System of Particles and Rotational Motion

SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

6.1 INTRODUCTION

**Motion of Extended Bodies**: Real bodies have finite size. A **rigid body** is defined as a body with perfectly definite and unchanging shape—the distance between any two particles remains constant. While no real body is truly rigid (all deform under force), many bodies like wheels, steel beams, and molecules can be treated as rigid when deformations are negligible.

**Types of Motion of Rigid Bodies**:

1. **Pure Translational Motion**: All particles of the body move with the same velocity at any instant. Example: a block sliding down an inclined plane without rotating (Fig. 6.1).

2. **Rotational Motion**: The body rotates about an axis. Every particle moves in a circle perpendicular to the axis with the axis passing through the center. Example: a ceiling fan or potter's wheel (Fig. 6.3).

3. **Combined Motion (Rotation + Translation)**: Rolling motion combines both. Example: a cylinder rolling down an inclined plane. The particles have different velocities—the contact point has zero velocity while the top has maximum velocity (Fig. 6.2).

**Axis of Rotation**: In rotation about a fixed axis, each particle at distance **r** from the axis moves in a circle of radius r in a plane perpendicular to the axis. Particles on the axis remain stationary.

**Important Distinction**:

Rigid bodies with fixed support undergo pure rotation

Free rigid bodies undergo pure translation or combined translation + rotation

In this chapter, we focus on rotation about a fixed axis

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6.2 CENTRE OF MASS

**Definition**: The centre of mass (CM) of a system is that point which moves as if the entire mass of the system were concentrated there, and all external forces were applied at that point.

**For Two Particles (1D)**:

Position of CM: **X = (m₁x₁ + m₂x₂)/(m₁ + m₂)** ... (6.1)

where x₁, x₂ are distances from origin O, and m₁, m₂ are masses.

**Key Point**: X is the mass-weighted mean of positions.

**Special Case**: For equal masses (m₁ = m₂ = m), the CM lies exactly midway: X = (x₁ + x₂)/2

**For n Particles (1D)**:

**X = Σ(mᵢxᵢ)/Σmᵢ = Σ(mᵢxᵢ)/M** ... (6.2)

where M = Σmᵢ is total mass.

**For Multiple Particles in 2D**:

**X = (m₁x₁ + m₂x₂ + m₃x₃)/(m₁ + m₂ + m₃)** ... (6.3a)

**Y = (m₁y₁ + m₂y₂ + m₃y₃)/(m₁ + m₂ + m₃)** ... (6.3b)

For equal masses, the CM coincides with the **geometric centroid** of the triangle.

**For n Particles in 3D**:

Position vector of CM:

**R = Σ(mᵢrᵢ)/M** ... (6.4d)

where **rᵢ** = position vector of ith particle.

Component form:

**X = Σ(mᵢxᵢ)/M**, **Y = Σ(mᵢyᵢ)/M**, **Z = Σ(mᵢzᵢ)/M** ... (6.4a, 6.4b, 6.4c)

If CM is chosen as origin: **Σ(mᵢrᵢ) = 0**

**For Continuous Mass Distribution**:

For a rigid body with continuous mass distribution:

**X = (1/M)∫x dm**, **Y = (1/M)∫y dm**, **Z = (1/M)∫z dm** ... (6.5a)

Vector form: **R = (1/M)∫r dm** ... (6.5b)

**Property of Homogeneous Bodies**: For bodies with uniform density (homogeneous bodies) having reflection symmetry, the CM coincides with the geometric center. Examples:

Uniform rod: CM at geometric center

Uniform ring, disc, sphere: CM at geometric center

Symmetric L-shaped or rectangular lamina: CM at geometric center (by symmetry)

**Example 6.1 (Worked)**: Three particles of masses 100g, 150g, 200g at vertices of equilateral triangle (side 0.5m).

Taking origin at first particle:

Particle 1: m₁ = 100g at (0, 0)

Particle 2: m₂ = 150g at (0.5m, 0)

Particle 3: m₃ = 200g at (0.25m, 0.25√3 m)

X = [100(0) + 150(0.5) + 200(0.25)]/(450) = [75 + 50]/450 = 125/450 = 5/18 m

Y = [100(0) + 150(0) + 200(0.25√3)]/(450) = 50√3/450 = √3/9 m

**Exam Important**: CM need not lie on the body itself or at geometric center when masses are unequal.

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6.3 MOTION OF CENTRE OF MASS

**Velocity of Centre of Mass**:

Differentiating **R = Σ(mᵢrᵢ)/M** with respect to time:

**MV = m₁v₁ + m₂v₂ + ... + mₙvₙ = Σ(mᵢvᵢ)** ... (6.8)

where **V = dR/dt** is velocity of CM and **vᵢ = drᵢ/dt** is velocity of ith particle.

