing, and how quickly? Exactly as linear acceleration is the time-derivative of linear velocity, angular acceleration is the time-derivative of angular velocity:

where: $\alpha$ is angular acceleration in $\text{rad/s}^2$, $\omega$ is angular velocity in $\text{rad/s}$, and $t$ is time in seconds. Read physically, $\alpha$ is large and positive when a wheel is spinning up rapidly, zero when it holds a steady spin (or sits at rest), and negative when it is winding down — the sign is always relative to whichever rotation sense you have chosen as positive, not an absolute statement about which way the body turns. Because a rigid body holds its shape, every particle in it sweeps through the same angle in the same time as every other particle — they all share one $\theta$, one $\omega$, and therefore one $\alpha$ at any instant, regardless of how far each particle sits from the axis. What does depend on distance from the axis is how that shared angular acceleration shows up as an ordinary, metres-per-second-squared acceleration for a particular point. A particle at radius $r$ picks up a tangential acceleration, exactly mirroring $v=r\omega$:

where: $a_t$ is the component of linear acceleration directed along the particle's circular path (tangent to the circle), $r$ is its distance from the axis, and $\alpha$ is the (shared) angular acceleration. A rim point on a large wheel therefore speeds up or slows down its linear motion faster than a point near the hub, even though both share the very same $\alpha$. It is essential not to confuse $a_t$ with the centripetal (radial) acceleration $a_c=\omega^2r$, which points toward the axis and exists for any particle moving in a circle, even one turning at perfectly constant $\omega$ with $\alpha=0$. Tangential acceleration exists only when the spin rate is changing; centripetal acceleration exists whenever there is any rotation at all, changing or not. A particle in non-uniform circular motion — the wheel speeding up on the bicycle stand — experiences both at once, and its total (resultant) linear acceleration is the vector sum of a tangential piece (along the motion) and a radial piece (toward the centre).

Once angular acceleration is constant, rotational motion picks up its own trio of kinematic equations, built on precisely the same logic as $v=u+at$, $s=ut+\tfrac12at^2$, and $v^2=u^2+2as$ for straight-line motion:

where: $\omega_0$ is the angular velocity at $t=0$, $\omega$ is the angular velocity at time $t$, and $\theta$ is the angle turned through in that interval. The correspondence linear-displacement $\leftrightarrow$ angle, linear-velocity $\leftrightarrow$ angular-velocity, linear-acceleration $\leftrightarrow$ angular-acceleration is exact and runs throughout this chapter's comparison of translational and rotational motion — any straight-line kinematics problem you can already solve, you can now solve in its rotational disguise by relabelling the symbols. For now $\alpha$ is purely kinematic — a description of how the spin rate changes — but it is worth flagging the physics that follows: once torque and moment of inertia are introduced later in this chapter, angular acceleration becomes the rotational effect of a net torque, $\tau=I\alpha$, in exact parallel with $F=ma$. That equation is quoted here only as a forward pointer; its derivation belongs to the torque topic, not this one.

Board level - a wheel spinning up from rest

A wheel starts from rest and speeds up with a constant angular acceleration of $2\ \text{rad/s}^2$. Find its angular velocity after $5\ \text{s}$, and the number of radians it has turned through in that time.

Step 1. Write down what is given and what is asked. Given $\omega_0=0$ (starts from rest), $\alpha=2\ \text{rad/s}^2$, $t=5\ \text{s}$. Required: $\omega$ at $t=5\ \text{s}$, and $\theta$ over that interval.

Step 2. Find the angular velocity using $\omega=\omega_0+\alpha t$. $\omega=0+2\times5=10\ \text{rad/s}$.

Step 3. Find the angle turned using $\theta=\omega_0t+\tfrac12\alpha t^2$. $\theta=0+\tfrac12(2)(5)^2=\tfrac12(2)(25)=25\ \text{rad}$.

Step 4. State the final answers with a physical check. The wheel reaches $\omega=10\ \text{rad/s}$ after turning through $\theta=25\ \text{rad}$ ($\approx 3.98$ revolutions, since $25/(2\pi)\approx3.98$). As a quick cross-check, $\omega^2=\omega_0^2+2\alpha\theta$ gives $10^2=0+2(2)(25)=100$, which matches — confirming the two earlier steps are consistent.

Competitive level - a decelerating disc and its rim point

A rotating disc of radius $0.4\ \text{m}$ slows from an angular velocity of $12\ \text{rad/s}$ to rest in $6\ \text{s}$ under a constant (negative) angular acceleration. Find the angular acceleration, the tangential (linear) deceleration of a point on the rim, and the total angle turned through before the disc stops.

