Linear–rotational analogy
Translation ↔ Rotation: F ↔ τ, m ↔ I, v ↔ ω, a ↔ α, p = mv ↔ L = Iω, KE = ½mv² ↔ KE_rot = ½ I ω². Newton's 2nd Law analogue: τ = I α.
-- NCERT Class 11 Physics, Ch. 6, p. 15Every linear-motion quantity has a rotational twin. NCERT Class 11 Physics Chapter 7 (System of Particles and Rotational Motion), page 15, presents the comparison table that maps displacement → angular displacement, velocity → angular velocity, acceleration → angular acceleration, mass → moment of inertia, force → torque, momentum → angular momentum, and kinetic energy → rotational kinetic energy. This mapping is what NEET tests when it asks you to "write the rotational analogue of" a linear equation.
The structural rule is: replace every linear variable with its angular counterpart, and the equation's form stays identical.
| Linear | Symbol | Rotational | Symbol |
|---|---|---|---|
| Displacement | s | Angular displacement | θ |
| Velocity | v | Angular velocity | ω |
| Acceleration | a | Angular acceleration | α |
| Mass (inertia) | m | Moment of inertia | I |
| Force | F | Torque | τ |
| Momentum | p = mv | Angular momentum | L = Iω |
| Kinetic energy | ½mv² | Rotational KE | ½Iω² |
| Newton's 2nd law | F = ma | Rotational form | τ = Iα |
Two points where students lose marks:
1. The analogue of mass is I, not m. Moment of inertia depends on both mass AND its distribution about the axis. Two objects of the same mass can have different I values. When a question says "write the rotational analogue of ½mv²," the answer is ½Iω² — substituting m with I and v with ω.
2. Units shift but dimensional structure is preserved. Torque is N·m (not just N), angular momentum is kg·m²/s (not kg·m/s). If you write the rotational analogue but keep linear units, NEET marks it wrong.
The comparison table is a recall item — NEET can and does test it as a straightforward "which quantity is the analogue of..." question.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The rotational analogue of linear momentum (p = mv) is:
Which of the following is the rotational analogue of force?
In the rotational analogue of Newton's second law, mass (m) is replaced by:
The rotational kinetic energy of a body rotating about a fixed axis is given by ½Iω². This expression is the rotational analogue of:
A disc of moment of inertia 4.0 kg·m² rotates at angular velocity 3.0 rad/s about a fixed axis. Its rotational kinetic energy is:
The SI unit of angular momentum is:
Two bodies A and B have the same mass and the same angular velocity about a fixed axis. If A has a larger moment of inertia than B, which statement is correct?
The linear equation v = u + at has the rotational analogue ω = ω₀ + αt. In this analogy, the quantity that replaces linear acceleration 'a' is:
Given
A uniform solid cylinder of mass M = 2.0 kg and radius R = 0.10 m rolls without slipping on a horizontal surface. Its centre of mass moves at v = 4.0 m/s.
Required
Find the ratio of rotational kinetic energy to translational kinetic energy.
Concept
For a rolling body, KE_total = KE_trans + KE_rot = ½Mv² + ½Iω². The linear–rotational analogy gives us the rotational KE term by replacing m with I and v with ω. For rolling without slipping, v = Rω, so ω = v/R.
Formula
KE_trans = ½Mv² KE_rot = ½Iω² For a solid cylinder about its symmetry axis: I = ½MR² Rolling constraint: ω = v/R
Substitution
KE_rot = ½ × (½MR²) × (v/R)² = ½ × ½MR² × v²/R² = ¼Mv² KE_trans = ½Mv²
Calculation
Ratio = KE_rot / KE_trans = (¼Mv²) / (½Mv²) = (¼)/(½) = 1/2 Note on exact constants: The coefficients ½ (in KE formulas and in the cylinder's MOI formula ½MR²) are exact mathematical/geometric constants. They do not limit significant figures. The given values M = 2.0 kg, R = 0.10 m, v = 4.0 m/s each have 2 significant figures, but since M, R, and v all cancel in the ratio, the answer is an exact fraction.
Final answer
KE_rot / KE_trans = **1/2** (or equivalently, rotational KE is one-half of translational KE for a rolling solid cylinder). This means one-third of the total KE is rotational and two-thirds is translational.
Common trap
A common error is using the wrong MOI coefficient — plugging in I = MR² (ring) instead of I = ½MR² (solid cylinder). With the ring formula, the ratio would come out as 1 instead of ½. Always verify the geometry before picking I.
