AP Physics 1 Unit 5 Notes: Torque, Rotational Motion, and Angular Momentum

Torque: the rotational “push” that causes spinning

What torque is (and what it is not)

Torque is a measure of how effective a force is at causing an object to rotate about a chosen axis (or pivot). In everyday life you feel torque when you push on a door: the same force that barely budges the door near the hinge easily swings it open when applied at the handle.

A common early confusion is to treat torque as “the same thing as force.” Force causes linear acceleration; torque causes angular acceleration. They are related because the same force can produce both translation and rotation depending on where and how it is applied.

Torque depends on three ideas at once:

• how big the force is,
• where the force is applied relative to the axis,
• and the direction of the force relative to the line from the axis to the point of application.

The torque magnitude formula and the lever arm idea

For a force applied at a point a distance r from the axis, the magnitude of the torque is

\tau = rF\sin(\theta)

where:

• \tau is the torque magnitude (units: N·m),
• r is the distance from the axis to the point where the force is applied (meters),
• F is the force magnitude (newtons),
• \theta is the angle between the radius vector (from axis to point) and the force.

The factor \sin(\theta) captures a key physical truth: only the component of the force that is perpendicular to the radius “tries to spin” the object. If you push directly toward the hinge of the door (force parallel to the radius), you produce essentially no torque.

A very useful reframe is in terms of the lever arm (also called the moment arm): the perpendicular distance from the axis to the force’s line of action. If the lever arm is r_\perp, then

\tau = F r_\perp

This is often the fastest way to compute torque in geometry-heavy problems: draw the line of action of the force, then measure the perpendicular distance from the pivot to that line.

Direction (sign) of torque: clockwise vs counterclockwise

Torque is a vector in full physics, but in AP Physics 1 you usually treat rotation about a fixed axis, so torque can be positive or negative depending on whether it tends to cause counterclockwise (CCW) or clockwise (CW) rotation.

A consistent convention is:

• CCW torques are positive,
• CW torques are negative.

What matters is consistency within a problem. Many mistakes come from switching sign conventions mid-solution.

Net torque and rotational inertia (preview)

Just like multiple forces combine into a net force, multiple torques combine into a net torque:

\tau_{\text{net}} = \sum \tau_i

Net torque is what determines angular acceleration. But an object’s resistance to angular acceleration is not its mass alone; it is its moment of inertia (introduced later). Conceptually, mass far from the axis “matters more” for rotation.

Worked example 1: torque on a wrench

You apply a 40\text{ N} force at the end of a 0.25\text{ m} wrench. The force is perpendicular to the wrench.

Because the force is perpendicular, \sin(\theta)=1, so

\tau = rF = (0.25)(40)=10\text{ N·m}

If you instead push at a 30^\circ angle relative to the wrench, then

\tau = rF\sin(30^\circ) = (0.25)(40)(0.5)=5\text{ N·m}

Same force, same distance, half the torque because the perpendicular component is smaller.

Worked example 2: choosing signs

A horizontal beam pivots about its left end. A downward force at the right end tends to rotate it clockwise about the left pivot. If CCW is positive, that torque is negative. If you write equilibrium equations later, this sign choice will matter.

Exam Focus

• Typical question patterns:
  • “A force is applied at some point/angle—find the torque about a pivot.”
  • “Several forces act on a rigid body—compute the net torque (with signs).”
  • “Find the lever arm or determine where to apply a force to achieve a required torque.”
• Common mistakes:
  • Using \tau=rF even when the force is not perpendicular (forgetting \sin(\theta)).
  • Using the distance to the point of application instead of the perpendicular lever arm to the line of action.
  • Inconsistent sign conventions (calling CW positive in one term and negative in another).

Moment of inertia: how mass distribution controls rotational response

What moment of inertia means

Moment of inertia is the rotational analog of mass. For translation, mass measures resistance to linear acceleration. For rotation about a fixed axis, moment of inertia measures resistance to angular acceleration.

