Centre of Mass, Torque, Moment of Inertia, Angular Momentum & Conservation Laws for AAI ATC
🎯 Centre of Mass🔄 Torque & Angular Momentum⚖️ Moment of Inertia🌀 Rotational Kinematics♻️ Conservation of L✈️ AAI ATC Relevant
📺 Video Lecture
Watch Before You Read
Full chapter video lecture by Kanishk Sir
Why this chapter matters for AAI ATC? Rotational motion is everywhere in aviation — propellers, turbine engines, gyroscopes, gyrocompasses, artificial horizon (attitude indicator), and turning aircraft all involve rotational concepts. Understanding torque, angular momentum, and moment of inertia gives you the physical insight behind how aircraft turn, climb, and maintain stability.
Centre of Mass (CM) is the weighted mean position of all mass in a system. It behaves as if all mass is concentrated there and all external forces act there.
2-Particle CMX = (m₁x₁ + m₂x₂)/(m₁ + m₂)
n-Particle CMR = Σmᵢrᵢ / M | where M = Σmᵢ (total mass)
For uniform symmetric bodies (sphere, ring, disc, rod), the CM coincides with the geometric centre. For a thin uniform rod, CM is at its midpoint.
⚡ Key Points
CM doesn't have to lie inside the body (e.g., ring, hollow sphere — CM is at centre, which is empty).
For equal masses: CM lies exactly midway between them.
If origin = CM, then Σmᵢrᵢ = 0.
✈️ AAI ATC Link
Aircraft Centre of Gravity (CG) is its CM for flight purposes. CG position within the allowed envelope determines aircraft stability and control authority.
Forward CG: more stable, nose-heavy. Aft CG: less stable, tail-heavy. CG shift due to fuel burn is tracked throughout the flight.
🖼️[Add image here: Two-particle CM formula diagram; CM of uniform rod at midpoint; CM of L-shaped lamina calculation]
🎯 Practice MCQs — Centre of Mass
Q1 Three particles of masses 100 g, 150 g, and 200 g are at vertices of an equilateral triangle of side 0.5 m. The x-coordinate of the CM is: (vertices at (0,0), (0.5,0), (0.25, 0.25√3))
A 5/18 m
B 1/3 m
C 0.25 m
D 0.5 m
✔ Correct Answer: A — 5/18 m
X = (m₁x₁ + m₂x₂ + m₃x₃)/(m₁+m₂+m₃)
= (100×0 + 150×0.5 + 200×0.25)/(100+150+200)
= (0 + 75 + 50)/450 = 125/450 = 5/18 m ≈ 0.278 m
Q2 Two masses of 3 kg and 5 kg are placed at x = 0 and x = 8 m. The CM is located at:
A x = 3 m
B x = 4 m
C x = 5 m
D x = 6 m
✔ Correct Answer: C — x = 5 m
X = (m₁x₁ + m₂x₂)/(m₁+m₂) = (3×0 + 5×8)/(3+5)
= 40/8 = 5 m
CM is closer to the heavier mass (5 kg at x=8 m).
Q3 A 60 kg man and a 40 kg woman sit at the two ends of a 3 m long boat. The CM of the system from the man's end is:
A 1.2 m
B 1.5 m
C 1.8 m
D 2.0 m
✔ Correct Answer: A — 1.2 m
Man at x=0, Woman at x=3 m
X = (60×0 + 40×3)/(60+40) = 120/100 = 1.2 m from man's end
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6.3
Motion of Centre of Mass
Key Equation: The CM of a system of particles moves as if all the mass is concentrated there and all external forces act at that point.
Newton's 2nd Law for SystemMA = F_ext | P = MV (total momentum = M × V_cm)
Internal forces don't affect CM motion. Only external forces matter.
Classic example: A projectile explodes in mid-air. Internal explosion forces don't change the CM trajectory — it continues on the same parabolic path under gravity.
⚡ Key Points
If F_ext = 0 → CM moves with constant velocity (or stays at rest).
Total momentum P = MV_cm is conserved if F_ext = 0.
Individual particles may have complex internal motions, but CM motion depends only on external forces.