Similar NEET-style question
A solid sphere of mass 3.0 kg rolls without slipping at 5.0 m/s. What fraction of its total kinetic energy is rotational? (Hint: I_sphere = 2MR²/5; use the same ratio method.) ---
Translation ↔ Rotation: F ↔ τ, m ↔ I, v ↔ ω, a ↔ α, p = mv ↔ L = Iω, KE = ½mv² ↔ KE_rot = ½ I ω². Newton's 2nd Law analogue: τ = I α.
-- NCERT Class 11 Physics, Ch. 6, p. 15For a particle: L = r x p. For a rigid body about its rotation axis: L = I omega. Vector quantity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| L | angular momentum | kg*m^2/s |
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
The position of the centre of mass equals the mass-weighted average of particle positions. For continuous bodies use integral form.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R_cm | CoM position | m |
| m_i | mass of i-th particle | kg |
| r_i | position of i-th particle | m |
Standard moments of inertia about the symmetry axis. For other axes use parallel/perpendicular axes theorems.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | mass | kg |
| R | radius | m |
| L | length | m |
| I | moment of inertia | kg*m^2 |
Moment of inertia about any axis = moment about parallel axis through CM + Md^2.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | MOI about given axis | kg*m^2 |
| I_cm | MOI about parallel CM axis | kg*m^2 |
| M | total mass | kg |
| d | perpendicular distance | m |
For planar lamina: MOI about axis perpendicular to plane = sum of MOI about two perpendicular in-plane axes through same point.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I_z | MOI perp to plane | kg*m^2 |
| I_x, I_y | MOI in plane | kg*m^2 |
Rotational analogues of linear kinematic equations under constant angular acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| omega | angular velocity | rad/s |
| alpha | angular acceleration | rad/s^2 |
| theta | angular displacement | rad |
| t | time | s |
Energy of rotation about an axis. Adds to translational KE for rolling bodies.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
Cross product of position vector and force vector. Magnitude r F sin(theta).
| Symbol | Quantity | SI Unit |
|---|---|---|
| tau | torque | N*m |
| r | position from pivot | m |
| F | force | N |
| theta | angle between r and F | rad |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student answers L/2 for two-particle CoM regardless of mass ratio.
Question gives two masses on rigid rod and asks for CoM distance.
R_cm from m1 = m2*L/(m1+m2). Heavier mass pulls CoM closer to it.
Category: Similar Terms
Student conserves rotational KE when angular momentum is conserved (or vice versa). When I changes, L = Iω is conserved but KE = ½Iω² is NOT (it depends on I and ω together).
Question describes a body whose moment of inertia changes (skater pulling arms in, star collapsing).
L conservation requires zero external torque. KE conservation requires no work done — different criteria. When I changes via internal forces, L conserved, ω increases, KE increases.
Category: Similar Terms
Student confuses 2/5 (solid sphere) with 2/3 (hollow sphere) or 1/2 (disc) with 1 (ring).
Question gives a specific geometry and asks for I or radius of gyration.
Memorise: solid sphere 2/5, hollow sphere 2/3, disc/cylinder 1/2, ring/hoop 1, rod-centre 1/12, rod-end 1/3.
Category: Unit Conversion
Student plugs rpm directly into formulas requiring rad/s. 1 rpm = 2π/60 rad/s.
Question gives ω in rpm and asks for kinematic quantities in SI units.
Convert: ω(rad/s) = (2π/60) × rpm. Always check units before substituting.
Root cause: concept gap
L = Iω is conserved when external torque is zero. KE = ½Iω² is NOT conserved when I changes (since ω changes too). When skater pulls arms in, L conserved, ω increases, KE increases (work done by muscles).
Root cause: concept gap
Memorise standard moments of inertia. Solid sphere has more mass near axis (smaller MOI = 2MR²/5); hollow sphere has all mass at radius R (larger MOI = 2MR²/3).
Root cause: unit error
Convert: ω(rad/s) = (2π/60) × ω(rpm). For example, 1200 rpm = 1200 × 2π/60 = 125.66 rad/s.
The angular acceleration of a body, moving along the circumference of a circle, is
forgets conversion rpm to rad s
Treats rpm as rad/s without 2*pi/60 conversion
uses equal distribution
Default to L/2 regardless of mass ratio
uses energy conservation instead
Confusing L conservation with KE conservation
swap solid hollow coefficients
Confusing 2/5 (solid sphere) with 2/3 (hollow sphere)
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