**Acceleration of Centre of Mass**:

Differentiating velocity equation:

**MA = m₁a₁ + m₂a₂ + ... + mₙaₙ = Σ(mᵢaᵢ)** ... (6.9)

where **A = dV/dt** is acceleration of CM and **aᵢ = dvᵢ/dt** is acceleration of ith particle.

**Significance**: The acceleration of CM depends only on **net external force**, not on internal forces (which act in equal and opposite pairs by Newton's third law).

**Key Result**: The CM of a system moves as if all the external force acts on the entire mass concentrated at the CM. Internal forces between particles do not affect CM motion.

**Example**: In a firework explosion, the CM of fragments continues in the parabolic trajectory it had before explosion (only affected by gravity, an external force).

**Condition for Zero Acceleration of CM**:

If **Fₙₑₜ,ₑₓₜ = 0**, then **A = 0** and CM moves with constant velocity.

**Conservation Implication**: If no external force acts, the velocity and momentum of CM remain constant.

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6.4 LINEAR MOMENTUM OF A SYSTEM OF PARTICLES

**Definition**: Linear momentum of a particle is **p = mv**.

For a system of n particles:

**P = p₁ + p₂ + ... + pₙ = Σ(mᵢvᵢ)** ... (6.10)

**Relation with CM Velocity**:

From equation (6.8): **P = MV** ... (6.11)

The total linear momentum equals the total mass times the velocity of the centre of mass.

**Newton's Second Law for a System**:

Differentiating **P = MV**:

**dP/dt = M(dV/dt) = MA = Fₙₑₜ,ₑₓₜ** ... (6.12)

**Statement**: The rate of change of total linear momentum of a system equals the net external force acting on it. Internal forces do not contribute as they cancel in pairs.

**Conservation of Linear Momentum**:

If **Fₙₑₜ,ₑₓₜ = 0**, then **dP/dt = 0**, so **P = constant** ... (6.13)

**Total linear momentum is conserved when no net external force acts on the system.**

**Applications**:

1. Explosion: Initial momentum = sum of momenta of fragments

2. Collision: Total momentum before = total momentum after

3. Rocket motion: Thrust from expelled gas conserves momentum in isolated system

**Exam Important**:

Internal forces (friction between parts, explosion forces) do NOT change total momentum

Only external forces change system momentum

CM always accelerates in direction of net external force

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6.5 VECTOR PRODUCT OF TWO VECTORS

**Definition**: The vector product (cross product) of vectors **A** and **B** is:

**A × B = AB sin θ n̂** ... (6.14)

where:

θ is angle between **A** and **B** (0 ≤ θ ≤ π)

**n̂** is unit vector perpendicular to both **A** and **B**, determined by **right-hand rule**

**|A × B| = AB sin θ** (magnitude)

**Right-Hand Rule**: Curl fingers of right hand from **A** toward **B** through angle θ; thumb points in direction of **A × B**.

**Properties**:

1. **Non-commutative**: **A × B = −(B × A)**

2. **Not associative**: **(A × B) × C ≠ A × (B × C)** in general

3. **Distributive**: **A × (B + C) = A × B + A × C**

4. **Scalar multiplication**: **(kA) × B = k(A × B)**

5. **Parallel vectors**: If **A || B**, then **A × B = 0**

6. **Perpendicular vectors**: If **A ⊥ B**, then **|A × B| = AB** (maximum)

**Unit Vector Products**:

**î × î = ĵ × ĵ = k̂ × k̂ = 0**

**î × ĵ = k̂**, **ĵ × k̂ = î**, **k̂ × î = ĵ**

**ĵ × î = −k̂**, **k̂ × ĵ = −î**, **î × k̂ = −ĵ**

**Component Form**:

If **A = Aₓî + Aᵧĵ + Aᵤk̂** and **B = Bₓî + Bᵧĵ + Bᵤk̂**:

**A × B = |î ĵ k̂|**

|Aₓ Aᵧ Aᵤ|

|Bₓ Bᵧ Bᵤ|

= **î(AᵧBᵤ − AᵤBᵧ) − ĵ(AₓBᵤ − AᵤBₓ) + k̂(AₓBᵧ − AᵧBₓ)**

**Geometric Interpretation**: **|A × B|** equals the area of parallelogram with sides **A** and **B**.

**Physical Significance**: Vector product appears in torque (**τ = r × F**), angular momentum (**L = r × p**), and magnetic force (**F = q(v × B)**).

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6.6 ANGULAR VELOCITY AND ITS RELATION WITH LINEAR VELOCITY

**Angular Displacement**: For a particle at distance r from axis, moving through arc length s:

**θ = s/r** (in radians) ... (6.15)

**Angular Velocity** (ω):

**ω = dθ/dt** (rate of change of angular displacement) ... (6.16)

SI unit: **rad/s** or **s⁻¹**

**Angular Acceleration** (α):

**α = dω/dt = d²θ/dt²** ... (6.17)

SI unit: **rad/s²**

**Relation Between Linear and Angular Quantities**:

For a particle at perpendicular distance r from rotation axis:

**v = ωr** ... (6.18)

where v is linear (tangential) speed.

**Proof**: Arc length ds = r dθ, so v = ds/dt = r(dθ/dt) = rω

**Linear acceleration**:

**a = αr** (tangential acceleration) ... (6.19)

**Centripetal acceleration**: **aᶜ = v²/r = ω²r** ... (6.20)

**Vector Form of Angular Velocity**:

Angular velocity is a **vector** directed along the axis of rotation (using right-hand rule): if fingers curl in direction of rotation, thumb points in direction of **ω**.

**Relation with Linear Velocity** (vector form):

**v = ω × r** ... (6.21)

where **r** is position vector from axis to particle. This automatically gives correct magnitude v = ωr sin(90°) = ωr and correct direction (perpendicular to both **ω** and **r**).

**Exam Important**:

Angular displacement, velocity, and acceleration are pseudovectors (axial vectors)

All particles of rigid body rotating about fixed axis have same ω, but different v

v ∝ r: particles farther from axis move faster

**Example**: Wheel of radius 0.5m rotating at ω = 10 rad/s.

Point at rim: v = ωr = 10 × 0.5 = 5 m/s

Point at radius 0.25m: v = 10 × 0.25 = 2.5 m/s

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6.7 TORQUE AND ANGULAR MOMENTUM

**Torque (τ)**: The turning effect of force about an axis.

**Definition**: **τ = r × F** ... (6.22)

where **r** is position vector from axis to point of application of force **F**.

**Magnitude**: **τ = rF sin θ = r⊥F = Fr⊥** ... (6.23)

where:

θ is angle between **r** and **F**

r⊥ = r sin θ (perpendicular distance from axis to line of action of force) = **moment arm**

F⊥ = F sin θ (component of force perpendicular to **r**)

SI unit: **N⋅m** (newton-metre)

**Direction**: Determined by right-hand rule for **r × F**. Thumb points along axis in direction of torque. Positive torque causes counterclockwise rotation (by convention).

**Conditions for Zero Torque**:

1. F = 0 (no force)

2. r = 0 (force passes through pivot)

3. sin θ = 0 (force parallel to **r**, passes through axis)

**Angular Momentum (L)**: Rotational equivalent of linear momentum.

**Definition for a particle**: **L = r × p = r × (mv)** ... (6.24)

**Magnitude**: **L = mrv sin θ = mrω sin(90°) = mrω = mr²ω** (for perpendicular motion)

SI unit: **kg⋅m²/s**

**For a System of Particles**:

**L_total = Σ(rᵢ × pᵢ) = Σ(rᵢ × mᵢvᵢ)** ... (6.25)

**Relation Between Torque and Angular Momentum**:

**τ = dL/dt** ... (6.26)

**Proof**:

L = r × p = r × (mv)

dL/dt = d(r × mv)/dt = (dr/dt × mv) + (r × m dv/dt)

= (v × mv) + (r × ma)

= 0 + (r × F) [since v × v = 0 and F = ma]