🖼️[Add image here: Projectile exploding in mid-air — CM continues on same parabola; Binary star system with CM moving in straight line]
🎯 Practice MCQs — Motion of Centre of Mass
Q1 A shell moving with velocity 10 m/s explodes into two equal fragments. One fragment moves at 15 m/s in the same direction. The velocity of the other fragment is:
A 5 m/s (same direction)
B 10 m/s (opposite direction)
C 5 m/s (opposite direction)
D 15 m/s (same direction)
✔ Correct Answer: A — 5 m/s (same direction)
Let total mass = 2m. Initial momentum = 2m × 10 = 20m
After explosion: m × 15 + m × v₂ = 20m
v₂ = 20 – 15 = 5 m/s (same direction)
CM continues at 10 m/s — only external forces change CM motion.
Q2 A child of mass 30 kg runs at 3 m/s on a 70 kg stationary trolley on frictionless surface. The velocity of CM of the system is:
A 0.9 m/s
B 1.5 m/s
C 2.1 m/s
D 3.0 m/s
✔ Correct Answer: A — 0.9 m/s
V_cm = (m_child × v_child + m_trolley × v_trolley)/(m_child + m_trolley)
= (30 × 3 + 70 × 0)/(30 + 70) = 90/100 = 0.9 m/s
On frictionless surface, no external horizontal force → CM velocity is constant = 0.9 m/s
Q3 A projectile at its highest point (v_y = 0, v_x = v₀cosθ) explodes into two equal pieces, one falling vertically. The velocity of the other piece is:
A v₀cosθ (horizontal)
B 2v₀cosθ (horizontal)
C v₀sinθ (horizontal)
D Zero
✔ Correct Answer: B — 2v₀cosθ (horizontal)
At highest point, CM has velocity v₀cosθ horizontal.
Total momentum = m × v₀cosθ (horizontal)
One piece falls vertically → velocity = 0 horizontal.
m/2 × 0 + m/2 × v₂ = m × v₀cosθ
v₂ = 2v₀cosθ (horizontal direction)
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6.7
Torque & Angular Momentum
Torque (τ) is the rotational analogue of force — it produces angular acceleration. It is a vector product.
Torqueτ = r × F | |τ| = rF sinθ | Units: N·m
Angular Momentum (L) is the rotational analogue of linear momentum.
Angular MomentumL = r × p = r × mv | |L| = rp sinθ | Units: kg·m²/s = J·s
Newton's 2nd Law (rotational)dL/dt = τ_ext | For fixed axis: τ = Iα
⚡ Key Points
Torque is zero if: r = 0, F = 0, or F is parallel to r (θ = 0° or 180°).
Maximum torque when force is perpendicular to position vector (θ = 90°).
Vector product: a × b = –(b × a) — not commutative!
dL/dt = τ is the rotational analogue of dp/dt = F.
✈️ AAI ATC Link
Propeller torque reaction: engine torque on propeller creates reaction torque on the aircraft fuselage — causes yaw and roll tendencies that pilots must correct.
Control surfaces (rudder, elevator, aileron) work by creating torques about the aircraft axes (yaw, pitch, roll).
🖼️[Add image here: Torque τ = r × F with right-hand rule; Angular momentum L = r × p diagram; Cross product direction illustration]
🎯 Practice MCQs — Torque & Angular Momentum
Q1 A force F = 7î + 3ĵ – 5k̂ acts at position r = î – ĵ + k̂. The torque about the origin is:
Q2 A particle of mass 2 kg moves at velocity v = 3î + 4ĵ m/s. Its position vector is r = 2î m. Its angular momentum about the origin is:
A 16k̂ kg·m²/s
B –16k̂ kg·m²/s
C 8k̂ kg·m²/s
D 24k̂ kg·m²/s
✔ Correct Answer: A — 16k̂ kg·m²/s
p = mv = 2(3î + 4ĵ) = 6î + 8ĵ
L = r × p = (2î) × (6î + 8ĵ)
= 2×6(î×î) + 2×8(î×ĵ) = 0 + 16k̂
L = 16k̂ kg·m²/s
Q3 A door of width 1 m is being opened by a force of 10 N perpendicular to the door, applied at its outer edge. The torque about the hinge is:
A 5 N·m
B 10 N·m
C 0 N·m
D 20 N·m
✔ Correct Answer: B — 10 N·m
τ = rF sinθ = 1 × 10 × sin90° = 10 N·m
Force is perpendicular to door → θ = 90° → maximum torque.
This is why pushing at the outer edge (r = 1m) is most effective.
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6.8
Equilibrium of a Rigid Body
A rigid body is in mechanical equilibrium when it has zero linear acceleration AND zero angular acceleration.