The essential idea is distribution: a kilogram close to the axis contributes much less to moment of inertia than a kilogram far away. That is why it is easier to spin a bicycle wheel without its tire than with its tire: more mass is at larger radius.

Definition for point masses and the big idea behind all formulas

For a system of point masses, the moment of inertia about a chosen axis is

I = \sum m_i r_i^2

where:

• I is moment of inertia (units: kg·m²),
• m_i is the mass of the ith particle,
• r_i is the perpendicular distance from the axis to that particle.

The square on r is important: doubling distance increases contribution by a factor of four.

For extended rigid objects (like a disk or rod), the full definition uses an integral, but AP Physics 1 typically gives you the standard results or expects you to use a provided formula.

Common rigid-body moments of inertia (about standard axes)

Below are frequently used results. Always check that the axis in the problem matches the axis in the formula.

Two conceptual checks help you avoid memorizing blindly:

• If most mass is far from the axis, I should be larger.
• Units must be kg·m².

Why choosing the axis matters

Moment of inertia depends on the axis. The same object can be easy to spin about one axis and hard to spin about another. A rod spun about its center has a smaller I than the same rod spun about its end, because on average the mass is farther from the axis when the axis is at the end.

Worked example 1: comparing hoop vs disk

Two objects have the same mass M and radius R. One is a hoop and one is a solid disk. About the central axis:

I_{\text{hoop}} = MR^2

I_{\text{disk}} = \tfrac{1}{2}MR^2

So the hoop has twice the moment of inertia. If the same torque is applied, the hoop’s angular acceleration will be smaller (this becomes precise in rotational Newton’s second law later).

Worked example 2: point masses on a light rod

Two masses are attached to a very light rod (rod mass negligible). One mass m is at r=0.20\text{ m} and another mass 2m is at r=0.10\text{ m} from the axis. Then

I = m(0.20)^2 + 2m(0.10)^2 = m(0.04) + 2m(0.01)=0.06m

Even though one mass is double, the one farther out can still dominate because of the square.

Exam Focus

• Typical question patterns:
  • “Compute I for a system of point masses about a given axis.”
  • “Compare angular accelerations/energies for different shapes (hoop vs disk vs sphere if provided).”
  • “An object changes axis (center vs end)—determine which has larger I and predict effects.”
• Common mistakes:
  • Plugging into a rigid-body formula with the wrong axis or wrong shape.
  • Forgetting the square on distance in I=\sum mr^2.
  • Mixing radius and diameter, or using centimeters without converting to meters.

Rotational kinematics: describing angular motion

Angular position, velocity, and acceleration

Rotational motion uses angular variables that mirror linear ones.

• Angular displacement \Delta\theta describes how much an object rotates (radians).
• Angular velocity \omega describes how fast it rotates (radians per second).
• Angular acceleration \alpha describes how quickly \omega changes (radians per second squared).

Definitions:

\omega = \frac{\Delta\theta}{\Delta t}

\alpha = \frac{\Delta\omega}{\Delta t}

For many AP Physics 1 problems, you treat the rotation as about a fixed axis with constant angular acceleration.

Why radians matter

Radians make rotational relationships clean because an arc length is directly proportional to angle:

s = r\theta

where s is the arc length along a circle of radius r.

This leads immediately to the most-used connections between linear and rotational motion:

v = r\omega

a_t = r\alpha

Here v is tangential speed and a_t is tangential acceleration (the part that changes the speed along the circular path).

A major misconception is to use degrees in these formulas. In AP Physics 1, assume angles in rotational kinematics are in radians unless explicitly told otherwise.

Constant angular acceleration equations

When angular acceleration is constant, the equations look exactly like constant-acceleration linear kinematics, but with x \to \theta, v \to \omega, and a \to \alpha:

\omega = \omega_0 + \alpha t

\Delta\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2

\omega^2 = \omega_0^2 + 2\alpha\Delta\theta

These are powerful because they let you relate rotational motion without needing torque or energy.

