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Rotational Motion

Part of the NEET UG study roadmap. Physics topic phy-006 of Physics.

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Rotational Motion

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Rotational motion describes a rigid body turning about a fixed axis, where every particle sweeps the same angular displacement θ in the same time. The master relation is τ = Iα, the angular form of Newton’s second law, with τ as torque (N·m), I the moment of inertia (kg·m²) about that axis, and α the angular acceleration (rad/s²).

  • Moment of inertia: I = Σ mᵢrᵢ² for a system of particles; depends on both mass and its distance from the axis.
  • Angular momentum: L = Iω for a rigid body, conserved when net external torque is zero.
  • Rolling without slipping: v_cm = Rω and total KE = ½mv² + ½Iω².
  • For NEET, expect 1–2 questions: numericals on rolling inclines, I via parallel-axis theorem, or angular momentum conservation in collisions.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Torque and the Angular Form of Newton’s Laws

Torque is the rotational analogue of force, defined as τ = r × F, with magnitude τ = rF sinθ. A force produces maximum torque when applied perpendicular to the position vector (θ = 90°). The angular form of Newton’s second law, τ = dL/dt, reduces to τ = Iα for a rigid body rotating about a fixed axis where I is constant.

Moment of Inertia and Radius of Gyration

Moment of inertia I = Σ mᵢrᵢ² is the rotational analogue of mass; it depends on how mass is distributed about the axis, not just the total mass. The radius of gyration k is defined by I = Mk², giving a characteristic length scale of the mass distribution.

The Two Axis Theorems

  • Parallel axis theorem: I = I_cm + Md², where d is the distance between the chosen axis and a parallel axis through the centre of mass.
  • Perpendicular axis theorem: I_z = I_x + I_y, valid only for planar laminas (2-D bodies), not for 3-D solids.

Translating Translational to Rotational Quantities

TranslationalRotationalRelation
Mass mMoment of inertia II = Σmr²
Velocity vAngular velocity ωω = v/r
Force FTorque ττ = rF sinθ
Momentum p = mvAngular momentum L = IωL = r × p
KE = ½mv²KE_rot = ½Iω²Both add in rolling

NEET typically tests I-computation for standard shapes (ring, disc, solid sphere, hollow sphere) using these theorems, often combining them with a rolling-on-an-incline numerical.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Rolling on an Inclined Plane — The Classic NEET Setup

A body of mass m and radius R with radius of gyration k rolling down an incline of angle θ (no slipping) has acceleration:

a = g sinθ / (1 + k²/R²)

Using energy conservation, mgh = ½mv² + ½Iω², and the no-slip condition v = Rω, gives the same result. The smaller the k²/R² ratio, the faster the body rolls — which is why a solid sphere beats a hollow sphere and both beat a ring in a race down the same incline.

Angular Momentum in Collisions

For a point particle, L = mvr sinθ; for a rigid body about a symmetry axis, L = Iω. The two coincide only when I is evaluated about the instantaneous axis. In a perfectly inelastic collision where a point mass sticks to a rotating disc, L is conserved about the rotation axis if no external impulsive torque acts — a frequent NEET assertion-reason trap.

Work-Energy Theorem in Rotation

The work done by a constant torque through angle θ is W = τθ, and this equals the change in rotational kinetic energy Δ(½Iω²). Power delivered by torque is P = τω, the rotational analogue of P = Fv.

Common Traps to Avoid

  • Treating I as a fixed property of a body — it changes the moment the axis changes.
  • Applying the perpendicular axis theorem to 3-D bodies like a solid cylinder (invalid).
  • Forgetting the ½Iω² term in rolling energy, which gives the wrong acceleration and the wrong time-to-reach-bottom on an incline.
  • Equating τ with F, or assuming the largest force gives the largest torque.

Practice Prompts

  1. A solid sphere and a ring of equal mass and radius roll down the same incline. Find the ratio of their translational speeds at the bottom, starting from rest.
  2. A disc of moment of inertia I is rotating at ω₀. A second identical disc, initially at rest, is placed coaxially on top and they reach a common angular speed under friction. Calculate the energy dissipated and explain why it differs from ½Iω₀² even though angular momentum is conserved.

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