Definition
- Property of a body by virtue of which it opposes change in its rotational state of motion
- Analogue of mass in translational motion: I = Σmr² = ∫r²dm
- Scalar (or tensor); unit = kg·m²; dimensional formula [ML²]
- Always ≥ 0; depends on mass, distribution of mass, and choice of axis
- Does NOT depend on: torque, angular acceleration, angular velocity
For point mass: I = mr²
For n masses: I = m₁r₁² + m₂r₂² + m₃r₃² + ...
Radius of gyration K: I = MK² → K = √(I/M)
K = equivalent distance from axis at which all mass can be considered to be concentrated
Theorems on MOI
① Parallel Axis Theorem
I = ICM + Md²
Valid for ALL objects (1D, 2D, 3D). Both axes must be parallel; one axis MUST pass through COM.
ICM = MOI about axis through COM; d = perpendicular distance between axes; M = total mass
② Perpendicular Axis Theorem
Iz = Ix + Iy
Valid ONLY for planar (2D) objects. z-axis is perpendicular to the plane; x and y are in the plane.
MOI of Common Bodies (About Axis Through COM)
| Body | Axis | ICM | K² |
|---|
| Thin Rod (length L) | ⊥ to rod, through centre | ML²/12 | L²/12 |
| Thin Rod (length L) | ⊥ to rod, through end | ML²/3 | L²/3 |
| Ring (radius R) | ⊥ to plane (through centre) | MR² | R² |
| Ring (radius R) | Diameter | MR²/2 | R²/2 |
| Ring (radius R) | Tangential (⊥ to plane) | 2MR² | 2R² |
| Disc (radius R) | ⊥ to plane (through centre) | MR²/2 | R²/2 |
| Disc (radius R) | Diameter | MR²/4 | R²/4 |
| Solid Cylinder (R) | Geometric axis | MR²/2 | R²/2 |
| Hollow Cylinder (R) | Geometric axis | MR² | R² |
| Solid Sphere (R) | Diameter | 2MR²/5 | 2R²/5 |
| Hollow Sphere (R) | Diameter | 2MR²/3 | 2R²/3 |
| Square plate (side a) | ⊥ to plate, through centre | Ma²/6 | a²/6 |
| Rectangular plate (a×b) | ⊥ to plate, through centre | M(a²+b²)/12 | (a²+b²)/12 |
| Annular disc (R₁, R₂) | ⊥ to plane | M(R₁²+R₂²)/2 | (R₁²+R₂²)/2 |
⚡ Order of MOI (Same Mass, Same Radius)
IRing > IHollow Cyl > IHollow Sphere > IDisc > ISolid Cyl > ISolid Sphere
In terms of MR² factor: 1 > 1 > 2/3 > 1/2 > 1/2 > 2/5
More hollow → more I (mass farther from axis)
Mass Removed from Body
- Iremaining = Ioriginal − Iremoved
- First calculate I of whole body, then subtract I of removed part (about same axis)
Q: A uniform disc of mass M, radius R has a circular hole of radius R/2 at its centre. Find MOI about axis through centre ⊥ to plane.
I = Idisc − Ihole = MR²/2 − (M/4)(R/2)²/2... use correct mass: Mhole = M × (R/2)²/R² = M/4; Ihole = (M/4)(R/2)²/2 × 2... actually = (M/4)R²/8 = MR²/32; I = MR²/2 − MR²/32 = 15MR²/32
Ans: 15MR²/32 ✓ (use Idisc = ½mr²)
Definition
- Torque = turning effect of force = τ = r × F
- |τ| = rF sinθ = F × d (d = perpendicular distance from axis to line of action of F)
- Vector quantity; direction: perpendicular to plane of r and F (right-hand rule)
- Unit = N·m; Dimensional formula [ML²T⁻²] (same as energy, but NOT energy)
- Analogue of force in rotational motion: τnet = Iα
- Clockwise torque → negative; Anticlockwise → positive
τ = Iα (Newton's 2nd law for rotation)
For equilibrium: ΣF = 0 (translational) AND Στ = 0 (rotational)
τnet = 0 → angular acceleration α = 0 → ω = constant
Translational vs Rotational Analogues
- Displacement s ↔ Angular displacement θ
- Velocity v ↔ Angular velocity ω
- Acceleration a ↔ Angular acceleration α
- Mass m ↔ Moment of Inertia I
- Force F ↔ Torque τ
- Linear momentum p = mv ↔ Angular momentum L = Iω
- KE = ½mv² ↔ KE = ½Iω²
- Power P = Fv ↔ Power P = τω
- Impulse = FΔt ↔ Angular Impulse = τΔt = ΔL
| Translational | Rotational |
|---|
| v = u + at | ω = ω₀ + αt |
| s = ut + ½at² | θ = ω₀t + ½αt² |
| v² = u² + 2as | ω² = ω₀² + 2αθ |
| snth = u + a(2n−1)/2 | θnth = ω₀ + α(2n−1)/2 |
| S1st:S2nd:S3rd = 1:3:5 (u=0) | θ1st:θ2nd:θ3rd = 1:3:5 (ω₀=0) |
Definition
- L = r × p = r × mv → vector quantity
- |L| = rp sinθ = mvr sinθ
- About axis: L = Iω
- τnet = dL/dt (angular impulse-momentum theorem)
- Depends on frame of reference (choice of origin)
- For object moving in straight line: L = mvd (d = perpendicular distance from axis to line of motion)
- If θ = 90° (object moving in circular path): L = mvr (maximum)
- If θ = 0° or 180°: L = 0
For projectile at max height: L = m × u cosθ × H = m u cosθ × u²sin²θ/(2g)
At time t: L = m × u cosθ × [u sinθ × t − ½gt²]
Angular momentum changes if there is a net torque; constant if τ_net = 0
5
Conservation of Angular Momentum
Law
- If τext = 0 → L = constant → I₁ω₁ = I₂ω₂
- When I decreases → ω increases (and vice versa)
- Examples: spinning ice skater pulling arms in (I↓ → ω↑), diver tucking in
Collision on Rotating System
Same direction: Lfinal = (I₁ω₁ + I₂ω₂)/(I₁ + I₂) → ωcommon = (I₁ω₁ + I₂ω₂)/(I₁ + I₂)
Opposite direction: ωcommon = (I₁ω₁ − I₂ω₂)/(I₁ + I₂)
KE lost in collision: ΔKE = I₁I₂(ω₁−ω₂)²/[2(I₁+I₂)]
Q: A disc I₁ = 4 kg·m² rotating at ω₁ = 10 rad/s has another disc I₂ = 2 kg·m² (at rest) placed on it. Find common angular velocity.
