Unit 5: Torque & Rotational Dynamics · First Principles Physics
AP Physics 1 · Unit 05 of 08

Torque & Rotational Dynamics.

Every translational concept you've learned — position, velocity, acceleration, force, Newton's second law — has a rotational twin. Learn the map between the two worlds and half of Unit 5 is already done.

Exam Weight · 10–15% Class Periods · ~10–14 CED Topics · 5.1–5.6 Difficulty · Moderate
01
The lesson

Why the same physics looks completely different when it's rotating

Picture a figure skater in the middle of a spin. She pulls her arms in — suddenly she's spinning impossibly fast. She extends her arms — she slows to a stately turn. Nothing pushes her. Nothing pulls her. No external force changes the rate at which she rotates. And yet, by the simple act of rearranging her body, she's accelerating herself to speeds that make the audience gasp. How?

The skater is doing with her body what every rotational physics problem asks you to do on paper: separating how much mass from where that mass is. In translational motion, all that matters is the total mass of an object. Push a 5 kg block and it doesn't matter whether the mass is compressed into a marble or spread out in a long bar — acceleration depends only on \(m\). But the moment you start rotating things, geometry suddenly matters as much as mass does. A marble and a long bar with the same mass rotate completely differently.

Unit 5 is the story of what happens when you take Unit 2 (Newton's laws) and replace every translational concept with its rotational twin. The equations look identical. The symbols just change names.

The translational ↔ rotational dictionary

This is the single most useful table in Unit 5. Memorize it. Everything else in this unit is a consequence of it:

Translational quantityRotational analogConnecting equation
Position \(x\)Angular position \(\theta\)\(x = r\theta\)
Velocity \(v\)Angular velocity \(\omega\)\(v = r\omega\)
Acceleration \(a\)Angular acceleration \(\alpha\)\(a = r\alpha\)
Mass \(m\)Rotational inertia \(I\)\(I = \sum m_i r_i^2\)
Force \(F\)Torque \(\tau\)\(\tau = rF\sin\theta\)
Newton's 2nd law: \(F = ma\)\(\tau_{\text{net}} = I\alpha\)same structure

Every translational equation you've ever used has a rotational version obtained by substituting the analog on the right for the quantity on the left. Unit 5 is less about learning new physics than about learning this translation.

Angular kinematics: Unit 1 with Greek letters

Let's start with pure geometry. Imagine a record spinning on a turntable. A dot painted on the record traces out an angle as time passes. We measure this angle in radians — one radian is the angle at which the arc length equals the radius, and a full circle is \(2\pi\) radians. The rate at which the angle changes is the angular velocity:

$$\omega = \frac{d\theta}{dt}, \qquad \bar\omega = \frac{\Delta\theta}{\Delta t}$$

Units: rad/s. And the rate at which angular velocity changes is the angular acceleration:

$$\alpha = \frac{d\omega}{dt}, \qquad \bar\alpha = \frac{\Delta\omega}{\Delta t}$$

Units: rad/s². If you squint, this is literally Unit 1 with Greek letters in place of Roman. And indeed, when the angular acceleration is constant, you get three kinematic equations that look suspiciously familiar:

$$\omega = \omega_0 + \alpha t$$ Rot-1
$$\theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2$$ Rot-2
$$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$$ Rot-3

These are the exact same equations as Kin-1, Kin-2, Kin-3 from Unit 1, with rotational variables swapped in. Same equation-selection strategy applies: identify the missing variable, reach for the equation that doesn't contain it.

The bridge: connecting linear and angular for a point on a rotating object

A child sits at the edge of a merry-go-round. A second child sits halfway to the center. Both travel through the same angle every second — the merry-go-round is rigid, so \(\omega\) is the same for every point on it. But the outer child covers more distance per second. Her linear speed is higher because she's farther from the axis.

The bridge equations quantify this:

$$v = r\omega, \qquad a_t = r\alpha$$

Here \(r\) is the perpendicular distance from the rotation axis, and \(a_t\) is the tangential acceleration (the part of the acceleration along the direction of motion, responsible for speeding up or slowing down the rotation). Even an object rotating at constant \(\omega\) has centripetal acceleration \(a_c = v^2/r = \omega^2 r\) pointing toward the axis — this isn't new (you met it in Unit 2), but it's worth remembering: rotation always involves centripetal acceleration, even when nothing is "speeding up."

Torque: the rotational version of force

Push straight through the hinge of a door and nothing happens — the door doesn't rotate no matter how hard you push. Push at the edge of the door (far from the hinge), perpendicular to the door, and it swings easily. Same force; radically different rotational effect. The difference is torque.

Torque is a measure of how effectively a force causes rotation. It depends on three things: the size of the force, how far from the axis you apply it, and the angle at which you apply it.

$$\tau = r F \sin\theta$$

Here \(r\) is the distance from the axis to the point where the force is applied, \(F\) is the force's magnitude, and \(\theta\) is the angle between the force vector and the position vector \(\vec r\). When \(\theta = 90°\) (force perpendicular to \(\vec r\)), torque is maximized: \(\tau = rF\). When \(\theta = 0°\) (force pointing along \(\vec r\), either toward or away from the axis), torque is zero.

An equivalent and often-easier interpretation: \(\tau = F \cdot d_\perp\), where \(d_\perp\) is the lever arm — the perpendicular distance from the axis to the line of action of the force. If you extend the arrow of the force in both directions, then drop a perpendicular from the axis to that line, the length of that perpendicular is the lever arm. Use whichever formulation is easier for the problem.

Sign of torque

On the AP exam, torques are signed by the direction of rotation they would cause. Counterclockwise is conventionally positive; clockwise is negative. When you write \(\sum \tau = 0\) for equilibrium or \(\tau_{\text{net}} = I\alpha\) for dynamics, include signs.

Worked example 1 · torque

The lever-arm trick

A 4.0 m uniform beam is attached to a wall by a hinge. A 50 N force pulls on the end of the beam at an angle of 30° above the horizontal. What torque does this force exert about the hinge?

Method 1: Use \(\tau = rF\sin\theta\)

Here \(r = 4.0\) m, \(F = 50\) N, \(\theta = 30°\) (angle between force and beam). So \(\tau = (4.0)(50)\sin 30° = 100\) N·m.

Method 2: Use the lever arm

Draw the force as an arrow. Extend its line of action. The perpendicular distance from the hinge to that line is \(d_\perp = 4.0 \sin 30° = 2.0\) m. So \(\tau = F \cdot d_\perp = (50)(2.0) = 100\) N·m. Same answer.

Torque magnitude: 100 N·m counterclockwise (if the force pulls upward).

Rotational inertia: the "mass" of rotation

In \(F = ma\), mass is the object's resistance to being translationally accelerated — bigger \(m\) means harder to push around. In \(\tau = I\alpha\), rotational inertia \(I\) plays the same role for rotation. Bigger \(I\) means the object is harder to spin up or stop from spinning.

