Rotational Motion & Torque
Practice JEE Main Rotational Motion PYQs online. Moment of inertia theorems, angular momentum conservation, rolling motion without slipping & solutions.
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Rotational Motion & Torque:
- Center of Mass (COM) Calculation & Motion of COM
- Moment of Inertia (Parallel & Perpendicular Axis Theorems)
- Torque τ = r × F = I α and Angular Work W = ∫ τ dθ
- Conservation of Angular Momentum (L = I ω = constant)
- Pure Rolling Without Slipping on Incline (a = g sin θ / (1 + k²/R²))
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Rotational Motion questions in under 45 seconds:
Rolling acceleration down incline: a = (g sin θ) / (1 + I / (mR²)). Smaller I means faster descent (Solid Sphere reaches bottom first).
Parallel Axis Theorem: I = I_cm + M d².
Angular momentum conservation: I₁ ω₁ = I₂ ω₂.
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: √2 : 1
- Step 1: For a ring of mass M and radius R, moment of inertia is I_ring = M R² ⇒ radius of gyration k_ring = R.
- Step 2: For a disc of mass M and radius R, moment of inertia is I_disc = ½ M R² ⇒ radius of gyration k_disc = R / √2.
- Step 3: The ratio k_ring / k_disc = R / (R / √2) = √2 / 1 = √2 : 1.
- Step 4: Ratio is √2 : 1.
⚡ Speed Tip: Gyration Ratio: k = √(I / M). k_ring / k_disc = √(MR² / (½MR²)) = √2 : 1.
💡 View Step-by-Step Solution & Answer
Correct Answer: 5/7 g sin θ
- Step 1: Acceleration of a body rolling without slipping down an incline is a = g sin θ / [1 + I / (M R²)].
- Step 2: For a solid sphere, the moment of inertia about its diameter is I = (2/5) M R².
- Step 3: I / (M R²) = 2/5.
- Step 4: a = g sin θ / [1 + 2/5] = g sin θ / (7/5) = 5/7 g sin θ.
⚡ Speed Tip: Rolling Acceleration Formula: a = g sin θ / (1 + 2/5) = 5/7 g sin θ.
💡 View Step-by-Step Solution & Answer
Correct Answer: 6 N·m
- Step 1: Angular acceleration α = (ω_final - ω_initial) / t = (0 - 30) / 10 = -3 rad/s².
- Step 2: Angular deceleration magnitude is α = 3 rad/s².
- Step 3: Torque relation is τ = I · α.
- Step 4: τ = 2 kg·m² × 3 rad/s² = 6 N·m.
⚡ Speed Tip: Torque = I · α = 2 × (30 / 10) = 2 × 3 = 6 N·m.
❓ Frequently Asked Questions (FAQs)
Which rolling object reaches bottom of an incline first?
Solid Sphere (k²/R² = 2/5 = 0.4) has the smallest rotational inertia ratio, so it has highest acceleration and reaches first.