Newton’s Laws for Rotation: Equilibrium and Dynamics (AP Physics 1 Unit 5)

Rotational Equilibrium and Newton's First Law in Rotational Form

What “rotational equilibrium” means

In linear motion, an object is in translational equilibrium when the net external force is zero, so its velocity doesn’t change. Rotation has an exactly parallel idea: an object is in rotational equilibrium when the net external torque about a chosen axis is zero, so its angular velocity doesn’t change.

Newton’s First Law says (for translation): if the net force is zero, velocity is constant. The rotational version is:

• If the net external torque about an axis is zero, the object’s angular velocity about that axis is constant.

That constant angular velocity could be zero (not rotating) or nonzero (spinning steadily). This is why a ceiling fan can spin at a steady rate even while many forces act on it: the forces may produce torques that cancel.

A crucial idea in AP Physics 1 is that “equilibrium” is not only “at rest.” Equilibrium means “not accelerating.” So you can have:

• Static equilibrium: not translating and not rotating.
• Dynamic equilibrium: translating at constant velocity and or rotating at constant angular velocity.

Torque: what it is and why it matters

Torque is the rotational “push” that tends to change an object’s rotation. A force can be large and still produce little torque if it acts close to the axis. Conversely, a modest force can produce a large torque if applied far from the axis.

In everyday terms, torque is why a door is easiest to open by pushing on the handle (far from the hinge) rather than near the hinge.

The magnitude of torque produced by a force depends on three things:

1. The force magnitude.
2. The distance from the axis to where the force is applied.
3. The direction of the force relative to the line from the axis to the application point.

Mathematically, for a force applied at distance r from the axis, with angle \theta between the radius line and the force direction:

\tau = rF\sin(\theta)

Where:

• \tau is the torque magnitude (units: N·m).
• r is the lever arm distance from axis to the point of application (m).
• F is force magnitude (N).
• \theta is the angle between \vec{r} and \vec{F}.

A very common and often clearer form uses the perpendicular lever arm r_\perp (the shortest distance from the axis to the force’s line of action):

\tau = F r_\perp

This highlights the physical meaning: only the component of force that “acts with a moment arm” produces torque.

Sign convention (clockwise vs counterclockwise)

Torque is a signed quantity in 2D problems. You choose a convention, typically:

• Counterclockwise torques are positive.
• Clockwise torques are negative.

Then you compute each torque with a sign and add them.

A common mistake is to assign signs based on where the force points (up or down) instead of whether it tends to rotate the object clockwise or counterclockwise about the axis.

Newton’s First Law in rotational form (net torque zero)

The rotational form can be summarized as:

\sum \tau = 0 \Rightarrow \alpha = 0

If angular acceleration \alpha is zero, angular velocity is constant. For rotational equilibrium problems, your job is usually to show that the torques cancel.

But AP Physics 1 equilibrium problems often involve objects that could both translate and rotate. In full static equilibrium (no linear acceleration and no angular acceleration), you typically need both conditions:

\sum F = 0

\sum \tau = 0

Even if your prompt is “rotational equilibrium,” the object is usually also not sliding, so the force balance matters too.

Why “axis choice” is powerful (and sometimes confusing)

A key skill is that you can compute torques about any axis you choose (as long as you’re consistent). In static equilibrium, if an object is not accelerating, then the net torque is zero about any point.

This matters because the “best” axis is often the one that eliminates unknown forces from your torque equation. If a force’s line of action passes through the axis, it produces zero torque about that axis, so you don’t have to include it.

For example, when a uniform beam is supported at a hinge with unknown hinge forces, choosing the hinge as the torque axis makes those hinge forces contribute zero torque, simplifying the math.

A subtle but important point: in non-equilibrium rotational dynamics, the statement “net torque depends on the axis” becomes more delicate because the relationship between torque and angular acceleration is simplest about the center of mass (or a fixed axis). But for AP Physics 1 static equilibrium situations, choosing a convenient axis is a standard and expected method.

Worked example 1: balancing a uniform beam (static equilibrium)

A uniform horizontal beam of length L = 4.0\ \text{m} and weight W_b = 200\ \text{N} is hinged to a wall at the left end. A cable attached to the right end pulls straight up with tension T. The beam is at rest.