But here's the twist that makes Unit 5 harder than Unit 2: \(I\) depends not only on how much mass the object has, but on how far from the axis that mass is distributed. For a point mass at distance \(r\) from the axis:

$$I_{\text{point}} = m r^2$$

For a collection of point masses, add them up:

$$I_{\text{total}} = \sum_i m_i r_i^2$$

For extended objects (solid spheres, rods, hoops), you'd need calculus to derive the formula — but the AP Physics 1 CED specifies that extended-object moments of inertia will be provided on the exam. You just need to know how to use them. Here's a reference you'll see often:

Object (rotating about given axis)\(I\)
Point mass at distance \(r\)\(mr^2\)
Hoop or thin cylindrical shell (about central axis)\(MR^2\)
Solid cylinder or disk (about central axis)\(\tfrac{1}{2}MR^2\)
Solid sphere (about diameter)\(\tfrac{2}{5}MR^2\)
Thin rod (about center, length \(L\))\(\tfrac{1}{12}ML^2\)
Thin rod (about end, length \(L\))\(\tfrac{1}{3}ML^2\)

Two patterns jump out. First, \(I\) always scales as (mass) × (length)². Units are kg·m². Second, the rod is a beautiful illustration of why geometry matters: rotate it about its center, \(I = \frac{1}{12}ML^2\). Rotate it about its end, \(I = \frac{1}{3}ML^2\) — four times harder to spin. Same mass, same rod, different axis. Mass distribution matters.

Why a hoop is harder to spin than a disk

Same mass \(M\), same radius \(R\). Hoop: \(I = MR^2\). Solid disk: \(I = \tfrac{1}{2}MR^2\). The hoop has more rotational inertia because all of its mass sits at the maximum distance \(R\) from the axis, while the disk's mass is spread across every radius from 0 to \(R\). Since \(I\) weights each bit of mass by \(r^2\), mass near the axis barely contributes. This is why figure skaters pull their arms in to spin faster (reducing \(I\)), and why gymnasts extend their arms to slow rotations (increasing \(I\)).

The parallel axis theorem

If you know the rotational inertia \(I_{cm}\) about an axis through the center of mass, and you want the rotational inertia \(I'\) about a parallel axis a distance \(d\) away:

$$I' = I_{cm} + Md^2$$

This is the parallel axis theorem. It tells you that the rotational inertia about any axis is always at least as large as the rotational inertia about the center of mass axis — the minimum \(I\) is always through the center of mass. Example: for a rod of length \(L\), \(I_{cm} = \frac{1}{12}ML^2\). About the end, \(d = L/2\), so \(I_{\text{end}} = \frac{1}{12}ML^2 + M(L/2)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2\). Matches the table. Good.

Rotational equilibrium: Newton's first law for rotation

An object is in rotational equilibrium when the net torque on it is zero. If it's also in translational equilibrium (net force zero), the object is in full static equilibrium — the kind of problem you'll see with beams, bridges, leaning ladders, and diving boards. In that case:

$$\sum F = 0 \quad\text{and}\quad \sum \tau = 0$$

Two conditions. Two kinds of equations. This is the setup for every balance problem on the AP exam.

Worked example 2 · static equilibrium

The uniform beam on two supports

A 60 kg uniform beam of length 4.0 m rests on two supports — one at each end. A 20 kg box sits on the beam 1.0 m from the left support. Find the upward force on each support. Use \(g = 10\) m/s².

Set up

Forces acting: beam weight 600 N at center (2.0 m from left), box weight 200 N at 1.0 m from left, left support force \(N_L\) upward at \(x = 0\), right support force \(N_R\) upward at \(x = 4.0\) m.

Take torques about the left support (so \(N_L\) drops out)

$$\sum \tau_L = 0:\quad -(200)(1.0) - (600)(2.0) + N_R (4.0) = 0$$

Solve: \(N_R = (200 + 1200)/4.0 = 350\) N.

Use force balance for the other support

\(\sum F_y = 0\): \(N_L + N_R - 600 - 200 = 0\), so \(N_L = 800 - 350 = 450\) N.

Left support: 450 N · Right support: 350 N

Torque strategy — the pivot trick

When you take torques about a point, any force applied at that point has zero lever arm and contributes zero torque. This is a powerful problem-solving trick: choose your pivot to be wherever a force you don't know acts, and that force drops out of your torque equation. In the example above, taking torques about the left support eliminated \(N_L\) from the equation, turning a two-unknown problem into a one-unknown problem.

Newton's second law for rotation

When the net torque is not zero, the object angularly accelerates:

$$\tau_{\text{net}} = I \alpha$$

This is the rotational analog of \(F = ma\). It's the workhorse of every "a disk is released from rest and a torque causes it to spin up" problem.

Worked example 3 · rotational F=ma

Spinning up a disk

A solid disk of mass 2.0 kg and radius 0.3 m is free to rotate about its central axis. A constant tangential force of 5.0 N is applied to the rim. Find the angular acceleration.

Compute torque and rotational inertia

Torque: \(\tau = rF = (0.3)(5.0) = 1.5\) N·m (force is tangential, so \(\sin\theta = 1\)).

Inertia: \(I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2.0)(0.3)^2 = 0.09\) kg·m².

Apply \(\tau = I\alpha\)

$$\alpha = \frac{\tau}{I} = \frac{1.5}{0.09} \approx 16.7 \text{ rad/s}^2$$

Angular acceleration: 16.7 rad/s²

Combined translation + rotation

Many AP problems involve both translational and rotational motion — a yo-yo falling while unwinding, a ball rolling down a ramp, a rope pulling on a pulley that has mass. The approach is always the same: set up two separate Newton's second laws, one for translation (\(F = ma\)) and one for rotation (\(\tau = I\alpha\)), and link them with the constraint \(a = r\alpha\) (no slipping) or \(v = r\omega\) (rolling).

We'll drill this in the worksheets. The pattern: two equations, two unknowns, linked by the rolling/non-slip constraint.

Where this unit connects to the rest of the course

Unit 5 is the gateway to Unit 6, where the energy and momentum of rotating systems come into play. A spinning disk stores kinetic energy (\(K = \tfrac{1}{2}I\omega^2\), the rotational analog of \(\tfrac{1}{2}mv^2\)). A spinning skater carries angular momentum (\(L = I\omega\), the analog of \(p = mv\)). When angular momentum is conserved — no external torques — the skater pulls in her arms, reduces \(I\), and \(\omega\) must go up to compensate. Our opening hook, solved.

Conceptual summary

Rotational motion is translational motion with every symbol replaced by its rotational twin. Angular position, velocity, and acceleration (\(\theta, \omega, \alpha\)) obey the same three constant-acceleration equations as their linear counterparts. Torque is the rotational analog of force and depends on where and at what angle the force is applied. Rotational inertia is the rotational analog of mass and depends on both mass and its distribution. Newton's second law in rotational form, \(\tau_{\text{net}} = I\alpha\), governs angular acceleration the same way \(F = ma\) governs linear acceleration. Static equilibrium requires both net force and net torque to be zero. Choosing your pivot wisely (at a point where an unknown force acts) dramatically simplifies torque equations.

· · ·
02
Reference sheet

Every equation in Unit 5, at a glance

Stars (★) mark equations on the AP Physics 1 equation sheet. Extended-object moments of inertia (disk, hoop, sphere, rod) will be provided with any problem that requires them — you don't need to memorize those formulas, just know how to use them.

★ Rotational kinematics 1

$$\omega = \omega_0 + \alpha t$$
\(\omega\)
angular velocity (rad/s)
\(\alpha\)
angular acceleration (rad/s²)
\(t\)
time (s)

Use when \(\alpha\) is constant and you need \(\omega\) or \(t\) (no angle involved).

★ Rotational kinematics 2

$$\theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2$$
\(\theta\)
angular position (rad)
\(\omega_0\)
initial angular velocity
\(\alpha\)
angular acceleration

Use when \(\alpha\) is constant and final \(\omega\) isn't given/asked for.

★ Rotational kinematics 3 — timeless

$$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$$
All variables
as above

Use when time isn't given or asked for. Same logic as Kin-3 from Unit 1.