= τ

**Statement**: Rate of change of angular momentum equals net torque.

**Conservation of Angular Momentum**:

If **τ_net = 0**, then **dL/dt = 0**, so **L = constant** ... (6.27)

**When net external torque is zero, angular momentum is conserved.**

**Examples**:

1. Spinning ice skater pulling arms inward: I decreases, ω increases (L constant, I = constant)

2. Planet in orbit: L conserved (gravity provides centripetal force, not torque about center of mass)

3. Freely rotating body: Angular momentum conserved when no external torque

**Exam Important**:

Torque is pseudovector (axial vector) like angular velocity

Even if ω changes, L can remain constant if moment of inertia changes (and vice versa)

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6.8 EQUILIBRIUM OF A RIGID BODY

**Definition**: A rigid body is in **equilibrium** when it does not accelerate (translational equilibrium) and does not rotate (rotational equilibrium).

**Conditions for Equilibrium**:

**1. Translational Equilibrium**:

**Fₙₑₜ = ΣF = 0** ... (6.28)

or in components: **ΣFₓ = 0**, **ΣFᵧ = 0**, **ΣFᵤ = 0**

The net force on the body must be zero.

**2. Rotational Equilibrium**:

**τₙₑₜ = Στ = 0** (about any axis) ... (6.29)

The net torque on the body must be zero about any point (or equivalently, about the axis of rotation).

**Both conditions must be satisfied simultaneously for complete equilibrium.**

**Types of Equilibrium**:

1. **Stable Equilibrium**: Body returns to original position after small displacement. Example: ball at bottom of bowl (COM raised when displaced).

2. **Unstable Equilibrium**: Body moves away from equilibrium position after small displacement. Example: pencil balanced on point (COM lowered when displaced).

3. **Neutral Equilibrium**: Body remains in equilibrium in new position after displacement. Example: ball on flat surface (COM height unchanged).

**Principle of Moments**: For a body in rotational equilibrium about a point:

**Sum of clockwise torques = Sum of anticlockwise torques** ... (6.30)

**Applications**:

Balancing seesaw: m₁r₁ = m₂r₂ (about pivot)

Ladder leaning against wall: torques about base balance

Door hinges: force distribution based on torque balance

**Exam Important**:

Check equilibrium about different points: easier calculation if point passes through multiple unknown forces

Both ΣF = 0 AND Στ = 0 must be checked

Στ = 0 must hold about ANY point, but typically choose point through which unknown forces act

**Example**: Uniform rod of mass M and length L supported at two points distance d and (L−d) from one end. Find normal forces N₁ and N₂.

Translation equilibrium: N₁ + N₂ = Mg

Rotation equilibrium (about left support): N₂(L−d) = Mg(L/2)

Solving: N₂ = MgL/[2(L−d)], N₁ = Mg − N₂

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6.9 MOMENT OF INERTIA

**Definition**: Moment of inertia is the rotational equivalent of mass—it measures resistance to change in rotational motion.

**For a Point Mass**:

**I = mr²** ... (6.31)

where m is mass and r is perpendicular distance from axis of rotation.

SI unit: **kg⋅m²**

**For a System of Particles**:

**I = Σ(mᵢrᵢ²) = m₁r₁² + m₂r₂² + ... + mₙrₙ²** ... (6.32)

where rᵢ is perpendicular distance of ith particle from axis.

**For Continuous Distribution (Rigid Body)**:

**I = ∫r² dm** ... (6.33)

where integration is over entire body, dm is mass element at distance r from axis.

**Physical Significance**:

Larger I means harder to change ω (rotational inertia)

I depends on mass distribution: all mass matters proportionally to its distance from axis

I is calculated about a specific axis—different axes give different I values

**Parallel Axis Theorem**:

If I_cm is moment of inertia about an axis through CM and I is moment of inertia about parallel axis at distance d from CM:

**I = I_cm + Md²** ... (6.34)

where M is total mass.

**Proof**: Consider axis through CM and parallel axis at distance d. For element dm at position r from axis through CM (distance x from parallel axis):

r² = (distance from parallel axis)² = x² = (r_cm ± d)²

I = ∫(r_cm + d)² dm = ∫r_cm² dm + d²∫dm + 2d∫r_cm dm

= I_cm + Md² + 0 (third term zero because ∫r_cm dm = 0 at CM)

**Perpendicular Axis Theorem** (for planar objects):

For a planar body in xy-plane:

**I_z = I_x + I_y** ... (6.35)

where I_z is moment of inertia about z-axis (perpendicular to plane), and I_x, I_y are moments about x, y axes in the plane.