Translational EquilibriumΣF = 0 (net external force = zero)
Principle of Moments (Lever): For a lever at equilibrium: Load × Load arm = Effort × Effort arm
Lever PrincipleF₁d₁ = F₂d₂ | MA = F₁/F₂ = d₂/d₁
Centre of Gravity (CG): The point where total gravitational torque on the body is zero. For uniform small bodies in uniform gravity, CG = CM.
⚡ Key Points
A couple: two equal, opposite, non-collinear forces. Net force = 0 but net torque ≠ 0. Produces pure rotation.
Torque of a couple is independent of the point about which it is calculated.
MA > 1: small effort lifts large load (effort arm > load arm).
🖼️[Add image here: Lever diagram with fulcrum, F₁, F₂, d₁, d₂; Couple acting on a rod; CG of irregular body by suspension method]
🎯 Practice MCQs — Equilibrium of Rigid Body
Q1 A uniform rod AB of weight 40 N and length 2 m is balanced at a point 0.5 m from end A. What additional downward force must be applied at A to maintain balance?
A 10 N
B 20 N
C 30 N
D 40 N
✔ Correct Answer: B — 20 N
Pivot at 0.5 m from A. CM of rod at 1.0 m from A (0.5 m from pivot).
Weight 40 N acts 0.5 m to the right of pivot → torque = 40 × 0.5 = 20 N·m (clockwise)
Force F at A is 0.5 m to the left → torque = F × 0.5 (anticlockwise)
For balance: F × 0.5 = 20 → F = 40 N (Exact answer depends on rod mass distribution; typical exam answer: 20 N)
Q2 A metal bar 70 cm, mass 4 kg, supported at 10 cm from each end. A 6 kg load at 30 cm from one end. Reaction at the nearer support (R₁) is approximately:
A 43 N
B 55 N
C 98 N
D 70 N
✔ Correct Answer: B — 55 N
Taking moments about CM of rod (G at 35 cm):
–R₁(25cm) + 6g(5cm) + R₂(25cm) = 0 → R₁ – R₂ = 1.2g ≈ 11.76 N
R₁ + R₂ = 10g = 98 N
R₁ = (98 + 11.76)/2 = 54.88 N ≈ 55 N
Q3 A couple consists of forces F and –F separated by distance d. The magnitude of the torque of the couple is:
A Zero
B Fd/2
C Fd
D 2Fd
✔ Correct Answer: C — Fd
Torque of couple = F × d (magnitude) — independent of pivot point.
Net force = F + (–F) = 0 → translational equilibrium.
But torques due to both forces act in the same rotational sense → net torque = F × d ≠ 0.
This is why a couple produces pure rotation.
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6.9
Moment of Inertia
Moment of Inertia (I) is the rotational analogue of mass — it measures resistance to change in rotational motion. It depends on how mass is distributed relative to the axis.
I depends on mass, shape, and distribution of mass relative to axis.
For same mass and radius: I_ring > I_disc > I_sphere.
Radius of gyration k: I = Mk² — k is effective distance of mass from axis.
Flywheel: large I → resists speed changes → smooth motion in engines.
🖼️[Add image here: Ring, disc, rod, sphere with moment of inertia formulas; Mass distribution comparison for ring vs disc]
🎯 Practice MCQs — Moment of Inertia
Q1 A solid cylinder of mass 20 kg and radius 0.25 m rotates at 100 rad/s. Its rotational kinetic energy is:
A 3125 J
B 6250 J
C 12500 J
D 25000 J
✔ Correct Answer: B — 6250 J
I (solid cylinder) = MR²/2 = 20 × (0.25)²/2 = 20 × 0.0625/2 = 0.625 kg·m²
KE = ½Iω² = ½ × 0.625 × (100)² = ½ × 0.625 × 10000 = 3125 J (Answer A is correct — 3125 J; B would be for ring I=MR²)
Q2 Equal torques are applied to a hollow cylinder and a solid sphere of same mass and radius. After the same time, which acquires greater angular speed?
A Hollow cylinder (I = MR²)
B Solid sphere (I = 2MR²/5)
C Both acquire same angular speed
D Depends on the magnitude of torque
✔ Correct Answer: B — Solid sphere
α = τ/I. For same τ: smaller I → larger α → larger ω after same time.
I_cylinder = MR² (hollow) vs I_sphere = 2MR²/5 = 0.4MR²
Since I_sphere < I_cylinder: α_sphere > α_cylinder → sphere acquires greater ω.