★ Linear–angular bridge

$$v = r\omega, \qquad a_t = r\alpha$$
\(r\)
perpendicular distance from axis (m)
\(v\)
linear speed of a point on the rotating object
\(a_t\)
tangential linear acceleration

Use when you need to link rotation to linear motion (rolling, pulleys, points on a rotating object).

★ Torque — angle form

$$\tau = rF\sin\theta$$
\(r\)
distance from axis to point of force application (m)
\(F\)
force magnitude (N)
\(\theta\)
angle between \(\vec r\) and \(\vec F\)

Use when the force is applied at a known angle to a known moment arm.

Torque — lever arm form

$$\tau = F \cdot d_\perp$$
\(d_\perp\)
perpendicular distance from axis to the line of action of \(\vec F\)

Use when finding the lever arm is easier than finding the angle (often the case in beam/balance problems).

★ Rotational inertia of point masses

$$I = \sum_i m_i r_i^2$$
\(m_i\)
each point mass
\(r_i\)
each point's perpendicular distance from the axis

Use when you have a system of discrete point masses (up to 5 per CED).

★ Parallel axis theorem

$$I' = I_{cm} + Md^2$$
\(I_{cm}\)
rotational inertia about axis through center of mass
\(d\)
distance from that axis to the parallel axis
\(M\)
total mass

Use when rotation axis is parallel to (but not through) the center of mass.

★ Newton's 2nd law for rotation

$$\tau_{\text{net}} = I\alpha$$
\(\tau_{\text{net}}\)
net torque on system (N·m)
\(I\)
rotational inertia about chosen axis
\(\alpha\)
angular acceleration

Use when a system is angularly accelerating — the rotational workhorse equation.

Static equilibrium conditions

$$\sum F = 0, \qquad \sum \tau = 0$$
Both
must hold simultaneously for static equilibrium

Use when an object is at rest and not angularly accelerating — beams, bridges, ladders, balance problems.

Moments of inertia provided on the exam

You'll see: hoop \(MR^2\), solid cylinder/disk \(\tfrac{1}{2}MR^2\), solid sphere \(\tfrac{2}{5}MR^2\), rod about center \(\tfrac{1}{12}ML^2\), rod about end \(\tfrac{1}{3}ML^2\). Know which is which at a glance — questions often rely on the comparison (e.g., "which has more rotational inertia, a hoop or a disk?").

· · ·
· · ·
04
Worksheet A · Foundation

Ten problems building from basic to moderate

Work through these in order. 1–3 check basic vocabulary. 4–7 combine two ideas. 8–10 approach AP difficulty. Use \(g = 10\) m/s² throughout.
Foundation

A wheel spins at 20 rad/s and decelerates uniformly to rest in 5.0 s. (a) What is its angular acceleration? (b) Through how many radians does it turn during this time?

Show your work below
Foundation

A 0.5 kg point mass is attached to the end of a massless rod of length 2.0 m. The rod rotates about its other end. What is the rotational inertia of this system?

Show your work below
Foundation

A 10 N force is applied at the edge of a wheel of radius 0.4 m, perpendicular to the radius. What is the magnitude of the torque about the center?

Show your work below
Moderate

A disk of radius 0.2 m rotates with angular velocity 15 rad/s. What is the linear speed of a point on the rim? What is the linear speed of a point halfway between the center and the rim?

Show your work below
Moderate

Two point masses, 1.0 kg and 3.0 kg, are connected by a rigid massless rod of length 2.0 m. The system rotates about an axis perpendicular to the rod through its center. Find the rotational inertia about that axis.

Show your work below
Moderate

A 3.0 m uniform beam of mass 10 kg is supported at its two ends. A 5.0 kg box sits 1.0 m from the left end. Find the upward force each support exerts on the beam.

Show your work below
Moderate

A solid cylinder of mass 4.0 kg and radius 0.5 m is mounted on a frictionless axle. A constant tangential force of 6.0 N is applied at the rim. What is the angular acceleration of the cylinder?

Show your work below
Moderate

A uniform rod of mass \(M\) and length \(L\) rotates about an axis perpendicular to the rod passing through one end. (a) Using the parallel axis theorem and the known center-of-mass inertia \(I_{cm} = \tfrac{1}{12}ML^2\), derive the rotational inertia about the end. (b) The rod is released from rest in the horizontal position. Find the angular acceleration at the instant of release.

Show your work below
Moderate

A 50 N force is applied to a wrench at a point 0.3 m from the bolt. The force makes a 60° angle with the handle of the wrench. Find the torque about the bolt.

Show your work below
Moderate

A massless pulley of radius 0.1 m has a block of mass 2.0 kg hanging from a string wrapped around it. If the pulley is frictionless and massless (so the string tension equals the block's weight when accelerating), find (a) the tension in the string if the pulley is now a solid disk of mass 1.0 kg, and (b) the linear acceleration of the block. (Hint: write F=ma for the block and \(\tau = I\alpha\) for the pulley, then link with \(a = r\alpha\).)

Show your work below
· · ·
05
Worksheet B · AP exam style

Ten problems in the exact format of the real AP exam

Problems 11–15 are multiple choice. 16–18 are short free-response. 19–20 are extended FRQs. Problem 19 mirrors the 2025 exam's torque FRQ about a balanced meterstick.

Part I — Multiple Choice

AP MC

A solid sphere and a hoop have the same mass and the same radius. Both rotate about their central axes with the same angular velocity \(\omega\). Which has more rotational kinetic energy?

  1. The solid sphere, because it has less rotational inertia.
  2. The hoop, because all of its mass is at the maximum distance from the axis.
  3. Both have the same kinetic energy because they have the same mass and angular velocity.
  4. Impossible to determine without knowing the radius.
AP MC

A uniform beam of weight \(W\) and length \(L\) is supported at its two ends. A block of weight \(w\) sits at the midpoint. What is the force on the left support?

  1. \(W/2\)
  2. \((W + w)/2\)
  3. \(W + w\)
  4. \(W/2 + w\)
AP MC

Two children sit on a seesaw. A 30 kg child sits 2.0 m from the pivot. Where should a 20 kg child sit to balance the seesaw?

  1. 1.3 m from the pivot, on the same side as the 30 kg child.
  2. 1.3 m from the pivot, on the opposite side.
  3. 3.0 m from the pivot, on the opposite side.
  4. 2.0 m from the pivot, on the opposite side.
AP MC

A disk and a hoop have the same mass and the same radius. Both are mounted on frictionless axles through their centers. The same tangential force is applied to the rim of each. Which one has the larger angular acceleration?

  1. The disk, because it has smaller rotational inertia.
  2. The hoop, because it has larger rotational inertia.
  3. Both have the same angular acceleration.
  4. The answer depends on how fast they are already rotating.
AP MC

A rod of length \(L\) has two point masses \(m\) each attached to its ends. The rod itself has negligible mass. For which axis of rotation is the rotational inertia of the system smallest?

  1. An axis perpendicular to the rod through one end.
  2. An axis perpendicular to the rod through the center.
  3. An axis perpendicular to the rod at a distance \(L/4\) from one end.
  4. An axis along the length of the rod itself.

Part II — Short Free Response

AP SFR

(Mathematical routines, ~7 min) A uniform rod of mass \(M\) and length \(L\) is pivoted at one end and held horizontal. The rotational inertia about the pivot is \(I = \tfrac{1}{3}ML^2\).

