M5L2 Torque and Rotational Dynamics

"Give me a lever long enough and a fulcrum on which to place it, and I shall move the world."
— Archimedes

In M5L1 you learned the kinematics of rotation — how to describe it. Now comes the dynamics: what causes rotation to change. The answer is torque (τ), the rotational analog of force. Just as Newton's Second Law says a net force produces linear acceleration (F = ma), the rotational version says a net torque produces angular acceleration (τ = Iα). The proportionality constant I is the moment of inertia — the rotational analog of mass. But here is where the analogy gets interesting: unlike mass, which is a single number, moment of inertia depends on how that mass is distributed relative to the axis of rotation. A hollow cylinder and a solid cylinder of the same mass and radius spin very differently when the same torque is applied. Understanding why is the central insight of this lesson.

The Analogy Scaffold Extended: Dynamics Edition

Everything from M5L1 still applies. This lesson adds the cause layer: force → torque, mass → moment of inertia, F = ma → τ = Iα.

Linear Dynamics (M2–M3)	Symbol
Force (cause of linear acceleration)	F
Mass (resistance to linear acceleration)	m
Newton's 2nd Law	F_net = ma
Work done by force	W = FΔx

Rotational Dynamics (this lesson)	Symbol
Torque (cause of angular acceleration)	τ
Moment of inertia (resistance to α)	I
Rotational 2nd Law	τ_net = Iα
Work done by torque	W = τΔθ

Torque definition: τ = rF sinθ • or as a cross product: τ = r × F

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

CC5.2 Apply Newton's Second Law for Rotation to analyze rotational dynamics

★ LO5.2.1 Define torque and calculate it using τ = rF sinθ and the cross product r × F

★ LO5.2.2 Identify the moment of inertia as the rotational analog of mass and explain why it depends on mass distribution

★ LO5.2.3 Apply τ_net = Iα to solve for unknown torque, moment of inertia, or angular acceleration

★ LO5.2.4 Use the parallel-axis theorem to find the moment of inertia about an axis displaced from the center of mass

★ LO5.2.5 Calculate rotational kinetic energy using K_rot = ½Iω²

★ LO5.2.6 Apply the work-energy theorem to rotation: W_net = ΔK_rot

Required Reading

Read these sections in order. Section 10.4 establishes torque; 10.5 derives τ = Iα; 10.6 builds your moment of inertia toolkit; 10.7 brings in rotational energy.

10.4 Moment of Inertia & Rotational KE 10.5 Calculating Moments of Inertia 10.6 Torque 10.7 Newton's Second Law for Rotation

Optional Reading

Explore More

Physics Classroom: Torque Khan Academy: Torque PhET: Torque Simulation Khan Academy: Rotational Inertia

Media

Video 1 establishes torque conceptually; Video 2 builds the moment of inertia toolkit; Video 3 applies τ = Iα to multi-body problems and introduces rotational kinetic energy.

Video 1: Torque — The Cause of Rotational Acceleration

What Makes Things Spin?

Torque is not just "twisting force" — it is force multiplied by its perpendicular distance from the axis. This video makes that geometric meaning precise.

Why pushing a door near the hinge is harder than pushing near the edge
Defining the moment arm (lever arm): perpendicular distance from axis to line of action
τ = rF sinθ — what each factor means and when θ is 0°, 90°, or 180°
Sign convention: CCW torques positive, CW torques negative
Net torque: how multiple torques about the same axis combine algebraically

Time: 9:30

Video 2: Moment of Inertia — Resistance to Rotation

Why Mass Distribution Matters

Two objects with identical mass can have very different moments of inertia. This video explains why — and gives you the standard table of results you'll use throughout M5.

I = Σmr² for a collection of point masses
Intuition: why hollow cylinders have higher I than solid cylinders
Standard results: solid disk (½MR²), hollow ring (MR²), solid sphere (⅖MR²), thin rod (&frac{1}{12}ML²)
The parallel-axis theorem: I = I_cm + Md²
A spinning figure skater — why pulling in the arms increases ω

Time: 11:00

Video 3: τ = Iα and Rotational Kinetic Energy

Newton's Second Law for Rotation in Action

Applying τ_net = Iα to real problems, including systems with both linear and rotational motion connected by a string or axle.