**Moments of Inertia of Standard Objects** (about axis through CM):

| Object | Axis | I |

|--------|------|---|

| Thin Rod (length L) | Perpendicular through CM | ML²/12 |

| Thin Rod | One end | ML²/3 |

| Ring (radius R) | Through center ⊥ to plane | MR² |

| Disc (radius R) | Through center ⊥ to plane | MR²/2 |

| Solid Sphere (radius R) | Through center | 2MR²/5 |

| Hollow Sphere (radius R) | Through center | 2MR²/3 |

| Solid Cylinder (radius R) | Along axis | MR²/2 |

| Hollow Cylinder (radius R) | Along axis | MR² |

**Exam Important**:

I depends on both mass and its distribution (distance from axis)

Two objects with same mass but different shapes have different I

Use parallel axis theorem when given I at CM but need I at different axis

Memorize standard formulas; some can be derived using integration

**Example**: Rod of mass M = 2 kg, length L = 1 m. Find I about:

(a) End: I_end = ML²/3 = 2(1)²/3 = 2/3 kg⋅m²

(b) Point 0.25 m from one end: Distance from CM = 0.5 − 0.25 = 0.25 m

I_cm = ML²/12 = 2(1)²/12 = 1/6 kg⋅m²

I = I_cm + Md² = 1/6 + 2(0.25)² = 1/6 + 0.125 = 0.267 kg⋅m²

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6.10 KINEMATICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS

**Rotational motion kinematics parallels linear motion kinematics**, with angular quantities replacing linear quantities.

**Linear Motion | Rotational Motion**

Position x | Angular position θ

Velocity v = dx/dt | Angular velocity ω = dθ/dt

Acceleration a = dv/dt | Angular acceleration α = dω/dt

**Kinematic Equations for Constant Angular Acceleration**:

**Equation 1**: **ω = ω₀ + αt** ... (6.36)

**Equation 2**: **θ = ω₀t + ½αt²** ... (6.37)

**Equation 3**: **ω² = ω₀² + 2αθ** ... (6.38)

**Equation 4**: **θ = (ω₀ + ω)t/2** ... (6.39)

where:

θ = angular displacement (radians)

ω₀ = initial angular velocity (rad/s)

ω = final angular velocity (rad/s)

α = angular acceleration (rad/s²)

t = time (s)

**Derivation of Equation 3** (Rotational v² = u² + 2as):

From equation 1: ω = ω₀ + αt, so t = (ω − ω₀)/α

From equation 2: θ = ω₀t + ½αt²

Substituting: θ = ω₀[(ω − ω₀)/α] + ½α[(ω − ω₀)/α]²

= (ω₀ω − ω₀²)/α + (ω − ω₀)²/(2α)

= (2ω₀ω − 2ω₀² + ω² − 2ωω₀ + ω₀²)/(2α)

= (ω² − ω₀²)/(2α)

Therefore: **ω² = ω₀² + 2αθ**

**Example**: Wheel starting from rest accelerates at α = 2 rad/s². After time t = 5 s:

ω = ω₀ + αt = 0 + 2(5) = 10 rad/s

θ = ½αt² = ½(2)(25) = 25 radians

Check: ω² = 0 + 2(2)(25) = 100, so ω = 10 rad/s ✓

**Relation Between Linear and Angular Kinematics**:

For a particle at distance r from axis:

| Linear | Angular | Relation |

|--------|---------|----------|

| v | ω | v = ωr |

| a_t (tangential) | α | a_t = αr |

| a_c (centripetal) | ω | a_c = ω²r |

**Total acceleration**: **a = √(a_t² + a_c²)** (tangential and centripetal perpendicular)

Direction: a_t along tangent, a_c toward center

**Exam Important**:

Use rotational equations when ω or α given

Convert between linear and angular quantities using v = ωr, a = αr

Remember α must be constant for these equations; if variable, use calculus (integration/differentiation)

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6.11 DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS

**Rotational Equation of Motion**:

For a rigid body rotating about fixed axis:

**τ = Iα** ... (6.40)

where:

τ = net torque about axis (N⋅m)

I = moment of inertia about axis (kg⋅m²)

α = angular acceleration (rad/s²)

**Analogy with F = ma**: Just as F = ma relates force, mass, and acceleration in linear motion, τ = Iα relates torque, moment of inertia, and angular acceleration in rotational motion. Moment of inertia is the rotational analog of mass.

