Dynamics of Rotational Motion

Newton's second law for rotation: τ = Iα

In linear motion, the cause of acceleration is force and the resistance to it is mass: $F = ma$. Rotation has an exact parallel. The cause of angular acceleration is torque, and the resistance to it is the moment of inertia. NCERT §6.11 derives the result by equating the rate at which a torque does work to the rate at which rotational kinetic energy grows, and arrives at

$$\tau = I\alpha$$

Read it the way you read $F = ma$. The angular acceleration $\alpha$ is directly proportional to the applied torque $\tau$ and inversely proportional to the moment of inertia $I$. A large torque on a body that resists rotation only weakly spins it up quickly; the same torque on a body of large moment of inertia barely changes its spin. NCERT calls this "Newton's second law for rotational motion about a fixed axis".

Why a fixed axis simplifies the problem

NCERT restricts §6.11 to rotation about a fixed axis, and that restriction is what makes the algebra tractable. Because the axis cannot move, only the components of torque along the axis can change the body's spin. A torque component perpendicular to the axis would try to tip the axis over; NCERT assumes that constraint forces at the bearings cancel exactly those perpendicular components, so we never write them down.

Two practical consequences follow, both stated in the text:

• Consider only forces lying in planes perpendicular to the axis. Forces parallel to the axis give torques perpendicular to it and are ignored.
• Use only the components of the position vectors that are perpendicular to the axis. The components along the axis again produce torques perpendicular to it.

For NEET problems — flywheels, pulleys, cylinders on fixed axles — this means you compute the single axial torque $\tau = rF$ for each force and add them algebraically, exactly as you would add forces along a line.

Work done by a torque: W = τθ

Consider a force $F_1$ acting on a particle of a rigid body in a plane perpendicular to the axis. As the body turns through a small angle $d\theta$, the particle (at radius $r_1$) moves along an arc $ds_1 = r_1\,d\theta$. The work done by the force on that particle works out to $dW_1 = \tau_1\,d\theta$, where $\tau_1$ is the axial torque of $F_1$. Summing over every force on the body, and noting that every particle turns through the same $d\theta$, NCERT obtains

$$dW = \tau\, d\theta \quad\Longrightarrow\quad W = \tau\,\theta \;\;(\tau \text{ constant})$$

Here $\tau$ is the total external torque about the axis. The likeness to the linear expression $dW = F\,dx$ is exact — replace force by torque and linear displacement by angular displacement. When the torque is not constant, the work is the area under the $\tau$–$\theta$ graph, just as linear work is the area under the $F$–$x$ graph.

Rotational power: P = τω

Dividing the work expression by the time taken gives the instantaneous power. Since $d\theta/dt = \omega$,

$$P = \frac{dW}{dt} = \tau\frac{d\theta}{dt} = \tau\,\omega$$

This is the rotational twin of $P = Fv$. A torque doing work on a body spinning at angular velocity $\omega$ delivers power $\tau\omega$. The unit is the watt, identical to linear power, because torque (N·m) times angular velocity (rad s⁻¹, dimensionless radian) is again the joule per second.

Rotational kinetic energy: ½Iω²

Where does the work go? In a perfectly rigid body there is no internal motion to dissipate it, so the work done by external torques becomes rotational kinetic energy. Add up the ordinary kinetic energy of every particle: a particle of mass $m_i$ at radius $r_i$ moves at speed $v_i = r_i\omega$, carrying $\tfrac{1}{2}m_i r_i^2\omega^2$. Summing over the body and factoring out $\tfrac{1}{2}\omega^2$,

$$K_{\text{rot}} = \tfrac{1}{2}\Big(\sum_i m_i r_i^2\Big)\omega^2 = \tfrac{1}{2}I\omega^2$$

The sum in brackets is precisely the moment of inertia $I$. So $\tfrac{1}{2}I\omega^2$ is not a new species of energy — it is plain kinetic energy, collected into the variable $I$, exactly as $\tfrac{1}{2}mv^2$ collects the linear kinetic energy of a body into its mass.

The rotational work-energy theorem

The whole of §6.11 hinges on one identification: the rate of work by the torque equals the rate of gain of rotational kinetic energy. Writing the kinetic energy as $\tfrac{1}{2}I\omega^2$ and differentiating (with $I$ constant), the rate of increase is $I\omega\,\alpha$. Setting this equal to the power $\tau\omega$ and cancelling $\omega$ returns $\tau = I\alpha$ — which is exactly how NCERT derives the second law. Integrated over a finite turn, the same statement reads as the work-energy theorem for rotation:

$$W = \tau\theta = \tfrac{1}{2}I\omega^2 - \tfrac{1}{2}I\omega_0^2$$

The work done by the resultant torque equals the change in rotational kinetic energy. For a body starting from rest, $W = \tfrac{1}{2}I\omega^2$. This is the rotational mirror of $W = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2$, and it is the fastest route through any "torque acts, find final $\omega$ or work done" problem.

The linear↔rotational correspondence

Everything above can be compressed into one substitution scheme. Wherever a linear-motion law appears, replace each linear quantity by its rotational analogue and the rotational law drops out. This is NCERT Table 6.2, the single most useful object in the chapter for fast NEET recall.

The discipline is mechanical: memorise the three primitive swaps $M \to I$, $F \to \tau$, $x \to \theta$ (from which $v \to \omega$ and $a \to \alpha$ follow), and every derived law in the right column is generated by substitution. The deeper relation — that torque is the rate of change of angular momentum, $\tau = dL/dt$, just as $F = dp/dt$ — is taken up under torque and angular momentum.

Worked example — rope on a hollow cylinder (NCERT 6.13)

This is also NEET 2017 Q.149 — the same numbers, asked verbatim. The takeaway is the two-step rhythm of every rotational dynamics problem: build the torque, build the matching moment of inertia, divide. Then convert to a linear quantity only at the end, through $a = R\alpha$.

Worked example — flywheel pulled by a cord (NCERT 6.12)

Part (d) is the work-energy theorem made visible: every joule the cord delivers reappears as rotational kinetic energy because nothing was dissipated. Had the bearings been rough, the work done would exceed the kinetic energy by the energy lost to friction.

Worked example — power to maintain a rotor (NCERT 6.14)