NEET · Physics
Teach official Unit 5 through center of mass, torque, angular momentum, moment of inertia, axis theorems, rigid-body equilibrium and fixed-axis rotational motion with exact body/axis conditions on inertia formulas.
In short
Rotational Motion translates linear-mechanics ideas into angular form, but moment of inertia is never just a property of the material or body name. It depends on the mass distribution and the specified axis. Define the axis, use vector torque/angular momentum where required, then apply rigid-body equations only under their stated conditions.
Subject faculty
Also in Physics: Kumar Gandharva Saxena
For a particle system, center of mass is the mass-weighted mean position. It provides a useful translational description of the whole system under external force. The page should distinguish center of mass from geometric center, which coincide only for appropriate symmetric uniform bodies.
Torque is the vector product r × F; its direction follows the right-hand rule. Angular momentum of a particle is r × p. About a fixed axis for a rigid body under the standard model, angular momentum and rotational kinetic energy can be expressed using the moment of inertia.
Every tabulated I value must name both body and axis. A thin ring about its symmetry axis, a disc about its central perpendicular axis and a rod about its center each have different formulas. The parallel-axis and perpendicular-axis theorems have specific geometric conditions and cannot be applied to arbitrary 3D axes.
Rigid-body equilibrium requires both zero net force and zero net torque under the chosen inertial-frame model.
Center of mass: R_cm = (Σ m_i r_i)/(Σ m_i). Vector position; m. Discrete particle system. Continuous form uses mass integration.
Center of mass
Use when — Vector position; m. Discrete particle system. Continuous form uses mass integration.
Torque: τ = r × F; magnitude τ=rF sinθ. Vector; N m. r from chosen origin/axis to force application point.
Torque
Use when — Vector; N m. r from chosen origin/axis to force application point.
Angular momentum particle: L = r × p. Vector; kg m^2 s^-1.
Angular momentum particle
Use when — Vector; kg m^2 s^-1.
Rotational dynamics fixed axis: τ_net = Iα. For rigid body about a fixed axis with constant I about that axis; I kg m^2, α rad s^-2.
Rotational dynamics fixed axis
Use when — For rigid body about a fixed axis with constant I about that axis; I kg m^2, α rad s^-2.
Rotational kinetic energy: K_rot = (1/2)Iω^2. Rigid body about a fixed axis; J.
Rotational kinetic energy
Use when — Rigid body about a fixed axis; J.
Radius of gyration: I = Mk^2. k in m about the specified axis.
Radius of gyration
Use when — k in m about the specified axis.
Parallel-axis theorem: I = I_cm + Md^2. Axis parallel to an axis through center of mass; d perpendicular distance between axes.
Parallel-axis theorem
Use when — Axis parallel to an axis through center of mass; d perpendicular distance between axes.
Perpendicular-axis theorem: I_z = I_x + I_y. For a planar lamina; x,y are perpendicular axes in its plane through same point and z perpendicular to plane.
Perpendicular-axis theorem
Use when — For a planar lamina; x,y are perpendicular axes in its plane through same point and z perpendicular to plane.
Angular momentum conservation: L_total = constant. Requires zero net external torque about the chosen origin/axis over the interval.
Angular momentum conservation
Use when — Requires zero net external torque about the chosen origin/axis over the interval.
Concept being tested: Rotational dynamics
Prompt: A uniform solid disc has mass 2 kg, radius 0.5 m, and rotates about its central axis perpendicular to its plane. A net torque of 1 N m acts. Find angular acceleration.
Decisive condition: The body and axis are specified, so use the solid-disc moment of inertia about the central perpendicular axis and τ=Iα.
Method choice: I=(1/2)MR^2=(1/2)×2×0.25=0.25 kg m^2; then α=τ/I.
Reasoning/calculation: α=1/0.25=4 rad s^-2.
Unit/sign/condition check: N m/(kg m^2)=s^-2; radian is dimensionless. Axis condition is essential.
PI if answered incorrectly: Decision / Selection Error if a ring/rod inertia formula is used; Execution Error if units are mishandled.
Using an I formula without checking the axis
Decision / selection errorWhy it happens
Preparation Intelligence classification: Decision / Selection Error.
How it is corrected
Name body geometry and axis before substitution.
Treating torque as always rF
Recall gapWhy it happens
Preparation Intelligence classification: Recall Gap.
How it is corrected
Use the perpendicular component: rF sinθ.
Using perpendicular-axis theorem for a non-planar 3D body
Knowledge gapWhy it happens
Preparation Intelligence classification: Knowledge Gap.
How it is corrected
Apply it only to a planar lamina with the stated axes.
Conserving angular momentum despite external torque
Decision / selection errorWhy it happens
Preparation Intelligence classification: Decision / Selection Error.
How it is corrected
Check net external torque about the chosen origin.
FAQ
Straight answers about how Rank Sarthi fits into serious exam preparation.
Yes. It depends on mass distribution relative to the specified axis.
For the standard rigid-body fixed-axis model with I defined about that axis.
When net external torque about the chosen origin/axis is zero.
Official scope is taken from the NEET (UG) 2026 syllabus and official NTA documents; concept structure follows current NCERT contents. No performance, weightage or question-frequency data is asserted.