Centripetal acceleration still exists for rotation

Even if an object rotates at constant angular speed, points on the object move in circles and have centripetal acceleration toward the axis:

a_c = \frac{v^2}{r}

Using v=r\omega gives

a_c = r\omega^2

This centripetal acceleration changes the direction of velocity, not its magnitude.

Worked example 1: spinning up a wheel

A wheel starts from rest and reaches \omega=12\text{ rad/s} in 3\text{ s} with constant angular acceleration.

Compute angular acceleration:

\alpha = \frac{\Delta\omega}{\Delta t} = \frac{12-0}{3}=4\text{ rad/s}^2

Angle turned in that time:

\Delta\theta = \tfrac{1}{2}\alpha t^2 = \tfrac{1}{2}(4)(3^2)=18\text{ rad}

If a point on the rim has radius 0.50\text{ m}, then its final tangential speed is

v=r\omega=(0.50)(12)=6\text{ m/s}

Worked example 2: relating tangential acceleration

A disk of radius 0.20\text{ m} has angular acceleration \alpha=5\text{ rad/s}^2. The tangential acceleration of a point on the edge is

a_t=r\alpha=(0.20)(5)=1.0\text{ m/s}^2

Notice the dependence on radius: points farther out have larger tangential acceleration for the same \alpha.

Exam Focus

• Typical question patterns:
  • “Given \omega_0, \omega, \alpha, t, find \Delta\theta (or similar).”
  • “Convert between rotational and tangential variables using v=r\omega and a_t=r\alpha.”
  • “Find centripetal acceleration at a given radius for a rotating rigid body.”
• Common mistakes:
  • Using degrees instead of radians in s=r\theta and kinematics equations.
  • Forgetting that different points on a rigid body share the same \omega and \alpha but not the same v and a_t.
  • Confusing centripetal acceleration a_c with tangential acceleration a_t (one changes direction, the other changes speed).

Rotational dynamics: connecting torque, angular acceleration, and forces

Rotational Newton’s second law

For translation, Newton’s second law is \sum F = ma. For rotation about a fixed axis, the analogous relationship is

\sum \tau = I\alpha

This equation is the core of rotational dynamics:

• \sum \tau is the net external torque about the axis,
• I is the moment of inertia about that axis,
• \alpha is the angular acceleration.

It tells you two deep ideas:

1. For a given torque, larger I means smaller angular acceleration.
2. For a given object (fixed I), net torque determines \alpha.

A frequent error is to treat I like mass and assume it is just MR^2 for everything. You must use the correct I for the object and axis.

Applying \sum\tau = I\alpha step by step

Rotational dynamics problems usually follow a consistent structure:

1. Choose an axis of rotation.
2. Identify forces that produce torque about that axis (forces through the axis produce zero torque).
3. Write torques with correct lever arms and signs.
4. Set \sum\tau = I\alpha.
5. If the object is also translating, simultaneously apply \sum F = ma to the center of mass.

The “both translate and rotate” part is where many AP problems live. For example, a yo-yo unwinding or a pulley accelerating with a rope involves both linear and angular dynamics connected by no-slip constraints.

The no-slip constraint: linking linear and angular acceleration

If a string does not slip on a rotating pulley (or an object rolls without slipping), the linear acceleration of the string (or the center of mass) relates to angular acceleration by

a = r\alpha

where r is the radius where the string contacts the pulley (or the rolling radius).

This is not an extra physics law; it is a geometric constraint that comes from equal arc length and linear displacement.

Worked example 1: torque causing angular acceleration

A solid disk with mass 2.0\text{ kg} and radius 0.30\text{ m} experiences a constant net torque of 1.8\text{ N·m} about its center.

First find the disk’s moment of inertia:

I=\tfrac{1}{2}MR^2=\tfrac{1}{2}(2.0)(0.30^2)=0.090\text{ kg·m}^2

Now apply rotational Newton’s second law:

\alpha=\frac{\tau}{I}=\frac{1.8}{0.090}=20\text{ rad/s}^2

Worked example 2: hanging mass driving a pulley (classic AP setup)

A mass m hangs from a light string wrapped around a solid disk pulley of radius R and moment of inertia I. The mass falls with acceleration a, and the pulley rotates with angular acceleration \alpha. Assume no slipping.