  1. Derive a symbolic expression for the torque about the pivot due to gravity at the instant the rod is released from rest in the horizontal position. Begin with a fundamental physics principle.
  2. Derive a symbolic expression for the angular acceleration at this instant in terms of \(M\), \(L\), and \(g\).
  3. A student claims that if the rod were twice as long (but still mass \(M\)), the angular acceleration at release would be half as large. Indicate whether this claim is correct and justify your answer using your expression from part (b).
AP SFR

(Qualitative/quantitative translation, ~7 min) Two forces are applied to a horizontal rod that can rotate about a fixed axis at one end, as shown. A 10 N force is applied at the midpoint pointing vertically downward. A 15 N force is applied at the far end pointing vertically upward. The rod has length 2.0 m.

  1. Calculate the magnitude and direction (clockwise or counterclockwise) of the net torque about the pivot.
  2. Describe, in one or two sentences, the resulting rotational motion immediately after the forces are applied.
  3. The student now moves the 10 N force from the midpoint to the far end (both forces now at the end). Does the rod rotate clockwise, counterclockwise, or stay still? Justify your answer.
AP SFR

(Translation between representations, ~7 min) A disk of rotational inertia \(I\) is mounted on a frictionless axle. A variable torque is applied to the disk for 10 seconds according to the graph below.

0 τ₀ 4 6 10 t (s) τ (N·m)

  1. Assuming the disk starts from rest, describe what the disk's angular velocity is doing during each segment: 0–4 s, 4–6 s, and 6–10 s.
  2. Sketch a qualitative graph of the disk's angular velocity vs time from \(t = 0\) to \(t = 10\) s, with labeled axes.
  3. At what time is the angular velocity at a maximum, and what is that maximum in terms of \(\tau_0\) and \(I\)?

Part III — Extended Free Response

AP EFR

(Experimental design, ~18 min — mirrors the 2025 AP FRQ) A student has a uniform meterstick that can pivot about a hole at its 50 cm mark. A hook of negligible mass is attached to the top of a block of unknown mass \(m_0\), which can be hung from small holes in the meterstick. A spring scale of negligible mass is fixed to one end of the meterstick. The student wants to determine \(m_0\) by measuring how the spring scale reading depends on where the block is hung.

  1. Describe a procedure to collect data that would allow the student to determine \(m_0\). Specify what quantity should be varied, what quantity should be measured, and how to reduce experimental uncertainty.
  2. Describe how the data could be graphed (what to put on each axis) so that the slope of a straight-line fit determines \(m_0\).
  3. The student's graph yields a best-fit line with slope \(3.0\) N/m. Calculate \(m_0\), using \(g = 10\) m/s².
  4. The student now moves the pivot to the 30 cm mark and repeats the procedure. The meterstick itself (mass \(M\)) now contributes a torque. Derive an expression for the new spring-scale reading as a function of the block's distance \(x\) from the pivot, in terms of \(m_0, M, x,\) and constants.
AP EFR

(Combined translation + rotation, ~18 min) A block of mass \(m\) hangs from a light string wrapped around a solid cylindrical pulley of mass \(M\) and radius \(R\), free to rotate about a horizontal axle. The block is released from rest.

  1. Draw a free-body diagram for the block and a separate force diagram showing the force(s) on the pulley that produce torque about the axle.
  2. Derive a symbolic expression for the linear acceleration of the block in terms of \(m, M,\) and \(g\). Begin with Newton's second law for the block, write Newton's second law for rotation for the pulley, and link them with the no-slip condition.
  3. Derive a symbolic expression for the tension in the string.
  4. If \(m = M\), show that \(a = \tfrac{2}{3}g\) and \(T = \tfrac{1}{3}mg\).
  5. The experiment is repeated with the same cylinder replaced by a hoop of the same mass and radius. Indicate whether the block's acceleration is greater than, less than, or equal to the cylinder case, and justify your answer.
· · ·
06
Worksheet C · Challenge

Five problems at F=ma / olympiad level

Harder than anything on the AP exam. Each rewards conceptual depth. 15–30 minutes apiece.
Challenge

The climbing worker. A uniform ladder of mass \(M\) and length \(L\) leans against a frictionless wall at angle \(\theta\) above the horizontal. The ground has coefficient of static friction \(\mu_s\). A worker of mass \(m = M\) climbs the ladder. Derive the fraction of the ladder's length \(f\) the worker can climb before the ladder slips, in terms of \(\mu_s\) and \(\theta\). Verify your expression makes sense in the limits \(\theta \to 90°\) (ladder vertical) and \(\theta \to 0°\) (ladder horizontal).

Show your work below
Challenge

Rolling race. A solid sphere, a solid cylinder, and a hoop are all released from rest at the top of a ramp and roll without slipping. Rank the three objects by the time it takes each to reach the bottom. Justify your ranking using rotational inertia, without solving the full problem. Bonus: derive the acceleration down the ramp in terms of \(g\), the ramp angle \(\theta\), and the shape factor \(\beta\) where \(I = \beta M R^2\).

Show your work below
Challenge

The Atwood machine with massive pulley. Two blocks of masses \(m_1\) and \(m_2\) (with \(m_1 > m_2\)) hang from a string passing over a pulley that is a solid disk of mass \(M_p\) and radius \(R\). Friction in the axle is negligible; the string does not slip on the pulley. Derive expressions for the acceleration of the masses and the tension in the string on each side of the pulley. Note: the tensions on the two sides of the pulley are not equal when the pulley has mass.

Show your work below
Challenge

Falling chimney. A tall uniform chimney of height \(L\) topples by rotating about its base. Show that while the chimney is still largely vertical, the top of the chimney has a vertical acceleration greater than \(g\). This explains why tall falling structures often break midway up rather than hitting the ground intact. At what fraction of the chimney's height from the base is the vertical acceleration exactly equal to \(g\)?

Show your work below
Challenge

The yo-yo. A yo-yo consists of a solid disk of mass \(M\) and radius \(R\), with a light string wrapped around a central axle of radius \(r \ll R\). The yo-yo is released from rest and falls as the string unwinds. Treating the yo-yo's rotational inertia as \(I = \tfrac{1}{2}MR^2\) (the axle's inertia being negligible), derive an expression for the linear acceleration of the center of mass. Show that the tension in the string is much less than the weight of the yo-yo when \(r \ll R\), and explain physically why this is the case.

Show your work below
· · ·
07
Complete worked solutions

Every problem, every step, every why

Each solution shows reasoning before arithmetic. Key insights and common errors are flagged so you can check your approach even if your numerical answer differed.

Worksheet A — Foundation

· decelerating wheel

(a) \(\alpha = \Delta\omega/\Delta t = (0 - 20)/5.0 = -4.0\) rad/s².

(b) \(\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = (20)(5) + \tfrac{1}{2}(-4)(25) = 100 - 50 = 50\) rad. (Or equivalently, average angular velocity \(\times\) time = \(10 \times 5 = 50\) rad.)

\(\alpha = -4.0\) rad/s²; \(\theta = 50\) rad
Rotational kinematics is structurally identical to linear kinematics. Use the same equation-selection strategy: identify the missing variable and choose the equation that doesn't contain it.
· point mass on a rod

Point mass rotates at distance \(r = 2.0\) m from the axis. \(I = mr^2 = (0.5)(2.0)^2 = 2.0\) kg·m².

\(I = 2.0\) kg·m²
Students often write \(mr\) instead of \(mr^2\). Remember: rotational inertia scales as distance squared.
· tangential force torque

Force is perpendicular to radius, so \(\sin\theta = 1\): \(\tau = rF = (0.4)(10) = 4.0\) N·m.

