Rotational Motion – JEE Main Physics Formula Sheet & Class 11 Notes | Moment of Inertia, Torque, Angular Momentum, Rolling Motion & All Rotational Formulas

Centre of Mass — Discrete, Continuous, and Standard Results

Why Centre of Mass Formulas for Standard Bodies Are Direct Rotational Motion JEE Main Questions

Centre of Mass — Discrete System (from Aakash PDF — Rotational Motion):

For a system of particles of masses m₁, m₂, …, mₙ at positions (x₁,y₁,z₁), (x₂,y₂,z₂), …:

x_cm = Σmᵢxᵢ/Σmᵢ; y_cm = Σmᵢyᵢ/Σmᵢ; z_cm = Σmᵢzᵢ/Σmᵢ

Centre of Mass — Continuous Body (from Aakash PDF — Rotational Motion JEE Main):

x_cm = ∫x dm / ∫dm = ∫x dm / M

Similarly for y_cm and z_cm. Here dm is the mass element at position x.

Centre of Mass of Standard Objects (from Aakash PDF — Rotational Motion JEE Main):

(1) Semicircular wire of radius R: OC = 2R/π (from the diameter)

(2) Semicircular disc of radius R: OC = 4R/3π (from the diameter)

(3) Non-uniform rod of length L with linear mass density varying from zero at O to maximum at B (linearly): OC = 2L/3 (from end O)

(4) Hemispherical shell: y_cm = R/2 from base

(5) Solid hemisphere: y_cm = 3R/8 from base

(6) Cone (solid): y_cm = h/4 from base

Velocity and Acceleration of Centre of Mass (from Aakash PDF — Rotational Motion):

v_cm = Σmᵢvᵢ / M = P_system / M

where P_system = total linear momentum of the system.

a_cm = Σmᵢaᵢ / M = F_ext / M

where F_ext = vector sum of all external forces on the system (internal forces cancel by Newton's 3rd Law).

Key implication: if F_ext = 0, then a_cm = 0, and the CM moves with constant velocity (or stays at rest). Conservation of momentum is equivalent to constant velocity of CM.

Download the Free PDF for Rotational Motion for all centre of mass examples for JEE Main.

Moment of Inertia — Complete Table and Standard Results

Why the Moment of Inertia Table Is the Most-Referenced Rotational Motion JEE Main Formula Set

Definition of Moment of Inertia (from Aakash PDF — Rotational Motion JEE Main):

I = Σmᵢrᵢ² (sum of mass × square of perpendicular distance from axis for each particle)

For continuous body: I = ∫r² dm. SI unit: kg·m². Dimensional formula: [ML²].

Radius of gyration k: I = Mk² → k = √(I/M). k represents the RMS distance of mass from axis.

Complete Moment of Inertia Table (from Aakash PDF — Rotational Motion JEE Main):

1. Ring or Hollow Cylinder (axis perpendicular to plane, through centre):

I_cm = MR²

K = R (radius of gyration = R for ring).

2. Solid Disc or Solid Cylinder (axis perpendicular to flat face, through centre):

I_cm = MR²/2

K = R/√2.

3. Thin Rod — axis perpendicular through CENTRE:

I_cm = ML²/12

K = L/√12 = L/2√3.

4. Thin Rod — axis perpendicular through ONE END:

I_end = ML²/3

(By parallel axis theorem: I_end = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3 ✓)

5. Rectangular Plate:

(a) Axis x–x (through CM, parallel to side b): I_xx = Mb²/12

(b) Axis y–y (through CM, parallel to side l): I_yy = Ml²/12

(c) Axis z–z (through CM, perpendicular to plate): I_zz = M(l²+b²)/12

6. Hollow Sphere (about diameter):

I_cm = 2MR²/3

K = R√(2/3).

7. Solid Sphere (about diameter):

I_cm = 2MR²/5

K = R√(2/5).

8. Thick Rod (Solid Cylinder) — axis through CM perpendicular to rod axis:

I_AA = ML²/12 + MR²/4

Axis through one end perpendicular to rod: I_BB = ML²/3 + MR²/4

K² values for rolling bodies (from Aakash PDF — Rotational Motion JEE Main):

Ring or hollow cylinder: K²/R² = 1; K² = R²

Hollow sphere (spherical shell): K²/R² = 2/3; K² = 2R²/3

Disc or solid cylinder: K²/R² = 1/2; K² = R²/2

Solid sphere: K²/R² = 2/5; K² = 2R²/5

Download the Free PDF for Rotational Motion for the complete MI table with all axes for JEE Main.

