Lesson 8: Rotational Motion

Week 8 — Spinning, turning, and rotating

🎯 Learning Objectives

Translate linear concepts to rotational equivalents
Calculate torque and angular momentum
Solve rotational equilibrium problems

🎯 Intuition First: Spinning Things

Think about opening a door. It's easier to push near the handle (far from the hinge) than near the hinge. Why? Because torque depends on where you apply the force.

💡 Real-World Connection:

Long wrenches make it easier to loosen tight bolts. A seesaw balances when the torques match, not just the weights. The Earth's day length is affected by how mass is distributed in our planet (rotational inertia).

📋 The Rotational Analogues

Rotational motion mirrors linear motion. Every linear quantity has a rotational partner:

Linear	Rotational	Relationship
Position: x	Angle: θ (radians)	x = rθ
Velocity: v	Angular velocity: ω (rad/s)	v = rω
Acceleration: a	Angular acceleration: α (rad/s²)	a = rα
Mass: m	Moment of inertia: I	—
Force: F	Torque: τ	—
Momentum: p = mv	Angular momentum: L = Iω	—
KE = ½mv²	KE = ½Iω²	—

📝 Radians:

Angles in physics are usually in radians. 1 revolution = 2π radians = 360°

ω = 2πf (where f is frequency in Hz)

📋 Torque

Torque: τ = r × F = rF sin(θ)

Units: N·m (newton-meters)

Newton's Second Law for rotation: τ_net = Iα

Where r is the distance from the pivot, F is the applied force, and θ is the angle between r and F.

📝 Worked Example: Torque on a Wrench

You apply 50 N perpendicular to a 0.3-m wrench. What torque does this produce?

τ = rF sin(90°) = (0.3)(50)(1) = 15 N·m

✅ Torque = 15 N·m

📋 Moment of Inertia

Moment of inertia measures how hard it is to spin an object. It depends on both mass and how that mass is distributed:

I = Σm_ir_i² (discrete)

I = ∫r² dm (continuous)

Common Moments of Inertia:

Object	Axis	Moment of Inertia
Point mass	distance r from axis	I = mr²
Uniform rod	through center, perpendicular	I = ⅑₁₂ML²
Uniform rod	through end, perpendicular	I = ⅓ML²
Solid sphere	through center	I = ⅖MR²
Hollow sphere	through center	I = ⅔MR²
Hollow cylinder (hoop)	through center	I = MR²
Solid cylinder/disk	through center	I = ½MR²

💡 Key Insight:

The farther the mass is from the axis of rotation, the larger the moment of inertia. A tightrope walker carries a long pole to increase their rotational inertia, making them harder to tip over!

📋 Rotational Kinematics

Same form as linear kinematics, just with angular variables:

1) ω = ω₀ + αt

2) θ = θ₀ + ω₀t + ½αt²

3) ω² = ω₀² + 2α(θ − θ₀)

📝 Worked Example: Spinning Disk

A solid disk (M = 5 kg, R = 0.4 m) spins from rest. A constant torque of 2 N·m is applied. How many revolutions after 5 seconds?

Step 1 — Moment of inertia:

I = ½MR² = ½(5)(0.4)² = 0.4 kg·m²

Step 2 — Angular acceleration:

α = τ/I = 2/0.4 = 5 rad/s²

Step 3 — Angular displacement:

θ = ½αt² = ½(5)(5²) = 62.5 rad

Step 4 — Convert to revolutions:

N = 62.5/(2π) = 9.95 revolutions

✅ Approximately 10 revolutions

📋 Rotational Equilibrium

An object is in equilibrium when both forces and torques balance:

ΣF = 0 (translational equilibrium)

Στ = 0 (rotational equilibrium)

📝 Worked Example: Beam on Two Supports

A 4-m beam (mass 20 kg) rests on two supports 1 m from each end. A 60-kg person stands 1.5 m from the left end. Find the normal forces on each support.

Step 1 — Draw FBD with forces:

Beam weight (W_B = 196 N) at center (2 m from left)
Person weight (W_P = 588 N) at 1.5 m
Support forces N₁ at 1 m, N₂ at 3 m

Step 2 — ΣF_y = 0:

N₁ + N₂ = 196 + 588 = 784 N

Step 3 — Στ = 0 (take torque about N₁):

−588(0.5) + 196(1) + N₂(2) = 0

−294 + 196 + 2N₂ = 0

2N₂ = 98

N₂ = 49 N

N₁ = 784 − 49 = 735 N

✅ N₁ = 735 N (left support), N₂ = 49 N (right support)