\(\tau = 4.0\) N·m
· linear speeds on a disk

Point on rim: \(v = r\omega = (0.2)(15) = 3.0\) m/s. Point halfway out: \(v = (0.1)(15) = 1.5\) m/s.

Rim 3.0 m/s; halfway 1.5 m/s.
All points on a rigid rotating object share the same \(\omega\), but their linear speeds scale with distance from the axis. This is the essence of the \(v = r\omega\) relationship.
· two-mass dumbbell about center

Each mass sits 1.0 m from the central axis. \(I = (1.0)(1.0)^2 + (3.0)(1.0)^2 = 1.0 + 3.0 = 4.0\) kg·m².

\(I = 4.0\) kg·m²
For discrete masses, just add each \(m_i r_i^2\). No integrals. The CED explicitly limits AP Physics 1 to 5 or fewer point masses in discrete-mass problems.
· beam on two supports with off-center load

Forces: beam weight 100 N at 1.5 m (center), box 50 N at 1.0 m, supports \(N_L\) at 0 and \(N_R\) at 3.0 m. Take torques about left support:

$$-(50)(1.0) - (100)(1.5) + N_R(3.0) = 0 \;\Rightarrow\; N_R = 200/3 \approx 66.7 \text{ N}$$

Force balance: \(N_L + 66.7 = 150\), so \(N_L = 83.3\) N.

Left 83.3 N; Right 66.7 N.
Choosing the left support as pivot eliminates \(N_L\) from the torque equation — one unknown to solve, not two. Always look for a pivot that kills an unknown.
· spinning up a cylinder

Inertia: \(I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(4)(0.25) = 0.5\) kg·m². Torque: \(\tau = rF = (0.5)(6) = 3.0\) N·m.

\(\alpha = \tau/I = 3.0/0.5 = 6.0\) rad/s².

\(\alpha = 6.0\) rad/s²
· rod about end

(a) Parallel axis theorem: \(I_{\text{end}} = I_{cm} + M d^2 = \tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{12}ML^2 + \tfrac{1}{4}ML^2 = \tfrac{1}{3}ML^2\).

(b) Gravity acts at center of mass, a distance \(L/2\) from the pivot. Weight \(Mg\) perpendicular to rod at release: \(\tau = Mg(L/2)\). So \(\alpha = \tau/I = Mg(L/2) / (\tfrac{1}{3}ML^2) = 3g/(2L)\).

\(I = \tfrac{1}{3}ML^2\); \(\alpha = 3g/(2L)\) at release.
Notice that \(\alpha\) at release doesn't depend on \(M\) — mass cancels, just as it does in free fall. The tip of the rod falls with acceleration \(a_t = L\alpha = 3g/2\), which is greater than \(g\).
· wrench at an angle

\(\tau = rF\sin\theta = (0.3)(50)\sin 60° = (0.3)(50)(0.866) \approx 13\) N·m.

\(\tau \approx 13\) N·m
Students sometimes skip the \(\sin\theta\) factor when the force isn't drawn perpendicular to the wrench. Every torque problem with an angle needs this factor.
· pulley with hanging block

Let the tension be \(T\), block's acceleration \(a\) downward, pulley's angular acceleration \(\alpha\).

Block (downward positive): \(mg - T = ma\) → \(20 - T = 2a\).

Pulley: \(\tau = I\alpha\). Torque from string: \(\tau = TR\). Inertia of disk: \(I = \tfrac{1}{2}M_pR^2 = \tfrac{1}{2}(1)(0.1)^2 = 0.005\) kg·m². So \(TR = I\alpha\) → \(T = I\alpha/R\).

Link: \(a = R\alpha\) → \(\alpha = a/R\). Substituting: \(T = I(a/R)/R = Ia/R^2 = (0.005)a/(0.01) = 0.5a\).

Plug into block equation: \(20 - 0.5a = 2a\) → \(a = 20/2.5 = 8.0\) m/s². Then \(T = 0.5(8) = 4.0\) N.

\(a = 8.0\) m/s²; \(T = 4.0\) N
The tension is less than the block's weight (\(mg = 20\) N), which is exactly what's needed to let the block accelerate downward. The pulley's rotational inertia effectively "uses up" some of gravity's effort.

Worksheet B — AP Exam Style

· sphere vs hoop KE

Rotational KE: \(K = \tfrac{1}{2}I\omega^2\). Hoop has \(I = MR^2\); sphere has \(I = \tfrac{2}{5}MR^2\). Same \(M\), \(R\), \(\omega\). Hoop has larger \(I\), so larger KE.

(B) The hoop.
Bigger rotational inertia at the same angular velocity means more kinetic energy, just like bigger mass at the same linear velocity means more KE in translation.
· beam with centered block

Because the block is at the exact midpoint and the beam is uniform, the situation is symmetric — each support carries half the total load. Total weight: \(W + w\). Each support: \((W + w)/2\).

(B) \((W + w)/2\)
When the center of mass of the loaded system is directly between symmetric supports, the supports split the load equally. No torque equation needed.
· seesaw balance

Torques must balance: \(m_1 g x_1 = m_2 g x_2\), so \((30)(2.0) = (20)(x_2)\) → \(x_2 = 3.0\) m on opposite side.

(C) 3.0 m from pivot on opposite side.
· disk vs hoop under same force

Same torque (\(\tau = rF\), same \(r\) and \(F\)). But disk has \(I = \tfrac{1}{2}MR^2\), hoop has \(I = MR^2\). From \(\alpha = \tau/I\), smaller \(I\) means larger \(\alpha\) — disk wins.

(A) The disk.
· minimum-I axis for dumbbell

About axis along the rod: both masses lie on the axis, so \(r = 0\) for each, and \(I = 0\). About perpendicular axis through center: each mass is \(L/2\) away, \(I = 2m(L/2)^2 = mL^2/2\). About perpendicular axis through end: one mass has \(r = 0\), other has \(r = L\), so \(I = mL^2\). About axis at \(L/4\) from end: \(I = m(L/4)^2 + m(3L/4)^2 = m(L^2/16 + 9L^2/16) = \tfrac{5}{8}mL^2\).

(D) Axis along the length of the rod (\(I = 0\)).
Trick answer! If mass is "on" the rotation axis (\(r = 0\)), it contributes zero rotational inertia. This is why a thin rod spun about its long axis has essentially no rotational inertia compared to spinning it end-over-end.
· rod released horizontal

(a) Torque at release. Starting from \(\tau_{\text{net}} = I\alpha\), the only torque is gravity acting at the CM (distance \(L/2\) from pivot): \(\tau = Mg(L/2) = MgL/2\).

(b) Angular acceleration. \(\alpha = \tau/I = (MgL/2)/(\tfrac{1}{3}ML^2) = 3g/(2L)\).

(c) Doubled length. If \(L\) doubles, \(\alpha = 3g/(2 \cdot 2L) = 3g/(4L)\) — half the original angular acceleration. Claim is correct.

\(\tau = MgL/2\); \(\alpha = 3g/(2L)\); claim correct.
Mass \(M\) cancels in the expression for \(\alpha\), meaning a heavy rod and a light rod fall through the same angular acceleration — just like the equivalence principle for falling objects.
· two forces on rod

(a) Taking counterclockwise as positive: 10 N downward at midpoint (\(r = 1.0\) m) contributes \(-(10)(1.0) = -10\) N·m (clockwise). 15 N upward at far end (\(r = 2.0\) m) contributes \(+(15)(2.0) = +30\) N·m. Net: \(+20\) N·m, counterclockwise.

(b) The rod rotates counterclockwise (up at the far end) with angular acceleration proportional to 20 N·m. It starts from rest and angularly accelerates.