Parallel Axis Theorem, Perpendicular Axis Theorem — With Applications

Why Both Axis Theorems Are Essential Tools for All Rotational Motion JEE Main MI Calculations

Parallel Axis Theorem (from Aakash PDF — Rotational Motion JEE Main):

If I_cm is the moment of inertia of a body about an axis passing through its centre of mass, then the moment of inertia about any parallel axis at perpendicular distance d from the CM axis is:

I = I_cm + Md²

This theorem is valid for any rigid body (2D or 3D). The parallel axis must pass through the CM.

Important: I_cm is always the MINIMUM MI about any set of parallel axes.

Applications of Parallel Axis Theorem:

(a) Ring about a tangent: I = MR² + MR² = 2MR²

(b) Disc about tangent (in plane): I = MR²/4 + MR² = 5MR²/4

(c) Disc about tangent (perpendicular to plane): I = MR²/2 + MR² = 3MR²/2

(d) Solid sphere about tangent: I = 2MR²/5 + MR² = 7MR²/5

(e) Rod about perpendicular through one end: I = ML²/12 + M(L/2)² = ML²/3

(f) Hollow sphere about tangent: I = 2MR²/3 + MR² = 5MR²/3

Perpendicular Axis Theorem (from Aakash PDF — Rotational Motion JEE Main):

For a LAMINA (flat 2D body) with x and y axes in the plane of the lamina and z-axis perpendicular to it — all three axes passing through the same point:

I_z = I_x + I_y

This theorem applies ONLY to laminas (flat bodies), NOT to 3D bodies.

Applications of Perpendicular Axis Theorem:

(a) For a disc: I_z = MR²/2 (⊥ axis). I_x = I_y (by symmetry). So I_x = I_y = MR²/4 (disc about diameter).

(b) For a ring: I_z = MR² (⊥ axis). I_x = I_y = MR²/2 (ring about diameter).

(c) For a rectangular plate: I_z = M(l²+b²)/12 = I_x + I_y = Mb²/12 + Ml²/12 ✓.

Download the Free PDF for Rotational Motion for all axis theorem applications for JEE Main.

Torque, Angular Momentum, Conservation, and Angular Kinematics

Why τ=Iα, L=Iω, and Conservation of Angular Momentum Are the Core Rotational Motion JEE Main Equations

Torque (from Aakash PDF — Rotational Motion JEE Main):

τ = r × F (vector cross product of position vector and force)

|τ| = rF sinθ = rF⊥ (perpendicular component of F) = F × r⊥ (perpendicular distance from axis)

τ = Iα (Newton's 2nd Law for rotation: torque = moment of inertia × angular acceleration)

SI unit of torque: N·m. Dimensional formula: [ML²T⁻²] (same as energy, but NOT energy).

Angular Momentum (from Aakash PDF — Rotational Motion JEE Main):

L = Iω (angular momentum of a rotating body about the axis of rotation)

For a particle at position r moving with momentum p: L = r × p = mvr sinθ

Relation: τ = dL/dt (torque = rate of change of angular momentum)

This is the rotational analogue of F = dp/dt.

Conservation of Angular Momentum (from Aakash PDF — Rotational Motion JEE Main):

If net external torque on a system is zero (τ_ext = 0): L = constant (dL/dt = 0)

This means: I₁ω₁ = I₂ω₂ (if MI changes, angular speed changes reciprocally).

Examples: ice skater pulling arms in (I decreases → ω increases), diver tucking (I decreases → spins faster).

For an isolated system (no external torque): total L = sum of individual Iω values = constant.

Angular Momentum — General Formula (from Aakash PDF — Rotational Motion JEE Main):

L = Mv_cm × R_cm + I_cm × ω (about any point)

The first term is the "orbital" angular momentum (due to CM motion) and second is "spin" (rotation about CM).

For PURE TRANSLATION (no rotation): L_O = Mv_cm × h where h = perpendicular distance from O to line of v_cm; L_cm = 0.

For ROLLING BODY about contact point A: L_A = I_cm·ω + Mv_cm·R = I_A·ω.