τ_ext = 0 → L conserved; I₁ω₁ = (I₁+I₂)ω₂ → 4×10 = 6×ω₂ → ω₂ = 40/6 = 20/3 rad/s
Ans: ω = 20/3 ≈ 6.67 rad/s ✓
Pure Rolling Condition
- Rolling = Translational + Rotational motion combined
- Pure rolling condition: vCM = Rω (no slipping at contact point)
- Velocity of contact point A = 0 (for pure rolling on ground)
- Velocity of top point = 2vCM
KEtotal = ½MvCM² + ½Iω² = ½MvCM²(1 + K²/R²)
where β = K²/R² = I/(MR²) → KEtotal = ½Mv² × (1 + β)
Fraction of translational KE = 1/(1+β) = 1/(1+K²/R²)
Fraction of rotational KE = β/(1+β) = K²/R²/(1+K²/R²)
| Body | β = K²/R² | KEtrans/KEtotal | KErot/KEtotal |
|---|
| Ring / Hollow Cyl. | 1 | 50% | 50% |
| Hollow Sphere | 2/3 | 60% | 40% |
| Disc / Solid Cyl. | 1/2 | 67% | 33% |
| Solid Sphere | 2/5 | 71% | 29% |
Rolling on Inclined Plane
| Quantity | Formula |
|---|
| Acceleration of CM | aCM = g sinθ / (1 + K²/R²) = βg sinθ / (K²/R²+1) |
| Velocity at bottom (height H) | vCM = √(2gH / (1 + K²/R²)) = √(2gHβ ... use COME) |
| Time to reach bottom | t = √(2S/aCM) = √(2S(1+K²/R²)/g sinθ) |
| Max height on rough incline | H = v²(1+K²/R²)/(2g) |
| Static friction | fstatic = mg sinθ(K²/R²)/(1+K²/R²) |
⚡ Order on Inclined Plane (Same slope — which reaches bottom first?)
- Greater β → smaller acceleration → reaches bottom LATER
- Solid Sphere (fastest) > Disc > Hollow Sphere > Ring (slowest)
- This order is independent of mass and radius!
- On smooth incline → only translational; no rolling → a = g sinθ (fastest for all)
Slipping Conditions
- V > Rω → translational dominates → forward slipping → kinetic friction acts backward
- V < Rω → rotational dominates → backward slipping → kinetic friction acts forward
- V = Rω → pure rolling → static friction acts (no energy loss due to friction)
Q: A solid sphere rolls down a rough incline of angle 30°. Find acceleration.
For solid sphere: K²/R² = 2/5; a = g sinθ/(1+2/5) = 10×0.5/(7/5) = 5×5/7 = 25/7 ≈ 3.57 m/s²
Ans: a = 25/7 m/s² ✓
Q: A ring and a disc start from rest at top of incline (same height). Which reaches bottom first?
βring = 1, βdisc = 1/2; adisc > aring (smaller β → more acceleration) → Disc reaches first
Ans: Disc reaches bottom first ✓
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Velocity at Different Points in Rolling
Velocity at Points on a Rolling Body
- Contact point A (bottom): vA = 0 (for pure rolling)
- Centre (CM): vCM = v
- Top point: vtop = 2v
- At angle θ from bottom: vP = 2v sin(θ/2)
- At θ = 0°: v = 2v sin 0 = 0 ✓ (contact point)
- At θ = 90°: v = 2v sin 45° = v√2 (side)
- At θ = 180°: v = 2v sin 90° = 2v ✓ (top)
Toppling Condition
- A block topples (tips over) when torque due to applied force exceeds torque due to gravity about the tipping edge
- Toppling occurs when: F × h ≥ Mg × (a/2) → F ≥ Mg × a/(2h)
- Minimum force to topple: Fmin = Mga/(2h)
- If F (applied at height h) → for toppling: F/Mg ≥ a/(2h)
Q: A box of mass M, base a, height 2h on rough surface. Horizontal force applied at top. Find min force to topple.
Torque about bottom edge: F×2h ≥ Mg×a/2 → F ≥ Mga/(4h)
Ans: Fmin = Mga/(4h) ✓
Q: An ice skater spins with arms outstretched (I₁ = 5 kg·m², ω₁ = 2 rad/s). She pulls arms in (I₂ = 1 kg·m²). Find ω₂.
No external torque → L = const; I₁ω₁ = I₂ω₂ → 5×2 = 1×ω₂ → ω₂ = 10 rad/s
Ans: ω₂ = 10 rad/s ✓