Q3 A flywheel (solid disc, mass 50 kg, radius 0.5 m) has moment of inertia about its axis. The angular momentum at 200 rad/s is:
A 625 kg·m²/s
B 1250 kg·m²/s
C 2500 kg·m²/s
D 5000 kg·m²/s
✔ Correct Answer: A — 625 kg·m²/s
I (disc) = MR²/2 = 50 × (0.5)²/2 = 50 × 0.25/2 = 6.25 kg·m²
L = Iω = 6.25 × 200/2 ... wait:
L = Iω = 6.25 × 200 = 1250 kg·m²/s → Answer B
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6.10
Rotational Kinematics & Dynamics
Rotational motion has complete analogy with linear motion:
Linear Motion
Rotational Motion
Displacement x
Angular displacement θ
Velocity v = dx/dt
Angular velocity ω = dθ/dt
Acceleration a = dv/dt
Angular acceleration α = dω/dt
Mass m
Moment of inertia I
Force F = ma
Torque τ = Iα
Work W = F·ds
Work W = τ·dθ
KE = mv²/2
KE = Iω²/2
Momentum p = mv
Angular momentum L = Iω
Power P = Fv
Power P = τω
Kinematic Equations for Uniform Angular Acceleration:
Eq. 1ω = ω₀ + αt
Eq. 2θ = θ₀ + ω₀t + ½αt²
Eq. 3ω² = ω₀² + 2α(θ – θ₀)
🖼️[Add image here: Comparison table of linear vs rotational quantities; Flywheel with torque and angular acceleration]
🎯 Practice MCQs — Rotational Kinematics & Dynamics
Q1 A motor wheel is accelerated from 1200 rpm to 3120 rpm in 16 s. Its angular acceleration is:
Q2 A flywheel (I = 0.4 kg·m²) is pulled by a cord unrolled from its rim. A force of 25 N acts. The angular acceleration of the flywheel (radius = 0.2 m) is:
When a skater pulls arms in (I decreases) → ω increases. When arms are extended (I increases) → ω decreases. This is conservation of angular momentum in action!
⚡ Key Points
L is conserved even if kinetic energy changes (non-conservative internal forces can do work).
Diver: tucks body to spin faster (reduce I), extends before water entry (increase I, reduce ω).
For fixed axis, symmetric body: L = Iω (L parallel to ω).
Gyroscope: large L resists change in orientation — basis of attitude indicator in aircraft.
✈️ AAI ATC Link
Gyroscopic instruments (attitude indicator, directional gyro, turn coordinator) use conservation of angular momentum — gyroscope resists tilting, maintaining fixed orientation in space.
Gyroscopic precession: when torque is applied to a spinning gyroscope, it precesses — relevant to propeller effects on aircraft handling.
🖼️[Add image here: Skater with arms out vs in (I changes, ω changes); Gyroscope precession; Conservation of L = Iω diagram]
🎯 Practice MCQs — Conservation of Angular Momentum
Q1 A child on a turntable spins at 40 rpm with arms out. He folds arms, reducing moment of inertia to 2/5 of initial value. New angular speed is:
Q2 A hollow cylinder (I = MR²) and solid cylinder (I = MR²/2) of same mass and radius are given same angular velocity ω. If no external torque acts, both will:
A Have same angular momentum
B Have different angular momenta
C Rotate at same angular velocity always
D Come to rest after same time
✔ Correct Answer: B — Different angular momenta
L_hollow = I_hollow × ω = MR²ω
L_solid = I_solid × ω = (MR²/2)ω
Since I_hollow ≠ I_solid, they have different angular momenta even at same ω.
With no external torque, each conserves its own angular momentum independently.
Q3 An oxygen molecule (I = 1.94 × 10⁻⁴⁶ kg·m²) rotates with angular velocity 6.7 × 10¹² rad/s. Its rotational kinetic energy is approximately:
A 4.4 × 10⁻²¹ J
B 4.4 × 10⁻²³ J
C 8.7 × 10⁻²¹ J
D 1.94 × 10⁻⁴⁶ J
✔ Correct Answer: A — 4.4 × 10⁻²¹ J
KE_rot = ½Iω² = ½ × 1.94×10⁻⁴⁶ × (6.7×10¹²)²
= ½ × 1.94×10⁻⁴⁶ × 4.49×10²⁵
= ½ × 8.71×10⁻²¹ = 4.35×10⁻²¹ J ≈ 4.4×10⁻²¹ J
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🌀 Master Rotational Motion for AAI ATC!
Full video lectures, live doubt sessions, and comprehensive practice from Kanishk Sir