For the falling mass (downward positive):

mg - T = ma

The tension produces torque on the pulley:

\tau = TR

Rotational dynamics of pulley:

TR = I\alpha

No-slip constraint:

a = R\alpha

Substitute \alpha=a/R into the torque equation:

TR = I\frac{a}{R}

So

T = \frac{Ia}{R^2}

Plug into the mass equation:

mg - \frac{Ia}{R^2} = ma

Solve for a:

a = \frac{mg}{m + \frac{I}{R^2}}

This result is extremely informative: the pulley’s rotational inertia acts like “extra mass” that reduces acceleration.

Exam Focus

• Typical question patterns:
  • “Given torques and I, find \alpha (or vice versa).”
  • “Pulley/rope systems: combine \sum F=ma with \sum\tau=I\alpha and a=r\alpha.”
  • “Identify which forces create torque about a chosen axis and which do not.”
• Common mistakes:
  • Forgetting that forces whose lines of action pass through the axis create zero torque.
  • Using a=r\alpha when there is slipping, or forgetting it when there is no slipping.
  • Mixing up radius R (geometry) with displacement r (generic distance), leading to wrong lever arms.

Rotational work, energy, and power: an energy lens on spinning

Why energy methods are valuable

Rotational dynamics can be solved with forces and torques, but energy methods often make multi-step motion simpler because you can relate initial and final states without solving for time. Many AP problems reward recognizing when conservation of energy applies.

Rotational kinetic energy

A rotating rigid body has rotational kinetic energy:

K_{\text{rot}} = \tfrac{1}{2}I\omega^2

This mirrors translational kinetic energy \tfrac{1}{2}mv^2, with I playing the “mass role” and \omega playing the “speed role.”

If an object both translates and rotates (like a rolling wheel), its total kinetic energy is

K_{\text{total}} = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2

A common misconception is to double-count kinetic energy by using only \tfrac{1}{2}mv^2 for a rolling object. Rolling motion stores energy in both translation of the center of mass and rotation about the center of mass.

Work done by a torque

For a constant torque applied through an angular displacement \Delta\theta, the work done by that torque is

W = \tau\Delta\theta

This parallels linear work W=F\Delta x.

More generally, if torque varies with angle, work is the area under the \tau vs \theta graph. AP Physics 1 sometimes tests this graph interpretation.

Power in rotational motion

Power is the rate of doing work. In rotational form:

P = \tau\omega

This is useful for motors: at a given angular speed, producing more torque requires more power.

Conservation of energy with rotation

If nonconservative work (like kinetic friction doing net work) is negligible, you can use energy conservation:

E_i = E_f

For a common case: an object starts from rest and a torque does work to spin it up. If the torque is constant, you can combine

W = \tau\Delta\theta

with

W = \Delta K_{\text{rot}} = \tfrac{1}{2}I\omega^2 - \tfrac{1}{2}I\omega_0^2

Rolling without slipping down an incline (energy view)

For an object that rolls without slipping, static friction often does no net work (the contact point is instantaneously at rest relative to the surface), so mechanical energy can be conserved even though friction is present.

Suppose an object of mass m drops a vertical height h while rolling without slipping. If it starts from rest:

mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2

With the rolling constraint v=r\omega, you can solve for v:

mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\frac{v^2}{r^2}

Factor v^2:

mgh = \tfrac{1}{2}v^2\left(m + \frac{I}{r^2}\right)

So

v^2 = \frac{2mgh}{m + \frac{I}{r^2}}

Again you see I/r^2 acting like extra effective mass.

Worked example 1: which rolls faster (disk vs hoop)

A hoop and a solid disk (same M and R) roll without slipping down the same height h from rest.

Using the result above:

• For hoop, I=MR^2 so I/r^2=M.
• For disk, I=\tfrac{1}{2}MR^2 so I/r^2=\tfrac{1}{2}M.