Atwood machine with a massive pulley — the moment of inertia changes the answer
Finding angular acceleration of a disk given applied torque and friction
Rotational kinetic energy K_rot = ½Iω² alongside translational K = ½mv²
Work done by torque: W = τΔθ and the rotational work-energy theorem
Power delivered by a torque: P = τω

Time: 12:15

Deeper Look: Torque as a Cross Product and Rotational Kinetic Energy from First Principles

Torque as a Cross Product

τ = r × F — The Full Vector Picture

In 2D problems, torque is just a signed scalar. In 3D, it is a full vector, and the cross product formulation is essential.

The magnitude |τ| = rF sinθ where θ is the angle between r and F. The direction is given by the right-hand rule applied to r × F:

Point fingers along r (from pivot to where force is applied)
Curl them toward F
Thumb points in the direction of τ

Key cross product results for standard geometries:

Force perpendicular to r (90°): |τ| = rF (maximum torque)
Force along r (0° or 180°): |τ| = 0 (no torque — pushing directly toward or away from pivot)

Calculus Connection

Newton's Second Law for rotation in its most general form is τ_net = dL/dt, where L = Iω is angular momentum. You will use this form in M5L3. For constant I, this reduces to τ = Iα.

Deriving I = Σmr² and the Parallel-Axis Theorem

Where the Moment of Inertia Formulas Come From

Start from τ_net = Iα for a rigid body. Each mass element m_i at radius r_i contributes a tangential force F_i = m_ia_t,i = m_ir_iα. The torque from that element is τ_i = r_iF_i = m_ir_i²α. Summing over all elements:

                        τnet = (Σ miri²)α = Iα   ⇒   I = Σ miri²
                    

For continuous bodies, the sum becomes an integral: I = ∫ r² dm.

Parallel-Axis Theorem derivation intuition:

If you know I_cm (moment of inertia about the center of mass), the moment of inertia about any parallel axis at distance d from the CM is:

                        I = Icm + Md²
                    

Interpretation: moving the axis away from the CM always increases I. The CM axis is always the minimum moment of inertia for any axis in that direction.

Moment of Inertia Reference Table

Standard Results to Know

These are derived from I = ∫ r² dm. You should understand why the hollow objects have larger I than solid ones before memorizing the formulas.

Object	Axis	I
Point mass	At distance r	mr²
Thin ring / hollow cylinder	Central axis	MR²
Solid disk / solid cylinder	Central axis	½MR²
Solid sphere	Through center	⅖MR²
Hollow sphere (thin shell)	Through center	⅔MR²
Thin rod	About center	&frac{1}{12}ML²
Thin rod	About one end	⅓ML²

Pattern to Remember

All standard I formulas have the form I = c·MR² where c ≤ 1. The coefficient c tells you what fraction of the mass is effectively at the full radius. Hollow objects (c closer to 1) "feel" larger than solid ones (c closer to ½ or ⅖) because more mass is far from the axis.

Practice and Apply — Conceptual

Torque Anatomy

Match Each Torque Term to Its Meaning

Drag each definition on the right to the matching term on the left.

Moment arm (lever arm)

Maximum torque condition

Zero torque condition

Moment of inertia

Net torque

Rotational inertia of a hollow ring

Sum of all torques about an axis, using sign convention

Force is applied perpendicular to the position vector (sin90° = 1)

MR² — all mass at maximum distance from axis

Perpendicular distance from the axis of rotation to the line of action of the force

Force is directed through the axis (sin0° = 0); no lever arm

Rotational analog of mass; measures resistance to angular acceleration

What Changes Moment of Inertia?

Sort: Does This Change Increase or Decrease Moment of Inertia?

A solid disk is spinning about its central axis. Sort each change into the correct category. Assume the same axis of rotation unless otherwise stated.

Changes to Sort

Double the total mass (same radius)
Replace solid disk with hollow ring of same mass and outer radius
Halve the radius (same mass)
Move the rotation axis from the center to the rim (parallel-axis theorem)
Replace with a solid sphere of same mass and radius
A figure skater extends their arms outward
A figure skater pulls their arms inward
Double the radius (same mass)

Increases Moment of Inertia

Decreases Moment of Inertia

Common Misconceptions

Think It Through — Then Flip

Each card states a common student belief about torque or moment of inertia. Decide whether it is correct or incorrect before flipping to see the explanation.

"A larger force always produces a larger torque."

