PhysicsPHY 03
Rotational Motion
Moment of inertia, torque, angular momentum, rolling without slipping
Concept Core
Essential theory — everything NCERT tests on Rotational Motion
MOMENT OF INERTIA
I = Σmr². Depends on mass AND distribution about axis. Analogous to mass in linear motion.
Key values: Ring: I = MR². Disc: I = MR²/2. Solid sphere: I = 2MR²/5. Hollow sphere: I = 2MR²/3. Rod (centre): I = ML²/12. Rod (end): I = ML²/3.
THEOREMS
Parallel axis theorem: I = Icm + Md². Used when axis is parallel to CM axis.
Perpendicular axis theorem: Iz = Ix + Iy. Applies to plane laminas only.
TORQUE & ANGULAR MOMENTUM
τ = r × F = Iα. Angular momentum: L = Iω = r × p. Conservation of L: if τext = 0, L = const. Explains ice skater spinning faster when arms pulled in.
τ = dL/dt (analogue of F = ma)
ROLLING MOTION
Rolling without slipping: v = Rω (contact point has zero velocity).
Total KE = ½Mv² + ½Iω² = ½Mv²(1 + k²/R²). k = radius of gyration.
For ring: KE ratio = 1:1. For disc: 3:1 (translational:rotational = 3:1 of total... actually ½Mv² : ½Iω² = 1:½ for disc).
Fact & Formula Vault
High-yield facts, numbers, and formulas
MOI Values
Ring: MR²
Disc: MR²/2
Solid sphere: 2MR²/5
Hollow sphere: 2MR²/3
Rod (centre): ML²/12
Angular Analogues
α (angular accel) ↔ a
τ (torque) ↔ F
I (MOI) ↔ m
L = Iω ↔ p = mv
τ = Iα ↔ F = ma
Rolling KE
KE_total = ½Mv²(1 + k²/R²)
Ring: k²/R² = 1
Disc: k²/R² = 1/2
Solid sphere: k²/R² = 2/5
Hollow sphere: k²/R² = 2/3
Worked Examples
NEET-style questions solved step-by-step
EASYA ring and a disc start from rest and roll down a frictionless slope. Which reaches bottom first?▾
A ring and a disc start from rest and roll down a frictionless slope. Which reaches bottom first?
Disc. Disc has lower k²/R² (=0.5) vs ring (=1). Lower k²/R² → higher acceleration on slope → reaches bottom first. Solid sphere (2/5) would be even faster.
MEDIUMA dancer spins with arms out (I = 4 kg·m²) at ω = 2 rad/s. Arms pulled in: I = 1 kg·m². New ω?▾
A dancer spins with arms out (I = 4 kg·m²) at ω = 2 rad/s. Arms pulled in: I = 1 kg·m². New ω?
Conservation of angular momentum: I₁ω₁ = I₂ω₂. 4×2 = 1×ω₂. ω₂ = 8 rad/s.
HARDMOI of a uniform rod of mass M, length L about an axis through one end perpendicular to rod:▾
MOI of a uniform rod of mass M, length L about an axis through one end perpendicular to rod:
Using parallel axis theorem: Icm = ML²/12, d = L/2. I = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3.
Mistake DNA
Common NEET traps for this chapter
⚠ Perpendicular axis theorem scope
Perpendicular axis theorem (Iz = Ix + Iy) applies ONLY to plane laminas, NOT to 3D objects.
✓ Fix: Perpendicular axis = 2D laminas only. Parallel axis = all objects.
⚠ Rolling KE formula
Rolling KE is NOT just ½Mv². It includes rotational KE: ½Mv² + ½Iω². The fraction depends on shape.
✓ Fix: Total KE_rolling = ½Mv²(1 + k²/R²)
⚠ Conservation of L condition
Angular momentum conserved only if net external torque = 0, NOT if net force = 0.
✓ Fix: τ_ext = 0 → L conserved. F_ext = 0 → p conserved.
Chapter Intelligence
Exam data and last-minute strategy
NEET Frequency
2–3 Q/year. MOI standard shapes, parallel/perpendicular axis theorems, angular momentum conservation, rolling KE are common.
High-Yield
MOI table (ring/disc/sphere). I₁ω₁ = I₂ω₂. Rolling: KE = ½Mv²(1+k²/R²). Perp. axis = laminas only. Disc beats ring on slope.
Strategy
Memorise MOI for 5 standard shapes. Practice angular momentum conservation problems (skater-type). Rolling problems: always use total KE formula.
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