Thus the denominator m + I/r^2 is larger for the hoop, making v^2 smaller. The disk reaches the bottom with greater speed. This matches intuition: the hoop “stores” more of the energy in rotation because it has larger I.

Worked example 2: torque work to spin up

A constant torque \tau=6.0\text{ N·m} acts on a rigid object initially at rest. After rotating through \Delta\theta=4.0\text{ rad}, its angular speed is \omega=8.0\text{ rad/s}. Find its moment of inertia.

Work done by torque:

W=\tau\Delta\theta=(6.0)(4.0)=24\text{ J}

This becomes rotational kinetic energy:

24 = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}I(8.0^2)=32I

So

I=\frac{24}{32}=0.75\text{ kg·m}^2

Exam Focus

• Typical question patterns:
  • “Use energy to find speed at bottom for rolling objects; compare shapes.”
  • “Use W=\tau\Delta\theta and K_{\text{rot}}=\tfrac{1}{2}I\omega^2 to connect torque, angle, and final \omega.”
  • “Interpret a \tau vs \theta graph to find work.”
• Common mistakes:
  • Treating a rolling object’s kinetic energy as only translational (forgetting rotational term).
  • Assuming friction always dissipates energy; static friction in pure rolling often does no net work.
  • Mixing up \omega and \theta in work expressions (work uses angle, not angular speed).

Angular momentum and its conservation

What angular momentum represents

Angular momentum is the rotational analog of linear momentum. It measures “how much rotational motion” a system has and, crucially, it is conserved when there is no net external torque.

For a rigid body rotating about a fixed axis, angular momentum is

L = I\omega

where:

• L is angular momentum (units: kg·m²/s),
• I is moment of inertia about the axis,
• \omega is angular speed.

For a single particle moving in a circle of radius r with tangential momentum p=mv, the magnitude can be written as

L = mvr

This particle form helps you understand why pulling mass inward increases rotation rate: decreasing r tends to decrease I, so to keep L constant, \omega must increase.

The torque-angular momentum connection

Net external torque changes angular momentum. The rotational analog of “net force changes momentum” is

\tau_{\text{net}} = \frac{\Delta L}{\Delta t}

This is the statement that torque is the time rate of change of angular momentum.

For constant net torque over a time interval \Delta t, the angular impulse is

\tau_{\text{net}}\Delta t = \Delta L

AP Physics 1 sometimes uses this in situations like a brief applied torque spinning up a wheel.

Conservation of angular momentum

If the net external torque about an axis is zero (or negligible), then angular momentum about that axis is conserved:

L_i = L_f

For a rigid body with changing moment of inertia (like a figure skater pulling arms in):

I_i\omega_i = I_f\omega_f

This is one of the most conceptually rich results in the unit. It explains why spinning becomes faster when mass moves closer to the axis.

A common misconception is to think energy must also be conserved in that process. In many “pull arms in” examples, rotational kinetic energy increases because internal work is done by the person.

Worked example 1: skater pulls arms in

A skater reduces her moment of inertia from I_i=3.0\text{ kg·m}^2 to I_f=1.5\text{ kg·m}^2 while spinning. Initially \omega_i=2.0\text{ rad/s}. Neglect external torque.

Conserve angular momentum:

I_i\omega_i = I_f\omega_f

\omega_f = \frac{I_i\omega_i}{I_f} = \frac{(3.0)(2.0)}{1.5}=4.0\text{ rad/s}

Angular speed doubles because moment of inertia halves.

If you check energy:

K_{\text{rot}}=\tfrac{1}{2}I\omega^2

Initial:

K_i=\tfrac{1}{2}(3.0)(2.0^2)=6.0\text{ J}

Final:

K_f=\tfrac{1}{2}(1.5)(4.0^2)=12\text{ J}

Energy increased, meaning the skater did internal work.

Worked example 2: rotational collision (sticky putty on a disk)

A small blob of putty of mass m drops onto the edge of a rotating disk and sticks. If external torque about the disk axis is negligible during the collision, angular momentum is conserved even though mechanical energy is not.

