Rotational Motion

Why This Chapter Matters

Rotation is a high-weightage JEE topic — 8-12 marks. Moment of inertia, torque, angular momentum, and rolling motion are all tested. This chapter has many analogies with linear motion.

Core Concepts

Linear vs Rotational Analogies

|---|---|

Moment of Inertia (I)

I = Σmᵢrᵢ² (sum of mass × distance² from axis)

Common values (mass M, dimension given):

Ring (radius R, about center): I = MR²

Disc (radius R, about center): I = MR²/2

Solid sphere (about diameter): I = 2MR²/5

Hollow sphere (about diameter): I = 2MR²/3

Rod (length L, about center): I = ML²/12

Rod (length L, about end): I = ML²/3

Parallel axis theorem: I = I_cm + Md²

(I about any axis = I about parallel axis through CM + Md² where d = distance between axes)

Perpendicular axis theorem (for flat plates only):

I_z = I_x + I_y (I about axis ⊥ to plate = sum of I about two axes in plane)

Torque

τ = r × F = rF sinθ (in 2D)

τ = Iα (Newton's second law for rotation)

Net torque = 0 → angular acceleration = 0 → constant angular velocity

Angular Momentum

L = Iω = r × p

τ = dL/dt

Conservation of angular momentum: if τ_net = 0, L = constant

Applications: ice skater pulling arms in (I decreases → ω increases to keep L constant)

Rolling Motion

For rolling without slipping: v = Rω (contact point has zero velocity)

Total KE = ½mv² + ½Iω² = ½mv²(1 + I/mR²)

For solid sphere: total KE = ½mv²(1 + 2/5) = 7mv²/10

For disc: total KE = ½mv²(1 + 1/2) = 3mv²/4

For ring: total KE = ½mv²(1 + 1) = mv²

Acceleration down incline: a = g sinθ/(1 + I/mR²)

Solid sphere rolls faster than disc which rolls faster than ring (least I fraction → most a)

PYQs

2024: Solid cylinder (mass M, radius R) rolls without slipping. Ratio of rotational to total KE?

KE_rot/KE_total = (½Iω²)/(½mv² + ½Iω²) = (I/mR²)/(1 + I/mR²) = (1/2)/(1+1/2) = 1/3

2023: Disc of mass M radius R. Moment of inertia about tangential axis in plane of disc?

I_diameter = MR²/4. I_tangent = I_diameter + MR² = 5MR²/4 (parallel axis theorem)

2022: Child of mass m sits at edge of rotating disc (I₀, ω₀). Disc + child: final ω?

L = I₀ω₀ = (I₀ + mR²)ω → ω = I₀ω₀/(I₀ + mR²)

Revision Notes

KEY MOMENTS OF INERTIA:
Ring: MR² | Disc: MR²/2 | Solid sphere: 2MR²/5
Hollow sphere: 2MR²/3 | Rod (center): ML²/12

PARALLEL AXIS: I = I_cm + Md²
PERPENDICULAR AXIS (flat bodies): I_z = I_x + I_y

TORQUE: τ = Iα = r×F
ANGULAR MOMENTUM: L = Iω, conserved if τ_net = 0

ROLLING (no slip): v = Rω
Total KE = ½mv²(1 + I/mR²)
Acceleration on incline = g sinθ/(1 + I/mR²)
Sphere > Disc > Ring (down an incline speed)