Rotational Motion

Rotational motion is the chapter where Newton's laws get a "twin" written in angular language — every year NEET asks 2–3 questions from it, typically on moment of inertia, angular momentum conservation, or rolling bodies on an incline. Master the parallels between linear and angular quantities and most of these become one-line calculations.

From Particles to Rigid Bodies: Centre of Mass

A rigid body is an idealised object in which the distance between any two constituent particles never changes, no matter what forces act. Real bodies deform slightly, but for exam purposes a wheel, rod, disc or sphere is treated as perfectly rigid.

A rigid body can execute two independent kinds of motion. In pure translation, every particle has the same velocity at a given instant. In pure rotation about a fixed axis, every particle traces a circle centred on that axis, and all particles share the same angular velocity ω but have different linear speeds (v = rω, where r is the perpendicular distance from the axis). General motion — a rolling wheel, a thrown bat — is a superposition: translation of the centre of mass plus rotation about the centre of mass.

The centre of mass (CM) is the mass-weighted average position of all the particles:

$$\vec{R}_{cm} = \frac{\sum m_i \vec{r}i}{\sum m_i} \quad \text{or} \quad \vec{R}{cm} = \frac{1}{M}\int \vec{r},dm$$

Key facts about the CM worth memorising:

• For symmetric, uniform bodies (ring, disc, sphere, cube, rod), the CM sits at the geometric centre. It may lie outside the material of the body, as with a ring.
• $M\vec{a}{cm} = \vec{F}{ext}$: the CM moves exactly as a single point particle of mass M would under the net external force. Internal forces never shift the CM.
• If $\vec{F}{ext} = 0$, total momentum $\vec{P} = M\vec{v}{cm}$ is conserved, so the CM either stays at rest or moves with constant velocity. A shell exploding in flight has fragments flying apart, but the CM continues along the original parabola.
• Two-particle result: the CM divides the line joining them internally in the inverse ratio of their masses, i.e. $m_1 r_1 = m_2 r_2$ measured from the CM.

Torque and Angular Momentum

The rotational analogue of force is torque (moment of force), defined about a chosen point or axis:

$$\vec{\tau} = \vec{r} \times \vec{F}, \qquad |\tau| = rF\sin\theta = F \times (\text{perpendicular distance from axis to line of action})$$

Its SI unit is N·m. Torque is what changes rotation; a force whose line of action passes through the axis produces zero torque no matter how large it is.

The rotational analogue of linear momentum is angular momentum:

$$\vec{L} = \vec{r} \times \vec{p}, \qquad \text{and for a rigid body about a fixed axis} \quad L = I\omega$$

with SI unit kg·m²·s⁻¹ (or J·s). The fundamental dynamical law is

$$\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$$

which for a rigid body of constant moment of inertia reduces to the familiar $\tau = I\alpha$ — the angular version of F = ma.

Conservation of angular momentum: if the net external torque about an axis is zero, L about that axis stays constant. This gives the standard problem type $I_1\omega_1 = I_2\omega_2$. A spinning skater pulling her arms in reduces I, so ω rises; her rotational kinetic energy $\tfrac12 I\omega^2 = L^2/2I$ actually increases, the extra energy coming from muscular work. Similarly, a planet moves fastest at perihelion because gravity exerts no torque about the Sun.

Equilibrium of a rigid body requires two conditions simultaneously: $\sum \vec{F} = 0$ (no translational acceleration) and $\sum \vec{\tau} = 0$ about any axis (no angular acceleration). This is the basis of lever, see-saw and ladder problems. A couple is a pair of equal, opposite, non-collinear forces: zero net force but non-zero torque, so it produces pure rotation.

Moment of Inertia and the Two Theorems

Moment of inertia measures a body's reluctance to change its rotational state, playing the role that mass plays in translation:

$$I = \sum m_i r_i^2 = \int r^2, dm$$

Its unit is kg·m². Unlike mass, I is not a fixed property of the body — it depends on the choice of axis, because r is measured perpendicular to that axis. Mass placed far from the axis contributes disproportionately.

The radius of gyration k is defined by $I = Mk^2$, i.e. $k = \sqrt{I/M}$. It is the distance from the axis at which the entire mass could be concentrated to give the same moment of inertia. Its unit is metre.

Two theorems let you avoid integration:

1. Perpendicular axis theorem (planar bodies only): if x and y are two perpendicular axes lying in the plane of a lamina and z is perpendicular to the plane through their intersection, then $I_z = I_x + I_y$.
2. Parallel axis theorem (any body): $I = I_{cm} + Md^2$, where d is the perpendicular distance between the given axis and a parallel axis through the CM. A corollary: of all parallel axes, the one through the CM gives the minimum moment of inertia.