Initial angular momentum (disk alone):

L_i = I_{\text{disk}}\omega_i

Final moment of inertia:

I_f = I_{\text{disk}} + mR^2

Final angular momentum:

L_f = I_f\omega_f

Conservation gives

I_{\text{disk}}\omega_i = (I_{\text{disk}} + mR^2)\omega_f

So

\omega_f = \frac{I_{\text{disk}}}{I_{\text{disk}} + mR^2}\,\omega_i

The angular speed decreases because you increased I without adding external angular momentum.

Exam Focus

• Typical question patterns:
  • “A rotating system changes shape (changing I)—use I_i\omega_i=I_f\omega_f.”
  • “Collision or sticking event on a rotating object—conserve angular momentum, then possibly discuss energy loss.”
  • “Relate torque and angular impulse to changes in angular momentum.”
• Common mistakes:
  • Conserving angular momentum when there is a significant external torque (for example, friction from an axle that is not negligible).
  • Assuming rotational kinetic energy is conserved in inelastic rotational events (it generally is not).
  • Forgetting to include added masses as mR^2 contributions to I in sticking problems.

Rotational equilibrium and static equilibrium (no angular acceleration)

What equilibrium means for rigid bodies

For a rigid object, equilibrium has two parts:

• Translational equilibrium: the center of mass is not accelerating.
• Rotational equilibrium: the angular acceleration is zero.

In equations:

\sum F = 0

\sum \tau = 0

If both are true, the object is in static equilibrium (not moving) or in dynamic equilibrium (moving at constant velocity and rotating at constant angular speed). In AP Physics 1, “equilibrium” problems are typically static: beams, ladders, sign supports, etc.

Why you can choose any pivot in torque equilibrium

When \sum\tau=0, you may compute torques about any point, not just the center of mass. This is a strategic advantage: choose a pivot that eliminates unknown forces.

For example, if a beam is supported at a hinge with unknown horizontal and vertical components, choosing the hinge as the torque pivot removes both hinge forces from the torque equation because their lever arm is zero.

A classic mistake is to choose a pivot and then still include a torque from a force applied at that pivot. If the line of action passes through the pivot, the lever arm is zero, so its torque is zero.

Center of mass and weight acting point

For uniform gravity near Earth’s surface, the object’s weight can be treated as a single force mg acting at the object’s center of mass.

For a uniform beam of length L, the center of mass is at its midpoint, L/2 from either end. In many equilibrium problems, recognizing where the weight acts is half the battle.

Two-equation (or three-equation) strategy for beams

Most static equilibrium problems in a plane use:

• one force balance in the vertical direction,
• (sometimes) one force balance in the horizontal direction,
• one torque balance about a chosen pivot.

You typically do not need more equations because AP problems usually have at most two or three unknowns.

Worked example 1: uniform beam with a hanging mass

A uniform horizontal beam of length L=4.0\text{ m} is hinged to a wall at the left end and supported by a cable at the right end. The beam has mass 20\text{ kg}. A 30\text{ kg} mass hangs from the beam at 3.0\text{ m} from the hinge. The cable is vertical (so it provides an upward tension T at the beam’s end). Find T.

Take torques about the hinge so hinge forces produce no torque.

Forces producing torque:

• Cable tension T upward at r=4.0\text{ m} gives CCW torque.
• Beam weight W_b = (20g) downward at midpoint r=2.0\text{ m} gives CW torque.
• Hanging weight W_m = (30g) downward at r=3.0\text{ m} gives CW torque.

Set net torque to zero (CCW positive):

T(4.0) - (20g)(2.0) - (30g)(3.0) = 0

Solve:

4.0T = 40g + 90g = 130g

T = \frac{130g}{4.0} = 32.5g

Using g=9.8\text{ m/s}^2:

T \approx 319\text{ N}

After finding T, you could use \sum F_y=0 to find the hinge’s vertical force if asked.

