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APEX CLASS ESTD 2018

IITJEE: Moment of Inertia, Torque & Rolling Motion Explorer

By

Ashish Kumar Bahekar

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Apex class • IITJEE Rotational Mechanics Lab

Moment of Inertia (I), Torque & Rolling Motion Explorer

Master Incline Rolling Races, Instantaneous Axis of Rotation (IAOR), and Axis Theorems!

🏁 Tap a Pure Rolling Concept to Inspect Velocities, Forces & Energy:
θ 3rd: Ring (β = 1) 2nd: Disc (β = 0.5) 🏆 1st: Solid Sphere (β = 0.4) a = g sinθ / (1 + β) 🥇 Sphere: (5/7) g sinθ 🥈 Disc: (2/3) g sinθ 🥉 Hollow Sph: (3/5) g sinθ 4️⃣ Ring: (1/2) g sinθ Smaller β = Faster Win! 🎯 JEE Race Rule: Acceleration & Final Speed depend ONLY on shape factor β = I/(MR²)—NEVER on Mass M or Radius R! Pure Translation (v) + Pure Rotation (ωR = v) = v_top = 2v! v_com = v ★ IAOR Contact: v_bottom = 0! ⚡ Speed at any point on rim at angle θ from bottom: v_net = 2v sin(θ/2) (So at θ=90° side point, v = √2 v)! Total Rolling Kinetic Energy: K_total = ½ Mv² (1 + β) | Ratio K_rot / K_trans = β 1. Ring (β = 1): K_trans = 50% (½ K_total) K_rot = 50% (½ K_total) 2. Solid Disc (β = ½): K_trans = 66.7% (⅔ K_total) K_rot = 33.3% (⅓) 3. Solid Sphere (β = ⅖): K_trans = 71.4% (⁵⁄₇ K_total) K_rot = 28.6% (²⁄₇) 🔋 Why Sphere Wins the Race: It wastes the LEAST energy in spinning (only 2/7) and puts 5/7 into forward speed! mg sinθ (Pulls COM) Static Friction f_s ↑ (Gives Torque τ = f_s R) ⚠️ Work Done by Friction in Pure Rolling = ZERO (because bottom contact point velocity v_contact = 0)!
Incline Acceleration (a)
a = g sinθ / (1 + β)
Sphere (5/7) > Disc (2/3) > Ring (1/2)
Bottom Velocity (v)
v = √[2gh / (1 + β)]
Less than slipping √(2gh)
Required Static Friction
f_s = mg sinθ / (1 + 1/β)
Zero on Smooth Incline (Slips!)
Min Coefficient (μ_min)
μ_min = tanθ / (1 + 1/β)
Sphere: (2/7)tanθ | Ring: ½tanθ
🎡 Select a Rigid Body or Axis Theorem to Inspect Moment of Inertia (I = βMR²):
R Thin Ring / Hollow Cylinder I_center = M R² (β = 1.0) Radius of Gyration: K = R 💡 All mass M sits at maximum distance R from the axis → Highest possible Moment of Inertia! R Uniform Solid Disc / Cylinder I_center = ½ M R² (β = 0.5) Radius of Gyration: K = R / √2 💡 Because mass is distributed from r = 0 to r = R, its average r² is half of R²! Solid: ⅖ MR² Hollow: ⅔ MR² 🌐 3D Sphere Comparison: Solid Sphere (β = 0.4) has lower inertia than Hollow Spherical Shell (β = 0.67)! I_com = ½ MR² d = R ★ Rim Axis: I = I_com + MR² = ³⁄₂ MR² Diameter Axis: I_d = ½ I_z = ¼ MR² 📐 Parallel Axis: I = I_com + Md² (All 3D/2D bodies) | Perpendicular Axis: I_z = I_x + I_y (Flat 2D Laminas ONLY!)
Central Axis (I_com)
M R² ½ M R² ⅖ MR² vs ⅔ MR² I_com (Must pass thru COM!)
Minimum Inertia Axis
About Diameter Axis
½ M R² (By I_z = 2I_d) ¼ M R² (By I_z = 2I_d) ⅖ MR² & ⅔ MR² I_d = ½ I_z (Symmetric 2D)
Perpendicular Axis Theorem
About Tangent (⊥ to Plane)
MR² + MR² = 2 M R² ½MR² + MR² = ³⁄₂ M R² ⁷⁄₅ MR² vs ⁵⁄₃ MR² I = I_com + M R²
Parallel Axis Shift d = R
Radius of Gyration (K)
K = R K = R / √2 √(2/5) R vs √(2/3) R K = √(I / M)
Since I = M K²
🔥 Angular Momentum Conservation Rule (τ_ext = 0 → I₁ω₁ = I₂ω₂):
When a skater pulls their arms in, or Earth shrinks to half its radius (R → R/2), Moment of Inertia drops to I/4, so Angular Velocity ω becomes 4× (and Length of Day becomes 24/4 = 6 hours)!
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