Incorrect — Moment Arm Matters Too

Torque = rF sinθ. A smaller force applied at a larger perpendicular distance can produce a greater torque than a large force applied near the axis or at a bad angle. This is why a long wrench is more effective than a short one — same force, larger r.

"The moment of inertia of an object is fixed — it cannot change."

Incorrect — I Depends on Axis and Configuration

I depends on both the mass distribution and the choice of rotation axis. A figure skater changes their I by moving their arms. The same rod has I = &frac{1}{12}ML² about its center but I = ⅓ML² about its end. I is a property of object + axis, not object alone.

"If a net torque acts on an object, it must be rotating."

Incorrect — Net Torque Causes α, Not ω

Net torque causes angular acceleration, not rotation per se. An object can have τ_net ≠ 0 while momentarily at rest (ω = 0) — it will then begin to rotate. Conversely, a rotating object (ω ≠ 0) with zero net torque will continue rotating at constant ω. This is exactly analogous to F_net causing a, not v.

"A heavier object always has a greater moment of inertia than a lighter object."

Incorrect — Distribution Beats Mass Alone

A 1 kg hollow ring (I = MR² = 1 × 0.5² = 0.25 kg·m²) has a greater moment of inertia than a 2 kg solid disk of the same radius (I = ½MR² = ½ × 2 × 0.5² = 0.25 kg·m²). Wait — they're actually equal in this case! But extend the radius or use a very light ring vs heavy disk: distribution wins. It is always I = cMR² where c is the shape factor.

Practice and Apply — Computational

Quick Reference: Torque and Rotational Dynamics

                        Torque: τ = rF sinθ

                        r = moment arm, θ = angle between r and F

                        Newton's 2nd (rotation): τnet = Iα

                        I in kg·m², α in rad/s²

                        Rotational KE: K = ½Iω²

                        Joules, exactly like ½mv²

                        Parallel-axis: I = Icm + Md²

                        d = distance from CM to new axis

Problem 1 — Torque from a Wrench

A mechanic applies a 45 N force to a wrench handle 0.30 m from the bolt. Find the torque if the force is applied (a) perpendicular to the handle, (b) at 60° to the handle, and (c) parallel to the handle.

(a) Force perpendicular (θ = 90°):

τ = rF sin90° = (0.30)(45)(1) = 13.5 N·m (maximum)

(b) Force at 60° to handle:

τ = rF sin60° = (0.30)(45)(0.866) = 11.7 N·m

(c) Force parallel (θ = 0°):

τ = rF sin0° = (0.30)(45)(0) = 0 N·m (force directed toward bolt — no rotational effect)

The 90° case is why proper wrench technique matters: pulling perpendicular gives maximum torque from the same effort.

Problem 2 — Moment of Inertia and Angular Acceleration

A solid disk of mass 4.0 kg and radius 0.20 m is free to rotate about its central axis. A net torque of 0.80 N·m is applied. Find: (a) the moment of inertia, (b) the angular acceleration, (c) the angular velocity after 5.0 s (starting from rest).

(a) Moment of inertia (solid disk):

I = ½MR² = ½(4.0)(0.20)² = ½(4.0)(0.04) = 0.080 kg·m²

(b) Angular acceleration:

τ = Iα ⇒ α = τ/I = 0.80/0.080 = 10 rad/s²

(c) Angular velocity after 5.0 s:

ω = ω₀ + αt = 0 + (10)(5.0) = 50 rad/s

Notice how we used the rotational 2nd law to get α, then fed that result directly into the M5L1 kinematic equation. The two lessons always work together.

Problem 3 — Comparing Solid Disk vs Hollow Ring

A solid disk and a hollow ring each have mass M = 2.0 kg and radius R = 0.25 m. The same torque τ = 1.5 N·m is applied to each. Compare their angular accelerations.

Solid disk:

I_disk = ½MR² = ½(2.0)(0.0625) = 0.0625 kg·m²

α_disk = τ/I = 1.5/0.0625 = 24 rad/s²

Hollow ring:

I_ring = MR² = (2.0)(0.0625) = 0.125 kg·m²

α_ring = τ/I = 1.5/0.125 = 12 rad/s²

Conclusion: The solid disk accelerates twice as fast as the hollow ring under the same torque, because the ring has twice the moment of inertia. Same mass, same radius, very different rotational behavior — entirely due to mass distribution.