Worked example 2: tipping condition (conceptual)

A box on a rough floor is pushed horizontally at some height. Whether it slides or tips depends on which requirement is met first:

• sliding begins when the needed friction exceeds maximum static friction,
• tipping begins when the normal force effectively shifts to the edge and torque about the edge causes rotation.

AP questions may ask qualitatively which happens first when you change the push height: pushing higher increases the torque about the bottom edge, making tipping more likely before sliding.

Static friction’s role in equilibrium

Static friction adjusts to whatever value is required (up to a maximum) to satisfy equilibrium:

f_s \le \mu_s N

A frequent mistake is to set f_s = \mu_s N automatically. That equality is only true at the threshold of slipping.

Exam Focus

• Typical question patterns:
  • “Beam/ladder/sign in static equilibrium—use \sum F=0 and \sum\tau=0 to find tensions and support forces.”
  • “Choose the best pivot to eliminate unknown forces and solve efficiently.”
  • “Qualitative tipping vs sliding: compare torques and friction limits.”
• Common mistakes:
  • Including torque from forces acting at the chosen pivot (lever arm is zero).
  • Using the wrong location for the weight (forgetting the center of mass of a uniform object is at its midpoint).
  • Setting f_s=\mu_s N when the object is not on the verge of slipping.

Putting it together: multi-concept rotational problems and modeling choices

How AP problems blend models

Unit 5 questions often require you to decide which physics “lens” is most efficient:

• Torque and angular acceleration for direct cause-and-effect rotation.
• Energy when forces are complicated but initial/final states are clean.
• Angular momentum for short interactions (collisions) or shape changes with negligible external torque.
• Static equilibrium for structures that do not accelerate.

The hardest part is usually not algebra; it is choosing the correct model and constraints.

Example 1: rolling object down an incline (dynamics vs energy)

If asked for acceleration down the ramp, you can use dynamics (forces + torque) or energy.

Dynamics approach (conceptual outline):

• Write \sum F = ma along incline (includes friction).
• Write \sum\tau=I\alpha about center of mass (friction produces the torque).
• Use a=r\alpha for rolling without slipping.

Energy approach:

• Use mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2 to find speed at a given height.
• If you need acceleration, you often still return to dynamics because energy gives speed vs position, not directly acceleration unless you relate to kinematics.

AP frequently rewards recognizing that speed comparisons (which is faster) are easiest with energy, while friction direction and magnitude are clearer with dynamics.

Example 2: when angular momentum conservation applies (and when it does not)

Angular momentum conservation requires negligible net external torque about the axis.

Good candidates:

• a person on a low-friction turntable pulling arms in,
• a blob sticking to a spinning disk during a brief collision,
• a spinning wheel with negligible axle friction over a short time.

Bad candidates:

• a wheel slowing down due to significant friction torque,
• a rolling object where external torque from static friction about the center of mass is not zero (angular momentum about the center of mass changes).

A subtle but important point: you can sometimes conserve angular momentum about a carefully chosen point even when external forces exist, if their net torque about that point is zero. AP questions occasionally test this with clever pivot choices.

Example 3: identifying the correct “system”

In conservation problems, your choice of system determines what is external vs internal:

• If you choose “disk + putty” as the system, the sticking forces are internal, so you can conserve angular momentum about the axis during the collision (if external torques are negligible).
• If you choose only the disk as the system, the putty’s interaction is external and angular momentum of the disk alone is not conserved.

Thinking clearly about system boundaries prevents many conservation-law errors.

Exam Focus

• Typical question patterns:
  • “Decide whether to use energy, torque, or angular momentum, and justify the choice.”
  • “Multi-step: find acceleration first, then use kinematics to find speed/angle/time.”
  • “Explain (in words) what assumptions allow conservation laws to apply.”
• Common mistakes:
  • Applying a conservation law without checking its condition (external torque or nonconservative work).
  • Treating rolling without slipping as if it always implies energy conservation (it depends on whether dissipative forces do work).
  • Writing correct equations but mixing incompatible variables (using v in rotational energy without linking to \omega, or using \tau=rF with wrong lever arm).