Rotational Motion

Misconceptions

Key Concepts

Angular Kinematics

Rotational motion uses angular quantities analogous to linear motion:

Key conversions:

• •1 revolution = 2π radians = 360°
• •ω (rad/s) = 2π × frequency (Hz) = 2π/T

Angular Velocity

• •Average: ω_avg = Δθ/Δt
• •Instantaneous: ω = dθ/dt
• •Direction: Right-hand rule (curl fingers in rotation direction, thumb points along ω)
• •Unit: rad/s

Angular Acceleration

• •Average: α_avg = Δω/Δt
• •Instantaneous: α = dω/dt
• •Positive α means speeding up (if ω positive) or slowing down (if ω negative)
• •Unit: rad/s²

Rotational Kinematic Equations

For constant angular acceleration (analogous to linear kinematics):

Torque

Torque is the rotational equivalent of force:

• •τ = rF sin(θ) where θ is the angle between r and F
• •Equivalently: τ = r × F (cross product)
• •τ = r_⊥ × F = r × F_⊥ (lever arm form)
• •Unit: N·m (not Joules, though same dimensions)
• •Sign: positive for counterclockwise, negative for clockwise

Key insight: Torque = (lever arm) × (force) where lever arm is perpendicular distance from axis to line of action of force.

Moment of Inertia

Moment of inertia (I) is the rotational analog of mass:

• •Measures resistance to angular acceleration
• •I = Σmᵢrᵢ² (discrete masses)
• •Depends on axis of rotation
• •Unit: kg·m²

Common moments of inertia (about central axis unless noted):

Parallel Axis Theorem: I = I_cm + Md² (d = distance from center of mass to new axis)

Newton's Second Law for Rotation

• •τ_net = Iα
• •Analogous to F_net = ma

Rotational Kinetic Energy

• •KE_rot = ½Iω²
• •Analogous to KE_trans = ½mv²
• •For rolling without slipping: KE_total = ½mv² + ½Iω² with v = ωR

Angular Momentum

• •L = Iω (for rigid body rotating about fixed axis)
• •Unit: kg·m²/s
• •Vector quantity (direction from right-hand rule)

Conservation of Angular Momentum

When net external torque is zero:

• •L_initial = L_final
• •I₁ω₁ = I₂ω₂
• •Examples: ice skater spin, collapsing star, diver's tuck

Rolling Motion

For rolling without slipping:

• •v_cm = ωR (constraint condition)
• •a_cm = αR
• •Friction provides torque but does no work (contact point instantaneously at rest)
• •Total KE = ½mv² + ½Iω² = ½mv²(1 + I/(mR²))

Equations

[1] ω = ω₀ + αt (angular velocity)
[2] θ = θ₀ + ω₀t + ½αt² (angular displacement)
[3] ω² = ω₀² + 2α(θ - θ₀) (angular velocity-displacement)
[4] τ = rF sin(θ) (torque)
[5] τ_net = Iα (Newton's 2nd law for rotation)
[6] I = Σmᵢrᵢ² (moment of inertia)
[7] KE_rot = ½Iω² (rotational kinetic energy)
[8] L = Iω (angular momentum)
[9] I₁ω₁ = I₂ω₂ (conservation of angular momentum)
[10] v = ωR (rolling constraint)

Worked Examples

Example 1: Angular Kinematics

Problem: A wheel accelerates uniformly from rest to 150 rad/s in 5 seconds. (a) What is the angular acceleration? (b) How many revolutions does it make?

Solution:

1. •Given: ω₀ = 0, ω = 150 rad/s, t = 5 s
2. •(a) α = (ω - ω₀)/t = (150 - 0)/5 = 30 rad/s²
3. •(b) θ = ω₀t + ½αt² = 0 + ½(30)(5)² = 375 rad
4. •Revolutions = 375/(2π) = 59.7 revolutions

Example 2: Torque Calculation

Problem: A 40 N force is applied at the end of a 0.5 m wrench at 60° to the handle. What torque is produced?

Solution:

1. •τ = rF sin(θ)
2. •τ = 0.5 × 40 × sin(60°) = 0.5 × 40 × 0.866
3. •τ = 17.3 N·m

Example 3: Rotational Dynamics

Problem: A solid disk of mass 4 kg and radius 0.2 m has a cord wrapped around it. If 10 N tension is applied, what is the angular acceleration?

Solution:

1. •Moment of inertia: I = ½MR² = ½(4)(0.2)² = 0.08 kg·m²
2. •Torque: τ = TR = 10 × 0.2 = 2 N·m
3. •τ = Iα → α = τ/I = 2/0.08 = 25 rad/s²

Example 4: Rolling Down an Incline

Problem: A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping down a 3 m high incline. What is its speed at the bottom?

Solution:

1. •For solid sphere: I = ⅖MR², so I/(MR²) = 2/5
2. •Energy conservation: Mgh = ½Mv² + ½Iω²
3. •With v = ωR: Mgh = ½Mv² + ½(⅖MR²)(v/R)² = ½Mv² + ⅕Mv² = (7/10)Mv²
4. •v² = (10/7)gh = (10/7)(9.8)(3) = 42
5. •v = 6.48 m/s

Example 5: Conservation of Angular Momentum

Problem: An ice skater with arms extended has I = 4.5 kg·m² and spins at 2 rev/s. When she pulls in her arms, I = 1.5 kg·m². What is her new spin rate?

Solution:

1. •Conservation: I₁ω₁ = I₂ω₂
2. •ω₁ = 2 rev/s = 4π rad/s
3. •ω₂ = I₁ω₁/I₂ = (4.5 × 4π)/1.5 = 12π rad/s
4. •ω₂ = 12π/(2π) = 6 rev/s

Explanation Patterns

1. •Identify the axis of rotation - All angular quantities are defined relative to this axis
2. •Draw an extended free body diagram - Show where forces act, not just at center of mass
3. •Calculate torques - Use τ = rF sin(θ) or lever arm method; include signs
4. •Apply τ_net = Iα for rotational dynamics
5. •For rolling problems, use both F = ma and τ = Iα with the constraint v = ωR
6. •For energy problems, include both translational and rotational KE
7. •Check for angular momentum conservation - Is external torque zero?

Common Problem Types

1. •Angular kinematics: Use rotational equations analogous to linear kinematics
2. •Torque calculation: Find τ = rF sin(θ) for individual forces
3. •Rotational dynamics: Apply τ_net = Iα with correct I for the shape
4. •Rotational energy: Include KE_rot = ½Iω² in energy conservation
5. •Rolling motion: Combine translation and rotation with v = ωR
6. •Angular momentum conservation: I₁ω₁ = I₂ω₂ when τ_external = 0
7. •Atwood machine with pulley: Include pulley's I in the analysis