Rolling Without Slipping and Angular Momentum Conservation
Systems of Particles and Rotational Motion: Rolling Without Slipping and Angular Momentum Conservation
Rolling Without Slipping and Angular Momentum Conservation
Rolling Without Slipping and Angular Momentum Conservation
What you'll learn
- Rolling condition: v_cm = Rω — translation and rotation linked by no-slip constraint.
- Total KE of rolling body: KE = ½mv_cm² + ½Iω² = ½mv_cm²(1 + I/mR²).
- Angular momentum: L = Iω (rigid body), L = r × p (particle); SI unit kg·m²·s⁻¹.
- Conservation of angular momentum: If τ_ext = 0, then L = constant.
- Application: rolling on incline (acceleration depends on I), ice skater spin, planetary orbits.
Key concepts
Level 1 — Foundations
Verbal: A body rolls without slipping when the contact point has zero velocity relative to ground — rotation and translation are perfectly matched.
Rolling condition: v_cm = Rω; equivalently a_cm = Rα.
Total KE (rolling): KE_total = KE_translational + KE_rotational = ½mv² + ½Iω². Substituting ω = v/R: KE = ½mv²(1 + I/mR²).
KE fractions for common shapes:
| Shape | I | I/mR² | KE_rot / KE_total |
|---|---|---|---|
| Ring | mR² | 1 | 50% |
| Hollow sphere | ²⁄₃ mR² | 2/3 | 40% |
| Solid disc | ½mR² | 1/2 | 33.3% |
| Solid sphere | ²⁄₅ mR² | 2/5 | 28.6% |
Angular momentum: L = Iω for rotation; L = mvr (for particle moving in circle).
Conservation: If net external torque = 0, ΔL = 0 → L_initial = L_final.
Level 2 — JEE / NEET depth
Rolling on incline (smooth): Using energy conservation (no friction work in pure rolling): mgh = ½mv²(1 + I/mR²) → v = √(2gh / (1 + I/mR²)). Smaller I/mR² → faster rolling → solid sphere reaches bottom first.
Acceleration on incline: a = g sinθ / (1 + I/mR²). Solid sphere: a = 5g sinθ/7; Solid disc: a = 2g sinθ/3; Ring: a = g sinθ/2.
Condition for pure rolling on rough incline: Friction f = mI sinθ / (mR² + I) — friction causes rotation but does no net work.
Angular momentum of system: L_total = L_cm + L_about_cm = mv_cm r + I_cm ω (for rolling: both terms present).
Conservation examples:
- Ice skater pulls in arms: I decreases → ω increases (L constant).
- Diver in tucked position: smaller I → faster spin.
- Earth-Moon system: angular momentum conserved (Moon slowly recedes).
τ_ext = dL/dt: This is the fundamental rotational law (Newton's 2nd law, angular form).
Worked example
Solid sphere rolling down incline
A solid sphere (M = 2 kg, R = 0.1 m) rolls without slipping from rest
down a 30° incline of length L = 5 m. Find speed at bottom and time taken.
Step 1 — Height: h = L sinθ = 5 × 0.5 = 2.5 m
Step 2 — For solid sphere: I = ²⁄₅ MR², so I/MR² = 2/5.
Step 3 — Speed at bottom (energy conservation):
v = √(2gh / (1 + I/MR²)) = √(2 × 10 × 2.5 / (1 + 2/5))
= √(50 / 1.4) = √35.71 ≈ 5.98 m/s
Step 4 — Acceleration along incline:
a = g sinθ / (1 + I/MR²) = 10 × 0.5 / 1.4 = 5/1.4 ≈ 3.57 m/s²
Step 5 — Time from rest: v = at → t = v/a = 5.98/3.57 ≈ 1.67 s
Step 6 — Check with kinematics: v² = 2aL → v = √(2×3.57×5) = √35.7 ≈ 5.97 m/s ✓
Step 7 — KE partition:
KE_total = ½mv² = ½×2×35.7 = 35.7 J
KE_rot = (2/5)/(1+2/5) × 35.7 = (2/7) × 35.7 = 10.2 J (28.6%)
Ice skater angular momentum conservation
A skater spins at ω₀ = 2 rad/s with arms extended (I₀ = 4 kg·m²).
She pulls her arms in to I₁ = 1.5 kg·m². Find new ω.
Step 1 — No external torque (ice is frictionless for rotation here).
L = I₀ω₀ = I₁ω₁
Step 2 — Conservation:
4 × 2 = 1.5 × ω₁
ω₁ = 8/1.5 = 5.33 rad/s
Step 3 — KE check:
KE₀ = ½I₀ω₀² = ½×4×4 = 8 J
KE₁ = ½I₁ω₁² = ½×1.5×28.4 = 21.3 J
KE increased — muscle work done pulling arms in provides extra energy.
Common mistakes
| Mistake | Why it happens | Fix |
|---|---|---|
| Using only ½mv² for rolling KE | Forgetting rotational KE | Rolling KE = ½mv²(1 + I/mR²) |
| Same speed at bottom for all rolling shapes | I/mR² differs by shape | Solid sphere fastest; ring slowest on incline |
| L conservation when friction present | Friction can create torque | Check if τ_external = 0 before applying conservation |
| ω = v/r vs ω = v_cm/R confusion | Multiple points on body | Only CM speed relates to ω by v_cm = Rω |
Quick check
- State the rolling condition. What is the velocity of the contact point in pure rolling?
- A solid disc rolls at v_cm = 4 m/s (m=3 kg, R=0.2 m). Find total KE.
- Why does a solid sphere reach the bottom of an incline before a hollow sphere of same mass?
- A diver reduces I from 6 to 2 kg·m² during a somersault at ω=1 rad/s. Find new ω.
- Stretch: Compare the time for a ring and solid disc to roll down the same incline from rest.
NCERT Chapter 7 link: Rolling without slipping ties together translation (v_cm) and rotation (ω) through one constraint: v_cm = Rω. Angular momentum conservation is as powerful as linear momentum conservation — apply it whenever external torque is absent.
Exam connections: JEE tests: speed at bottom of incline for various shapes (sphere always fastest, ring always slowest); angular momentum conservation for spinning bodies changing I; total KE split between translation and rotation. Classic: a sphere, disc, and ring released from same height — sphere always wins.
Study strategy: Memorise the I/mR² ratio for each shape and the speed formula v = √(2gh/(1+I/mR²)). For L conservation problems, check first: is τ_net = 0? If yes, write L_i = L_f.
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Key Takeaways (TL;DR)
- What you'll learn
- Key concepts
- Worked example
- Common mistakes
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