**Proof of τ = Iα**:

Consider rigid body rotating about fixed axis. For a particle of mass mᵢ at distance rᵢ:

Tangential force: F_t,i = mᵢaᵢ = mᵢαrᵢ

Torque due to this force: τᵢ = rᵢF_t,i = rᵢ(mᵢαrᵢ) = mᵢrᵢ²α

Total torque: τ = Στᵢ = Σ(mᵢrᵢ²)α = (Σmᵢrᵢ²)α = Iα

**Energy Consideration in Rotation**:

**Rotational Kinetic Energy**: **KE_rot = ½Iω²** ... (6.41)

Analogy: Compare with linear KE = ½mv²

**Proof**: For particle at distance r moving with speed v = ωr:

KE = ½mv² = ½m(ωr)² = ½(mr²)ω²

Total KE = Σ½(mᵢrᵢ²)ω² = ½(Σmᵢrᵢ²)ω² = ½Iω²

**Work Done by Torque**:

**W = τθ** (for constant torque) ... (6.42)

**W = ∫τ dθ** (for variable torque)

**Work-Energy Theorem for Rotation**:

**W_net = ΔKE_rot = ½I(ω² − ω₀²)** ... (6.43)

**Rolling Motion (Pure Rolling Without Slipping)**:

For a body rolling without slipping on surface:

**v_cm = ωR** ... (6.44)

where v_cm is velocity of center of mass, ω is angular velocity, R is radius.

**Total Kinetic Energy**:

KE_total = KE_translational + KE_rotational = ½Mv_cm² + ½Iω²

For rolling (ω = v_cm/R):

**KE_total = ½Mv_cm² + ½I(v_cm/R)²** ... (6.45)

**Example**: Solid cylinder (I = MR²/2) rolling without slipping with v_cm = 4 m/s, M = 2 kg, R = 0.2 m:

KE_trans = ½(2)(4)² = 16 J

ω = v_cm/R = 4/0.2 = 20 rad/s

KE_rot = ½(2×0.04/2)(20)² = ½(0.04)(400) = 8 J

KE_total = 16 + 8 = 24 J

**Condition for Rolling Without Slipping**:

Contact point must have zero velocity: v_contact = 0

If rolling down incline with acceleration a:

v = at (translational)

ω = αt (rotational)

For no slip: v = ωR, so a = αR

**For body rolling down frictionless incline** (sliding, not rolling):

Pure translation only, no rotation.

**For body rolling down incline with friction**:

Friction provides torque causing rotation. Accelerations satisfy a = αR.

**Exam Important**:

τ = Iα is fundamental equation; use it like F = ma

Distinguish between rolling and sliding—rolling requires friction and satisfies v = ωR

In rolling problems, check constraint v = ωR when solving

Energy method often simpler than force/torque method for rolling

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6.12 ANGULAR MOMENTUM IN CASE OF ROTATION ABOUT A FIXED AXIS

**Angular Momentum for Rotating Body**:

For a rigid body rotating about fixed axis with angular velocity ω:

**L = Iω** ... (6.46)

where I is moment of inertia about rotation axis.

**Proof**: For particle at distance r from axis:

L_particle = mrv = mr(ωr) = mr²ω

Total angular momentum:

L = ΣL_particle = Σmᵢrᵢ²ω = (Σmᵢrᵢ²)ω = Iω

**Rotational Equation in Terms of Angular Momentum**:

**τ = dL/dt** ... (6.47)

For constant I (rigid body):

τ = d(Iω)/dt = I(dω/dt) = Iα

This is consistent with τ = Iα.

**Conservation of Angular Momentum**:

If **τ_net = 0**, then **dL/dt = 0**, so **L = constant**

For rotating system: **L = Iω = constant**

If moment of inertia changes: **I₁ω₁ = I₂ω₂** ... (6.48)

**Examples**:

1. **Ice Skater**: Spinning with arms extended (I₁, ω