Problem 4 — Parallel-Axis Theorem

A thin rod of mass 0.60 kg and length 1.2 m is pivoted at one end. Find its moment of inertia about that end using the parallel-axis theorem, and confirm with the direct formula.

Method 1 — Parallel-axis theorem:

I_cm (about center) = &frac{1}{12}ML² = &frac{1}{12}(0.60)(1.2)² = &frac{1}{12}(0.60)(1.44) = 0.072 kg·m²

d = distance from CM to end = L/2 = 0.60 m

I_end = I_cm + Md² = 0.072 + (0.60)(0.60)² = 0.072 + 0.216 = 0.288 kg·m²

Method 2 — Direct formula:

I_end = ⅓ML² = ⅓(0.60)(1.44) = 0.288 kg·m² ✓

The parallel-axis theorem is most useful when you know I_cm and need I about a different axis — you don't have to redo the integral from scratch.

Problem 5 — Rotational Kinetic Energy and Work

A flywheel (solid disk, M = 20 kg, R = 0.40 m) is accelerated from rest to 600 rpm by a motor. Find: (a) the rotational KE at 600 rpm, (b) the net work done by the motor, (c) the average torque if the wheel completes 50 revolutions during spin-up.

Convert: ω = 600 rpm × 2π/60 = 20π rad/s

(a) Rotational KE:

I = ½MR² = ½(20)(0.16) = 1.6 kg·m²

K = ½Iω² = ½(1.6)(20π)² = ½(1.6)(400π²) = 320π² ≈ 3158 J

(b) Net work done (work-energy theorem):

W_net = ΔK = K_f − K_i = 3158 − 0 = 3158 J ≈ 3.16 kJ

(c) Average torque over 50 revolutions:

Δθ = 50 rev × 2π = 100π rad

W = τ_avgΔθ ⇒ τ_avg = W/Δθ = 3158/(100π) = 10.05 N·m

Problem 6 — Atwood Machine with Massive Pulley

An Atwood machine has masses m₁ = 3.0 kg and m₂ = 5.0 kg connected by a massless string over a solid disk pulley of mass M = 2.0 kg and radius R = 0.15 m. Find the linear acceleration of the masses.

This is a coupled linear + rotational system. The string is inextensible, so a (linear) = Rα (rotational).

Free body analysis — net force equation (taking m₂ side as positive):

(m₂ − m₁)g − τ_{net on pulley}/R = (m₁ + m₂)a ... need to include pulley inertia

Full system approach using Newton's 2nd for each part:

m₂g − T₂ = m₂a …(1)

T₁ − m₁g = m₁a …(2)

(T₂ − T₁)R = I_pulleyα = ½MR²·(a/R) = ½MRa …(3)

Add (1) + (2) + (3)/R:

(m₂ − m₁)g = (m₁ + m₂ + ½M)a

a = (m₂ − m₁)g / (m₁ + m₂ + ½M)

a = (5.0 − 3.0)(9.8) / (3.0 + 5.0 + 1.0) = 19.6/9.0 = 2.18 m/s²

Compare to massless pulley: a = (2)(9.8)/(8.0) = 2.45 m/s² — the massive pulley slows the acceleration by ~11%, because the pulley's rotational inertia also needs to be accelerated.

This is a direct example of why ignoring rotational inertia in engineering calculations can lead to significant errors.

Ready to Move On?

Self-Check Before M5L3

M5L3 (Rolling Motion and Angular Momentum) uses both M5L1 kinematics and M5L2 dynamics simultaneously. Make sure you can do all of these:

Calculate the torque of a 30 N force applied at 45° at a distance of 0.4 m from the pivot.
A torque of 6 N·m causes an angular acceleration of 3 rad/s² — what is the moment of inertia?
Without looking at the table, state whether a solid sphere or a hollow sphere has a greater I for the same M and R. Explain why.
A disk's rotation axis is moved from its center to its rim. Does I increase, decrease, or stay the same? By how much (in terms of MR²)?
A flywheel stores 500 J of rotational kinetic energy at ω = 100 rad/s. What is its moment of inertia?

In M5L3, rolling without slipping will require you to apply τ = Iα and a = rα simultaneously. If the Atwood machine problem (Problem 6 above) gave you trouble, revisit it before moving on.