Angular Kinematics — Equations for Constant α (from Aakash PDF — Rotational Motion JEE Main):

Exact analogues of linear kinematic equations with ω↔v, α↔a, θ↔s:

ω = ω₀ + αt

θ = ω₀t + ½αt²

ω² = ω₀² + 2αθ

θ_nth = ω₀ + α(2n–1)/2 (angular displacement in nth second)

Also: v = ωR; at = αR; ac = ω²R (centripetal); a_net = √(aT²+ac²) = R√(α²+ω⁴) (for non-uniform circular). Download the Free PDF for Rotational Motion for all torque and angular momentum examples for JEE Main.

Rolling Motion — Pure Rolling, Slipping, KE Split, All Cases

Why Rolling Motion KE Formula and Slipping Conditions Are the Most-Tested Rotational Motion Topics in JEE Main

Pure Rolling (from Aakash PDF — Rotational Motion JEE Main):

A body rolls without slipping when every point of contact has zero relative velocity with the surface. Condition:

v_cm = ωR (speed of CM = angular speed × radius)

The contact point I has instantaneous velocity = 0 (instantaneous centre of rotation).

Velocity of any point in pure rolling = vector sum of v_cm (translation) and ωR (rotation about CM).

At top: v_top = 2v_cm (maximum, in direction of motion)

At contact: v_contact = 0

At CM: v_cm

Kinetic Energy of Rolling Body (from Aakash PDF — Rotational Motion JEE Main):

Total KE = Translational KE + Rotational KE:

KE = ½mv_cm² + ½I_cm·ω² = ½mv_cm² + ½(mK²)(v_cm/R)² = ½mv_cm²(1 + K²/R²)

KE = ½mv_cm²(1 + K²/R²) where K = radius of gyration, I_cm = mK².

Fractions:

Translational fraction: KE_T/KE = 1/(1+K²/R²) = R²/(R²+K²)

Rotational fraction: KE_R/KE = (K²/R²)/(1+K²/R²) = K²/(R²+K²)

For different bodies (from Aakash PDF — Rotational Motion JEE Main):

Ring (K²=R²): KE_T = 50%; KE_R = 50%; ratio KE_T:KE_R = 1:1

Spherical shell (K²=2R²/3): KE_T = 60%; KE_R = 40%; ratio = 2:3 (Wait: T/total=1/(1+2/3)=3/5=60%; R/total=2/5×(3/3)=(2/3)/(1+2/3)=(2/3)/(5/3)=2/5=40%) ✓

Disc/Solid cylinder (K²=R²/2): KE_T = 2/3=66.7%; KE_R = 1/3=33.3%; ratio = 2:1 (Wait: T/total=1/(1+1/2)=2/3=66.7%)

Solid sphere (K²=2R²/5): KE_T = 5/7=71.4%; KE_R = 2/7=28.6%; ratio = 5:2

Note from Aakash PDF: These KE_T and KE_R fractions are INDEPENDENT of mass and radius — they depend only on the type of body (K²/R² ratio).

Slipping Conditions (from Aakash PDF — Rotational Motion JEE Main):

Forward slipping: v_cm > ωR → kinetic friction acts BACKWARD (opposing sliding of contact point).

Backward slipping: v_cm < ωR → kinetic friction acts FORWARD (to speed up translation or slow rotation).

Pure rolling: v_cm = ωR → friction is STATIC (may or may not be zero — prevents tendency to slip).

Download the Free PDF for Rotational Motion for all rolling motion examples for JEE Main.

Rolling on Inclined Plane, Force Application, and Hinged Rod

Why Rolling on Inclined Plane Acceleration and Hinged Rod Formulas Are Direct Rotational Motion JEE Main Questions

Rolling on Inclined Plane (from Aakash PDF — Rotational Motion JEE Main):

A body of mass m, radius R, radius of gyration K rolls without slipping on an inclined plane of angle θ. By conservation of energy from height h:

mgh = ½mv_cm²(1+K²/R²) → v_cm = √[2gh/(1+K²/R²)]

Acceleration (from Newton's laws and rolling constraint): a = g sinθ/(1+K²/R²)

Time to reach bottom from height h: t = (1/sinθ)√[2h(1+K²/R²)/g]

Friction force in rolling on incline (from Aakash PDF): fr = mg sinθ/(1+R²/K²) = mg sinθ·K²/(K²+R²)

Maximum angle for pure rolling (from Aakash PDF): θ_max = tan⁻¹[μ(1+R²/K²)]

Specific values: Ring: θ_max = tan⁻¹(2μ); Spherical Shell: θ_max = tan⁻¹(2.5μ); Disc: θ_max = tan⁻¹(3μ); Solid sphere: θ_max = tan⁻¹(3.5μ).