Step 1. List the data and the sign convention. Take the initial sense of rotation as positive: $\omega_0=12\ \text{rad/s}$, $\omega=0$ (comes to rest), $t=6\ \text{s}$, rim radius $r=0.4\ \text{m}$.

Step 2. Find $\alpha$ from $\omega=\omega_0+\alpha t$. $0=12+\alpha(6)\implies\alpha=-\dfrac{12}{6}=-2\ \text{rad/s}^2$. The negative sign says the angular acceleration opposes the chosen positive sense of spin — it is a deceleration, of magnitude $2\ \text{rad/s}^2$.

Step 3. Find the tangential deceleration of the rim point using $a_t=r\alpha$. $a_t=0.4\times(-2)=-0.8\ \text{m/s}^2$, i.e. the rim point decelerates tangentially at $0.8\ \text{m/s}^2$ while the disc winds down — note this is not the rim point's total acceleration, since a (shrinking, but nonzero until the very last instant) centripetal component $\omega^2r$ also acts toward the axis throughout the slowdown.

Step 4. Find the total angle turned using $\theta=\omega_0t+\tfrac12\alpha t^2$ (or the shortcut $\omega^2=\omega_0^2+2\alpha\theta$). $\theta=12(6)+\tfrac12(-2)(6)^2=72-36=36\ \text{rad}$. Cross-checking with $\omega^2=\omega_0^2+2\alpha\theta$: $0=144+2(-2)(36)=144-144=0$, confirming consistency. So $\alpha=-2\ \text{rad/s}^2$, $a_t=-0.8\ \text{m/s}^2$ at the rim, and the disc turns through $36\ \text{rad}$ ($\approx5.73$ revolutions) before stopping. Timing note: this problem packs three sub-answers into one, which is typical of a competitive-exam question — budget about 90 seconds: roughly 20 seconds to find $\alpha$ from the first equation, 20 seconds for $a_t=r\alpha$, and 40–50 seconds for $\theta$ plus a quick cross-check with the third equation, since a sign slip in $\alpha$ silently corrupts every later step if it goes unchecked.

• Confusing tangential acceleration $a_t=r\alpha$ (present only while the rotation rate is changing) with centripetal acceleration $a_c=\omega^2r$ (present in any circular motion, changing or not, directed toward the axis) — a particle in non-uniform circular motion has both simultaneously, and an exam question asking for "the acceleration of the rim point" usually wants their vector sum, not either one alone.
• Forgetting that $\alpha$ is the same for every particle of a rigid body at a given instant, while $a_t=r\alpha$ still varies from particle to particle depending on each one's distance from the axis.
• Mixing up the sign convention: a negative $\alpha$ (relative to the chosen positive rotation sense) means the body is decelerating, not that it has started spinning the other way — the disc in the second worked example never reverses, it simply slows to rest.
• Applying the constant-$\alpha$ kinematic equations to a situation where $\alpha$ is not actually constant (a fan winding down under variable friction, for instance) — the three shortcut equations only hold when $\alpha=d\omega/dt$ is genuinely fixed over the interval.
• Treating "the body is rotating" as automatically meaning "the body has angular acceleration" — a wheel spinning steadily at constant $\omega$ has $\alpha=0$ even though it is very much in motion, exactly as a car cruising at constant velocity has zero linear acceleration.

• Define angular acceleration as $\alpha=d\omega/dt$ and state its SI unit, $\text{rad/s}^2$.
• Explain why every particle of a rigid body shares the same $\alpha$, while the tangential acceleration $a_t=r\alpha$ still varies with radius.
• Distinguish tangential acceleration $a_t=r\alpha$ from centripetal acceleration $a_c=\omega^2r$, and recognise that non-uniform circular motion has both at once.
• Derive, from $\alpha=d\omega/dt$ and $\omega=d\theta/dt$ alone, the three constant-$\alpha$ kinematic equations $\omega=\omega_0+\alpha t$, $\theta=\omega_0t+\tfrac12\alpha t^2$, and $\omega^2=\omega_0^2+2\alpha\theta$.
• Solve numerical problems for $\omega$, $\theta$, $\alpha$, or $t$ given the other quantities, including problems that also ask for a tangential acceleration at a specified radius.
• Correctly interpret the sign of a computed $\alpha$ as speeding up or slowing down relative to a chosen positive sense, not as a reversal of rotation.