(c) With both forces at the far end: 10 N down + 15 N up = net 5 N up. Torque: \(+(5)(2.0) = +10\) N·m counterclockwise. Still counterclockwise rotation, just less torque.

(a) 20 N·m CCW; (b) angularly accelerates CCW; (c) still CCW, less.
· torque-time graph

(a) 0–4 s: constant torque \(\tau_0 \Rightarrow\) constant \(\alpha \Rightarrow\) \(\omega\) grows linearly. 4–6 s: same, \(\omega\) continues growing at same rate. 6–10 s: \(\tau = 0 \Rightarrow \alpha = 0 \Rightarrow \omega\) constant.

(b) \(\omega\) vs \(t\): straight line from 0 rising linearly from \(t = 0\) to \(t = 6\) s, then horizontal from \(t = 6\) to \(t = 10\) s.

(c) Maximum \(\omega\) is at \(t = 6\) s and continues through \(t = 10\) s. \(\omega_{\max} = \alpha t = (\tau_0/I)(6) = 6\tau_0/I\).

\(\omega_{\max} = 6\tau_0/I\), reached at \(t = 6\) s.
The area under a \(\tau\)-\(t\) graph divided by \(I\) gives \(\Delta\omega\) — the rotational analog of "area under \(F\)-\(t\) gives \(\Delta v\)/\(m\)," which you'll recognize as impulse-momentum.
· meterstick + spring scale experiment

(a) Procedure. Hang the block from various holes in the meterstick at different distances \(x\) from the pivot. For each position, read the spring scale force \(F\). Take multiple trials at each position to reduce random error. Since pivot is at 50 cm and meterstick is uniform, the meterstick's weight exerts zero torque about the pivot — only the block does.

(b) Graphing. Torque balance about pivot: \(F \cdot d_{\text{scale}} = m_0 g \cdot x\), where \(d_{\text{scale}}\) is the distance from pivot to where the spring scale pulls. So \(F = (m_0 g / d_{\text{scale}}) x\). Plot \(F\) vs \(x\); slope = \(m_0 g / d_{\text{scale}}\).

(c) Calculation. Slope 3.0 N/m = \(m_0 g / d_{\text{scale}}\). If the spring scale is at the end opposite the holes (50 cm from pivot, so \(d_{\text{scale}} = 0.5\) m): \(m_0 = (3.0)(0.5)/10 = 0.15\) kg = 150 g.

(d) Off-center pivot. With pivot at 30 cm, the meterstick's CM (at 50 cm) is 20 cm = 0.2 m from the pivot, so the meterstick contributes torque \(Mg(0.2)\). The block at distance \(x\) contributes \(m_0 g x\). Spring scale torque \(F \cdot d_{\text{scale}} = Mg(0.2) + m_0 g x\), so \(F = [Mg(0.2) + m_0 g x]/d_{\text{scale}}\). This is a straight line in \(x\) with \(y\)-intercept \(0.2Mg/d_{\text{scale}}\) and slope \(m_0 g/d_{\text{scale}}\).

\(m_0 = 0.15\) kg; off-center expression \(F = (0.2Mg + m_0 g x)/d_{\text{scale}}\).
This problem is essentially identical to the 2025 AP Physics 1 FRQ Question 3. Linearization — recognizing that \(F = (\text{slope}) \cdot x + (\text{intercept})\) — is the experimental-design skill the AP exam tests again and again.
· block + pulley combined

(a) Block: weight \(mg\) down, tension \(T\) up. Pulley: axle force (upward, does no torque) and tension \(T\) tangential at rim (exerts torque \(TR\)).

(b) Block: \(mg - T = ma\). Pulley: \(TR = I\alpha = (\tfrac{1}{2}MR^2)\alpha\). No-slip: \(a = R\alpha\), so \(\alpha = a/R\). Substituting: \(TR = \tfrac{1}{2}MR^2(a/R) = \tfrac{1}{2}MRa\), giving \(T = \tfrac{1}{2}Ma\). Into block equation: \(mg - \tfrac{1}{2}Ma = ma\) → \(mg = a(m + M/2)\) →

$$a = \dfrac{mg}{m + M/2} = \dfrac{2mg}{2m + M}$$

(c) \(T = \tfrac{1}{2}Ma = \dfrac{mMg}{2m+M}\).

(d) If \(m = M\): \(a = 2mg/(2m+m) = 2g/3\). \(T = m^2 g/(3m) = mg/3\). Both match.

(e) Hoop vs cylinder. Hoop has \(I = MR^2\) (twice the solid cylinder's \(\tfrac{1}{2}MR^2\)). Following the same derivation: \(T = Ma\), and \(mg - Ma = ma\) gives \(a = mg/(m+M)\), less than the cylinder case. Block accelerates slower with the hoop — the hoop's larger inertia takes more torque to spin up at the same angular rate.

\(a = 2mg/(2m+M)\); \(T = mMg/(2m+M)\); hoop gives slower \(a\).
This is the canonical "massive pulley" problem and shows up in some form on nearly every AP exam. Memorize the template: (1) translational F=ma for the block, (2) rotational \(\tau = I\alpha\) for the pulley, (3) no-slip \(a = R\alpha\), (4) substitute to eliminate tension and \(\alpha\).

Worksheet C — Challenge

· climbing worker on a ladder

Forces on ladder: weight \(Mg\) at center, worker weight \(Mg\) (given \(m=M\)) at distance \(fL\) from bottom, normal \(N_w\) from wall (horizontal, frictionless wall), normal \(N_g\) and friction \(f_s\) from ground. Ladder at angle \(\theta\) to horizontal.

Equilibrium gives: \(N_g = 2Mg\), \(f_s = N_w\). Torque about base: \(N_w (L\sin\theta) = Mg(L/2)\cos\theta + Mg(fL)\cos\theta\), so \(N_w = Mg\cos\theta(1/2 + f)/\sin\theta\). At slip, \(f_s = \mu_s N_g = 2\mu_s Mg\). Setting \(N_w = f_s\): \(2\mu_s Mg = Mg\cot\theta(1/2 + f)\), so \(f = 2\mu_s \tan\theta - 1/2\).

\(f = 2\mu_s \tan\theta - \tfrac{1}{2}\)

Limit checks: As \(\theta \to 90°\) (vertical), \(\tan\theta \to \infty\), \(f \to \infty\) — worker can climb arbitrarily far (a vertical ladder doesn't slip). As \(\theta \to 0°\) (horizontal), \(\tan\theta \to 0\), \(f \to -1/2\) — worker can't even reach the ladder before it slips (negative \(f\) means the ladder would slip with no one on it, which makes sense for a horizontal ladder with weight at its CM).

· rolling race

For rolling down a ramp at angle \(\theta\) with \(I = \beta MR^2\): Newton's second law on the object center: \(Mg\sin\theta - f = Ma\). Torque about center from friction: \(fR = I\alpha = \beta MR^2 (a/R)\), giving \(f = \beta Ma\). Substituting: \(Mg\sin\theta - \beta Ma = Ma\) → \(a = g\sin\theta/(1+\beta)\).

Sphere (\(\beta = 2/5\)): \(a = g\sin\theta/1.4 \approx 0.714 g\sin\theta\). Cylinder/disk (\(\beta = 1/2\)): \(a = g\sin\theta/1.5 \approx 0.667 g\sin\theta\). Hoop (\(\beta = 1\)): \(a = g\sin\theta/2 = 0.5 g\sin\theta\).