Order of reaching bottom (for same h): Solid sphere (largest a) > Disc = Solid cylinder > Hollow sphere > Ring (smallest a). Heavier rotation fraction → smaller a → reaches bottom later.

Force Applied at Different Heights — Rolling Body (from Aakash PDF — Rotational Motion JEE Main):

Horizontal force F applied at height h above centre, on a rolling body on rough horizontal surface:

a_cm = F/(m(1+K²/R²)) × (1 + hR/K²) × correction factor

More precisely from Aakash PDF: a_cm = F[1 + hR/K²]/[m(1+R²/K²)] ... (using torque about contact point)

General formula: a_cm = F(K²+hR)/(m(K²+R²))

At h = 0 (force at CM): a_cm = F·K²/[m(K²+R²)]; friction = FK²/(K²+R²) forward

Wait — from PDF exactly: Case h=0: a = F/[m(1+K²/R²)] and fr = FK²/(K²+R²)

At h = R (force at top): a_cm = 2F/[m(1+K²/R²)] (twice the force at CM gives double acceleration)

Friction at h=R: fr = F(R²–K²)/[R²+K²] (direction depends on body type)

Hinged Rod (from Aakash PDF — Rotational Motion JEE Main):

A uniform rod of mass m, length L, hinged at one end A, released from horizontal position:

Torque about A: τ = mg(L/2) (weight acts at CM, which is L/2 from hinge)

MI about A: I_A = ML²/3

Angular acceleration: τ = I_A × α → mg(L/2) = (ML²/3)α → α = 3g/2L

Linear acceleration of CM: a_cm = α × (L/2) = (3g/2L)(L/2) = 3g/4

Reaction at hinge: N = mg – ma_cm = mg – m(3g/4) = mg/4

When rod reaches vertical (hanging down), using energy: ω² = 3g/L; a_cm = ω²(L/2) = 3g/2.

Unstable equilibrium (hinged at top, horizontal): when released, α = 3g/L at the horizontal; ω at bottom = √(6g/L) (by energy). Download the Free PDF for Rotational Motion for all rolling and hinged rod examples for JEE Main.

Angular Momentum of Rolling Body — All 5 Cases

Why Angular Momentum of Rolling Bodies About Different Points Is a Key Rotational Motion JEE Main Topic

General Angular Momentum Formula (from Aakash PDF — Rotational Motion):

For any rigid body: L = Mv_cm × R_perp + I_cm × ω

where R_perp = perpendicular distance from the reference point to the line of v_cm.

Case I — Pure Translation (no rotation, ω=0):

L_O = Mv_cm × h (where h = perpendicular distance from O to line of v_cm)

L_cm = 0 (no rotation about own CM)

Case II — Rolling Body, about Contact Point A (from Aakash PDF — Rotational Motion JEE Main):

Rolling to the right with v_cm, ω anticlockwise (standard rolling): taking anticlockwise negative:

L_A = –I_cm·ω + Mv_cm·R = –I_cm·ω + MωR·R = ω(MR²–I_cm)

Wait — from Aakash PDF exactly: for rolling body about different points:

L_cm = I_cm·ω (spin about CM)

L_A (contact point below) = I_cm·ω + Mv_cm·R = I_A·ω (using I_A = I_cm + MR²)

L_O (any point, distance b from CM) = I_cm·ω + Mv_cm·b

L_B (point above CM, distance a) = I_cm·ω – Mv_cm·a [orbital term negative if above CM and motion to right]

The 5 standard cases from Aakash PDF — Rotational Motion JEE Main:

Case II (rolling right, about contact A): L_A = I_cm·ω + Mv_cm·R

Case III (CM fixed, pure rotation): L_O = L_A = L_cm = I_cm·ω (v_cm=0)

Case IV (rolling left, about contact A): L_A = –I_cm·ω + Mv_cm·R [Note: signs depend on chosen positive direction]

Case V (rolling up, about contact): L_A = I_cm·ω – Mv_cm·b [if motion reduces orbital contribution]

In all cases, the general formula L = ±I_cm·ω ± Mv_cm × perpendicular distance applies. The sign depends on whether the orbital contribution (Mv_cm×d) is in the same or opposite direction to the spin (I_cm·ω). Download the Free PDF for Rotational Motion for all angular momentum examples for JEE Main.