C12 PHY CH1
Chapter 1
~5 Marks
Unit 1 — Mechanics
Rotational Dynamics
Torque, moment of inertia, angular momentum, and rotational kinetic energy with rolling motion — the foundation of Class 12 mechanics and a high-yield topic for IOE/CEE entrances.
Torque
Moment of Inertia
Angular Momentum
Rotational KE
Rolling Motion
Conservation Laws
Core Concepts & Definitions
Learn these precisely — 2-mark definition questions come directly from here
Torque (τ)
The rotational equivalent of force. It is the cross product of the position vector and applied force: τ = r × F = rF sinθ. SI unit: N·m. A body rotates when a net torque acts on it.
Moment of Inertia (I)
The rotational equivalent of mass — it measures a body's resistance to angular acceleration. Defined as I = Σmr². Depends on mass distribution relative to the axis of rotation.
Angular Momentum (L)
The rotational equivalent of linear momentum. L = Iω = r × p. It is conserved when no external torque acts on the system (Conservation of Angular Momentum).
Angular Velocity (ω) & Acceleration (α)
Angular velocity is rate of change of angular displacement: ω = dθ/dt. Angular acceleration: α = dω/dt. Relates to linear: v = rω, a = rα.
Rotational Kinetic Energy
Energy due to rotational motion: KEⱼⱼᵗ = ½Iω². For rolling body: total KE = translational KE + rotational KE = ½mv² + ½Iω².
Key Formulas
Memorize all — numericals are built around these
Torque
τ = r × F = rF sinθ = Iα
τ = torque (N·m), r = perpendicular distance, F = force, I = moment of inertia, α = angular acceleration
Moment of Inertia — Common Bodies
Solid Sphere: I = (2/5)mr²
Hollow Sphere: I = (2/3)mr²
Solid Cylinder: I = (1/2)mr²
Thin Rod (end): I = (1/3)ml²
Thin Rod (mid): I = (1/12)ml²
Hollow Sphere: I = (2/3)mr²
Solid Cylinder: I = (1/2)mr²
Thin Rod (end): I = (1/3)ml²
Thin Rod (mid): I = (1/12)ml²
Angular Momentum & Conservation
L = Iω = mvr (for point mass)
τ = dL/dt
If τ = 0 ⇒ L = constant ⇒ I₁ω₁ = I₂ω₂
τ = dL/dt
If τ = 0 ⇒ L = constant ⇒ I₁ω₁ = I₂ω₂
Rolling Motion (no slipping)
v = rω
KEᵀᵒᵗᵃᴸ = ½mv² + ½Iω²
= ½mv²(1 + I/mr²)
Solid sphere rolling: KE = (7/10)mv²
KEᵀᵒᵗᵃᴸ = ½mv² + ½Iω²
= ½mv²(1 + I/mr²)
Solid sphere rolling: KE = (7/10)mv²
For incline: acceleration a = g sinθ / (1 + I/mr²)
Rotational Equations of Motion
ω = ω₀ + αt
θ = ω₀t + ½αt²
ω² = ω₀² + 2αθ
θ = ω₀t + ½αt²
ω² = ω₀² + 2αθ
Exact analogue of v = u + at, s = ut + ½at², v² = u² + 2as
Moment of Inertia — Quick Reference Table
Frequently tested in both NEB board and entrance exams
| Body | Axis | MOI (I) |
|---|---|---|
| Solid Sphere | Diameter | (2/5)mr² |
| Hollow Sphere | Diameter | (2/3)mr² |
| Solid Cylinder / Disk | Central axis | (1/2)mr² |
| Hollow Cylinder / Ring | Central axis | mr² |
| Thin Rod | Through center | (1/12)ml² |
| Thin Rod | Through one end | (1/3)ml² |
| Rectangular plate | Through center | (1/12)m(a²+b²) |
Core Derivation — Rotational KE of Rolling Body
4–5 mark derivation — learn step by step
1
Total KE = Translational + Rotational
A rolling body has two types of kinetic energy simultaneously.
KE = ½mv² + ½Iω²
2
Substitute rolling condition: v = rω ⇒ ω = v/r
For pure rolling without slipping, the contact point has zero velocity.
KE = ½mv² + ½I(v/r)²
3
Factor out ½mv²
Rearrange to express as a single term multiplied by translational KE.
KE = ½mv²(1 + I/mr²)
4
For a Solid Sphere: I = (2/5)mr²
Substitute the moment of inertia for a solid sphere.
KE = ½mv²(1 + 2/5) = (7/10)mv²
5
Conclusion
A solid sphere rolling down an incline has 7/10 of its total energy as translational and 2/10 as rotational (since KEᵀᵒᵗ = (2/10)mv²).
Worked Numerical Examples
3–4 mark numericals — follow the format: formula → substitution → answer + unit
Example 1 — Torque
A force of 20 N acts at a perpendicular distance of 0.5 m from the axis of rotation. Find the torque.
Given: F = 20 N, r = 0.5 m, θ = 90° (perpendicular)
Formula: τ = rF sinθ
Solution: τ = 0.5 × 20 × sin 90° = 0.5 × 20 × 1
τ = 10 N·m
Example 2 — Angular Momentum Conservation
A skater has MOI = 4 kg·m² at ω = 2 rad/s. She pulls her arms in so I becomes 1 kg·m². Find new ω.
Using: I₁ω₁ = I₂ω₂ (no external torque)
Solution: 4 × 2 = 1 × ω₂
ω₂ = 8 rad/s
Example 3 — Rolling Motion
A solid sphere of mass 2 kg rolls with v = 5 m/s. Find its total kinetic energy.
For solid sphere: KE = (7/10)mv²
Solution: KE = (7/10) × 2 × 5² = (7/10) × 2 × 25
KE = 35 J
Quick Revision — Key Points
Read these the night before the exam
Torque & Newton's 2nd
τ = Iα is the rotational analogue of F = ma. Always check the axis when computing I.
L Conservation
When τₑₓᵗ = 0, angular momentum is conserved: I₁ω₁ = I₂ω₂. Classic example: spinning skater.
Rolling Condition
Pure rolling (no slip): v = rω. Smallest I/mr² rolls fastest down an incline — solid sphere wins.
Work-Energy (Rotation)
Work done by torque: W = τ × θ. Power: P = τω. Same structure as W = F·d, P = Fv.