Order of arrival: sphere first, cylinder second, hoop last. \(a = g\sin\theta/(1+\beta)\).
Mass and radius drop out entirely — only the shape factor \(\beta\) matters. A ping-pong ball and a bowling ball roll down the ramp at the same acceleration. Objects with more mass near the axis (smaller \(\beta\)) win. This is a famous demonstration problem.
· Atwood with massive pulley

Let \(T_1, T_2\) be tensions on the \(m_1, m_2\) sides; \(a\) the magnitude of acceleration; \(\alpha\) the pulley's angular acceleration. Since \(m_1 > m_2\): \(m_1 g - T_1 = m_1 a\) and \(T_2 - m_2 g = m_2 a\). Pulley: \((T_1 - T_2)R = I\alpha = \tfrac{1}{2}M_p R^2 (a/R)\), so \(T_1 - T_2 = \tfrac{1}{2}M_p a\).

Adding the two block equations: \((m_1 - m_2)g = (m_1 + m_2)a + (T_1 - T_2) = (m_1 + m_2)a + \tfrac{1}{2}M_p a\), so:

$$a = \dfrac{(m_1 - m_2)g}{m_1 + m_2 + M_p/2}$$

Then \(T_1 = m_1(g - a)\) and \(T_2 = m_2(g + a)\).

\(a = (m_1-m_2)g/(m_1+m_2+M_p/2)\); \(T_1 > T_2\) when pulley has mass.
When \(M_p \to 0\), this recovers the standard Atwood result \(a = (m_1-m_2)g/(m_1+m_2)\) and \(T_1 = T_2\) — limit check passed.
· falling chimney

Let \(\phi\) be the angle the chimney has rotated from vertical. Rotational inertia about base: \(I = \tfrac{1}{3}ML^2\). Gravity torque: \(\tau = Mg(L/2)\sin\phi\). So \(\alpha = \tau/I = (3g\sin\phi)/(2L)\).

Consider a point at distance \(\ell\) from the base. Its tangential acceleration is \(a_t = \ell \alpha = \ell (3g\sin\phi)/(2L)\). The vertical component is \(a_{\text{vert}} = a_t \cos\phi = (3g\ell \sin\phi \cos\phi)/(2L) = (3g\ell \sin(2\phi))/(4L)\).

This is maximized when \(\phi = 45°\). At that moment, \(a_{\text{vert}} = 3g\ell/(4L)\). Setting this equal to \(g\): \(3\ell/(4L) = 1 \Rightarrow \ell = 4L/3\). But this is beyond the top of the chimney — so no part of the chimney has \(a_{\text{vert}} = g\) except in the \(\phi \to 90°\) limit.

However, the key insight: the top of the chimney has vertical acceleration \(a_{\text{vert,top}} = (3g\sin(2\phi))/(4)\) (with \(\ell = L\)), which exceeds \(g\) once \(\sin(2\phi) > 4/3\) — impossible, since \(\sin\) maxes at 1. More careful analysis using angular + tangential acceleration combined shows the top can exceed \(g\) in total (not just vertical) once \(\phi\) is past some critical angle. The chimney breaks because the required tension to hold the top connected to the bottom exceeds the chimney's tensile strength.

Qualitative: top accelerates faster than \(g\) beyond a critical angle, which is why falling chimneys snap.
This is the classic "break the pencil" demo. The moment a rigid object of length \(L\) angularly accelerates about its base, every point has a different linear acceleration. Points far from the base must travel faster than a free-falling object to keep up with the rigid constraint — which requires internal forces larger than the material can provide.
· yo-yo

Let \(a\) be downward acceleration, \(T\) string tension, \(\alpha\) angular acceleration. Linear: \(Mg - T = Ma\). Rotational (string winds at axle, radius \(r\)): \(Tr = I\alpha = \tfrac{1}{2}MR^2 \alpha\). String doesn't slip: \(a = r\alpha\), so \(\alpha = a/r\). Substituting: \(Tr = \tfrac{1}{2}MR^2 (a/r) = MR^2 a/(2r)\), giving \(T = MR^2 a/(2r^2)\).

Into linear: \(Mg - MR^2 a/(2r^2) = Ma\) → \(g = a(1 + R^2/(2r^2))\) →

$$a = \dfrac{g}{1 + R^2/(2r^2)}$$

When \(r \ll R\): \(R^2/(2r^2) \gg 1\), so \(a \approx 2gr^2/R^2 \ll g\). The yo-yo falls much slower than free-fall. Tension: \(T = Mg - Ma \approx Mg(1 - 2r^2/R^2) \approx Mg\). So \(T \approx Mg\), meaning the string supports nearly all the weight.

Wait — let me check that. If \(a \ll g\), then \(T = M(g - a) \approx Mg\), which is close to the weight, not much less. My original statement was wrong. When \(r \ll R\), the string tension is very close to the full weight of the yo-yo. The small falling acceleration comes from only a tiny excess of weight over tension.

\(a = g/(1 + R^2/(2r^2))\); when \(r \ll R\), \(a \ll g\) and \(T \approx Mg\).
A yo-yo is essentially a very-low-friction free-fall: nearly all the weight is supported by the string tension, leaving only a tiny net force to accelerate the yo-yo. This is why you can watch a yo-yo fall for a full second or more — a full second of free-fall would cover 5 meters.
· · ·
08
Common mistakes

Seven errors that cost students points on this unit

These are the specific misreadings that show up repeatedly on student work in Unit 5. Wrong approach, right approach, why the mistake happens, plus a practice problem for each.

Writing \(\tau = I\omega\) instead of \(\tau = I\alpha\)

A disk is rotating at 10 rad/s with \(I = 2\) kg·m². Find the net torque.

\(\tau = I\omega = (2)(10) = 20\) N·m.
You can't determine the net torque from just \(I\) and \(\omega\). The rotational analog of \(F = ma\) is \(\tau = I\alpha\) (involving acceleration, not velocity). If the disk rotates at constant \(\omega\), then \(\alpha = 0\) and \(\tau_{\text{net}} = 0\). If \(\alpha\) is given, multiply that by \(I\).

The quantity \(I\omega\) is actually angular momentum \(L\) — a different concept you'll meet in Unit 6. Students mix up the two because both "involve \(I\)."

A wheel of rotational inertia 5 kg·m² is initially at rest. A net torque of 10 N·m is applied for 3 seconds. Find (a) the angular acceleration, (b) the angular velocity after 3 s.

Using \(mr\) instead of \(mr^2\) for rotational inertia

A 2 kg mass is 3 m from the axis. Find its rotational inertia.

\(I = mr = (2)(3) = 6\) kg·m.
\(I = mr^2 = (2)(3)^2 = 18\) kg·m².

The \(r^2\) factor comes from the fact that kinetic energy of a rotating mass is \(\tfrac{1}{2}mv^2 = \tfrac{1}{2}m(r\omega)^2 = \tfrac{1}{2}(mr^2)\omega^2\). The \(r^2\) emerges naturally from the \(v^2\) in kinetic energy. Students who don't see this derivation forget the exponent.

A 3 kg point mass is attached to a 2 m massless rod. Find the rotational inertia about (a) the far end of the rod, (b) the near end (the mass is at this end).

Forgetting the \(\sin\theta\) in torque when force is angled

A 20 N force is applied at a 30° angle to a wrench, 0.5 m from the bolt. Find the torque.

\(\tau = rF = (0.5)(20) = 10\) N·m.
\(\tau = rF\sin\theta = (0.5)(20)\sin 30° = 5\) N·m. Only the component of force perpendicular to the moment arm generates torque.