Standard results (M = mass, all bodies uniform):

• Thin ring/hollow cylinder, axis through centre ⊥ to plane: $MR^2$
• Ring, axis along a diameter: $\tfrac12 MR^2$
• Disc/solid cylinder, axis through centre: $\tfrac12 MR^2$
• Disc, axis along a diameter: $\tfrac14 MR^2$
• Solid sphere, axis through centre: $\tfrac25 MR^2$
• Hollow (thin) spherical shell: $\tfrac23 MR^2$
• Thin rod of length L, axis through centre ⊥ to rod: $\tfrac{1}{12}ML^2$; through one end: $\tfrac13 ML^2$
• Rectangular lamina (l × b), axis through centre ⊥ to plane: $\tfrac{1}{12}M(l^2+b^2)$

Rotational Kinematics, Work and Energy

Because θ, ω and α are defined exactly as x, v and a are, the constant-acceleration equations carry over unchanged in form:

• $\omega = \omega_0 + \alpha t$
• $\theta = \omega_0 t + \tfrac12 \alpha t^2$
• $\omega^2 = \omega_0^2 + 2\alpha\theta$
• angle turned in the nth second, average angular velocity, etc., follow the same patterns

Energy relations also mirror the linear case:

• Rotational kinetic energy: $K_{rot} = \tfrac12 I\omega^2 = \dfrac{L^2}{2I}$
• Work done by a constant torque: $W = \tau,\theta$
• Instantaneous power: $P = \tau\omega$
• Work–energy theorem: $W_{net} = \tfrac12 I\omega_2^2 - \tfrac12 I\omega_1^2$

The full correspondence table — x↔θ, v↔ω, a↔α, m↔I, F↔τ, p↔L, ½mv²↔½Iω² — is the single most useful thing to memorise, because any linear-motion formula you already know can be converted instantly.

Rolling Motion

A body rolls without slipping when the contact point is instantaneously at rest relative to the surface. The condition is

$$v_{cm} = R\omega \quad \text{and} \quad a_{cm} = R\alpha$$

The total kinetic energy splits into translational plus rotational parts:

$$K = \tfrac12 M v_{cm}^2 + \tfrac12 I \omega^2 = \tfrac12 M v_{cm}^2\left(1 + \frac{k^2}{R^2}\right)$$

Note the velocities of different points: the contact point has zero velocity, the centre has $v_{cm}$, and the topmost point has $2v_{cm}$.

For a body rolling down an incline of angle θ from height h, starting from rest, energy conservation gives

$$v = \sqrt{\frac{2gh}{1 + k^2/R^2}}, \qquad a = \frac{g\sin\theta}{1 + k^2/R^2}$$

The factor $k^2/R^2$ decides everything, and it depends only on the shape, not on mass or radius:

• Solid sphere: 2/5 → fastest
• Solid disc/cylinder: 1/2
• Hollow shell: 2/3
• Ring/hollow cylinder: 1 → slowest

So in a race down an incline the solid sphere wins and the ring loses, regardless of their masses and sizes. Compared with a body that simply slides down frictionlessly, every rolling body is slower, since part of the potential energy goes into rotation. The friction required for rolling without slipping is static friction; it does no work in ideal rolling, which is why mechanical energy is conserved.

Common Mistakes and Exam Traps

1. Treating I as a constant of the body. Always ask "about which axis?" A rod has $ML^2/12$ about its centre but $ML^2/3$ about its end — a factor-of-four error if you grab the wrong one. Similarly, apply the parallel axis theorem only from the CM axis, never from an arbitrary axis.
2. Misusing the perpendicular axis theorem. It works only for plane laminae (disc, ring, rectangular sheet), not for spheres, cylinders or cones.
3. Assuming energy is conserved when angular momentum is. In "clutch"/skater/mud-sticking-to-disc problems, use $I_1\omega_1 = I_2\omega_2$; kinetic energy changes (it drops in inelastic coupling, rises when the skater pulls in). Never equate the kinetic energies.
4. Forgetting the rotational term in rolling. Writing $v=\sqrt{2gh}$ for a rolling body, or forgetting that the top of a rolling wheel moves at $2v_{cm}$, are classic slips. Also remember that mass and radius cancel out in incline problems — if a question offers "the heavier one reaches first", it is a distractor.