When torque is first introduced with perpendicular forces (\(\theta = 90°\), \(\sin\theta = 1\)), students anchor on \(\tau = rF\) and forget the general version. Always check the geometry: is the force perpendicular to the radius, or is there an angle?

A 40 N force is applied to the end of a 0.6 m rod at 60° above the rod. Find the torque about the other end. Then solve it a second way using the lever arm formulation.

Choosing a bad pivot for torque equations

A uniform 10 kg beam is held horizontally by a hinge at the left and a rope at the right. Find the tension in the rope.

Take torques about the center. Now the rope tension and the hinge force both produce nonzero torques, leaving two unknowns in one equation.
Take torques about the hinge. The hinge force drops out (zero lever arm), leaving only the beam weight and rope tension. One unknown, one equation — solvable immediately.

Students instinctively pivot "at the center" because it feels symmetric. But the smart move is always: pivot wherever an unknown force is applied, so that force contributes zero torque and disappears from your equation.

A 60 kg painter stands 2 m from the left end of a 4 m uniform plank of mass 20 kg. The plank is supported at both ends. Find both support forces by choosing each end as the pivot in turn.

Placing the beam's weight at its end instead of its center

A uniform 3 m beam of weight 100 N is hinged at one end. Find the torque about the hinge due to gravity.

Weight acts at the far end (3 m from hinge): \(\tau = (100)(3) = 300\) N·m.
For a uniform beam, gravity effectively acts at the center of mass, which is the midpoint of the beam. Distance from hinge: 1.5 m. \(\tau = (100)(1.5) = 150\) N·m.

The weight of an extended object is distributed along its length. For torque purposes, we replace the distributed weight with a single force acting at the center of mass. For a uniform object, that's the geometric center.

A uniform 2 m rod of mass 4 kg is pivoted at one end. An additional 1 kg mass is glued to the rod 0.5 m from the pivot. What is the total gravitational torque about the pivot when the rod is horizontal?

Mixing degrees and radians in the same calculation

A wheel accelerates from rest at 2 rad/s² for 10 seconds. Find the angle it has turned through.

\(\theta = \tfrac{1}{2}\alpha t^2 = \tfrac{1}{2}(2)(100) = 100\) degrees.
\(\theta = 100\) radians. If you want the answer in degrees, multiply by \(180/\pi\): about 5730°, or equivalently about 16 full revolutions.

Angular acceleration is in rad/s², so the angle from \(\theta = \tfrac{1}{2}\alpha t^2\) comes out in radians. Confusion enters when problems casually refer to rotations as "degrees" or "revolutions" without clarifying the units.

A disk at rest is given an angular acceleration of 4 rad/s². After 2 seconds, how many full revolutions has it made? (1 revolution = \(2\pi\) radians.)

Forgetting that tension differs on two sides of a massive pulley

In an Atwood machine with a massive pulley, is the tension in the string the same on both sides?

Yes — tension is always the same throughout a light string.
Only when the pulley is massless. If the pulley has rotational inertia, then the net torque on the pulley is \((T_1 - T_2)R = I\alpha\). If \(I \neq 0\) and \(\alpha \neq 0\), then \(T_1 \neq T_2\) — the tensions must differ to provide the net torque that spins the pulley up.

The "tension is the same everywhere in a light string" shortcut works for massless/frictionless pulleys because there's no torque needed to change the pulley's rotation. Once the pulley has mass, that shortcut breaks.

Two blocks of mass \(m\) and \(2m\) hang from a massive pulley (solid disk, mass \(m\), radius \(R\)). Derive an expression for the tension in the string on each side and confirm \(T_1 \neq T_2\).

· · ·
09
Tutor's corner

What I learned going through this unit — and from helping others through it

What every student struggles with (in order)

The #1 sticking point in Unit 5 is that students don't trust the translational-rotational analogy. They see a new set of symbols and convince themselves rotation is entirely new physics. It isn't. Every equation you wrote in Units 1–2 has an exact rotational twin. When you're stuck on a rotational problem, write out the translational version of the same problem first — then swap \(x \to \theta\), \(v \to \omega\), \(a \to \alpha\), \(m \to I\), \(F \to \tau\). The algebra is identical.

The #2 struggle is pivot choice. Students compute torques about whatever point "feels" central, and end up with equations containing multiple unknowns. The entire purpose of choosing a pivot is to eliminate an unknown force — pivot right at that force's application point so it contributes zero torque. This single trick turns half of Unit 5's problems from nightmares into one-line calculations.

What separates a 4 from a 5

Fluency with the translational ↔ rotational dictionary. Students who score a 5 can look at \(\tau = I\alpha\) and immediately recognize its structural identity to \(F = ma\) — same equation, different costume. They can predict without calculation that "if the rotational inertia doubles and torque stays the same, angular acceleration halves." Students who score a 4 can use the equation but treat it as a standalone fact divorced from everything they learned earlier. Build the bridges between concepts; that's what gets you the 5.

The question that reveals true understanding

"A figure skater pulls her arms in during a spin. She speeds up. No external torque acts on her. What changed?" The shallow answer is "she reduced her rotational inertia" — which is true but doesn't explain why \(\omega\) went up. The deep answer requires Unit 6: angular momentum \(L = I\omega\) is conserved in the absence of external torques. When she reduces \(I\), \(\omega\) must increase to keep \(L\) constant. Students who grasp this before even reaching Unit 6 have understood the architecture of rotational mechanics.

Exam-day strategy

On static equilibrium FRQs: always compute torques about the point with the most unknown forces acting on it. This is not always the "most natural" pivot — sometimes it's at a weird location that happens to eliminate two forces at once. Read the problem, identify the unknowns you don't care about, and pivot there.

On combined translation + rotation problems (pulleys, rolling, yo-yos): the template is always two Newton's second laws — one for translation, one for rotation — linked by \(a = R\alpha\). Write them mechanically before attempting to solve. Partial credit is generous for setting up the equations correctly even if you make an algebra mistake at the end.

On multiple choice: for "which has more rotational inertia" questions, remember the hierarchy from most-concentrated-at-axis to most-concentrated-at-rim: solid sphere (2/5) < solid cylinder/disk (1/2) < hoop (1). Same mass, same radius — that ordering of \(I\) values is worth memorizing cold.

A surprising real-world connection

Helicopters have a tail rotor. Why? Newton's third law in rotational form: when the main rotor spins one way, an equal and opposite torque is exerted on the helicopter body, which would spin it in the opposite direction. The tail rotor provides an external torque that cancels this out, keeping the body stable. Without it, helicopters would just spin. Same principle: submarines use contra-rotating propellers, quadcopters alternate rotor directions, and figure skaters use tiny hand-movements to adjust orientation mid-jump. Angular momentum and torque are the hidden grammar of every piece of rotating technology in your life.

One piece of advice before tomorrow's test

If you only have time to drill one type of problem tonight, make it static equilibrium with a uniform beam. Every AP exam has at least one. The template: draw the beam, mark every force with its location, pick a pivot that kills an unknown, write \(\sum \tau = 0\) and \(\sum F_y = 0\), solve. Do three of these from Worksheet A (problem 6) and Worksheet B (problems 12, 13, 19). You'll develop a reflex for pivot choice that carries you through whatever tomorrow throws at you.

Also: if combined translation+rotation appears (pulley problem), remember the template — two F=ma's linked by \(a = r\alpha\). Problem 10 on Worksheet A and the extended FRQ 20 on Worksheet B drill this. Master those and you've covered the bulk of what Unit 5 tests.

End of Unit 5 · Torque & Rotational Dynamics   ·  First Principles Physics