Step 1: Choose the axis.
Pick the hinge as the torque axis. The hinge forces (unknown horizontal and vertical components) produce zero torque about the hinge.

Step 2: Identify torques about the hinge.

• The upward tension T at distance L produces counterclockwise torque: +TL.
• The beam’s weight acts at its center at distance L/2, producing clockwise torque: -W_b(L/2).

Step 3: Apply rotational equilibrium.

\sum \tau = TL - W_b\frac{L}{2} = 0

Solve for T:

TL = W_b\frac{L}{2}

T = \frac{W_b}{2} = 100\ \text{N}

Interpretation: the cable only needs to support half the beam’s weight because it has twice the lever arm compared with the beam’s center of mass.

What could go wrong: A frequent error is to set T = W_b because “upward must equal downward.” But vertical force balance involves hinge forces too. Torque balance isolates the rotation condition.

Worked example 2: when a force produces zero torque

You push on a door at the hinge line with a force perpendicular to the door. Even if the force is large, r_\perp = 0 because the line of action passes through the hinge axis.

\tau = F r_\perp = 0

Meaning: the push does not tend to rotate the door; it might compress the hinge or cause other stresses, but it won’t open the door.

Common misconceptions to watch for (within equilibrium)

• “If forces balance, torques must balance.” Not necessarily. You can have \sum F = 0 but \sum \tau \ne 0, which would cause pure rotation without translation (think of a couple: equal and opposite forces separated by a distance).
• “Torque uses the whole force.” Only the perpendicular component matters. Using \sin(\theta) incorrectly is a major source of errors.
• “The center of mass is always the pivot.” The pivot (axis) is where you choose to compute torque; the center of mass is where weight effectively acts for uniform gravitational fields.

Exam Focus

• Typical question patterns
  • A beam, sign, ladder, or seesaw in static equilibrium where you must use \sum \tau = 0 (often combined with \sum F = 0) to find a tension, normal force, or location of a support.
  • “Choose the best pivot” prompts: selecting an axis to eliminate unknown forces and simplify solving.
  • Conceptual questions comparing torques when changing force location, angle, or pivot.
• Common mistakes
  • Using r instead of r_\perp (or forgetting \sin(\theta)) and overcounting torque.
  • Assigning torque signs based on up/down instead of clockwise/counterclockwise tendency.
  • Forgetting that weight acts at the center of mass of the object (for a uniform beam, at L/2).

Newton's Second Law in Rotational Form

From “net force causes acceleration” to “net torque causes angular acceleration”

Newton’s Second Law for translation links net force to linear acceleration. For rotation, the analogous statement is that net torque about an axis causes angular acceleration about that axis.

To make that precise, you need one new idea: rotational inertia.

Moment of inertia: the “mass” of rotation

Moment of inertia measures how hard it is to change an object’s rotational motion. It plays the same role in rotation that mass plays in translation.

What determines moment of inertia?

• How much mass the object has.
• How far that mass is from the axis.

Mass farther from the axis contributes much more strongly. This is why a figure skater spins faster when pulling arms in: the distribution of mass moves closer to the axis, decreasing moment of inertia and allowing angular speed to increase if angular momentum is conserved (a later topic, but the intuition is useful).

For point masses, the definition is:

I = \sum m r^2

Where:

• I is moment of inertia (units: kg·m²).
• m is each mass.
• r is the distance from the axis to that mass.

For solid objects (like disks and rods), AP Physics 1 typically provides or expects you to use standard moments of inertia for common shapes rather than derive them from calculus. What you must understand is the dependence on axis and geometry: same object, different axis, different I.

Newton’s Second Law in rotational form

The rotational version of Newton’s Second Law is:

\sum \tau = I\alpha

Where:

• \sum \tau is the net external torque about the axis (N·m).
• I is the moment of inertia about that axis (kg·m²).
• \alpha is the angular acceleration (rad/s²).

This equation is the backbone of rotational dynamics in AP Physics 1. It tells you exactly how strongly an object responds (angular acceleration) to a given net torque, given its rotational inertia.

Why it matters: connecting forces to rotational motion

In many AP problems, you’re given forces (tension, friction, applied pushes) and asked how an object rotates. Torque is the bridge: you translate forces into torques, add them with signs, and use \sum \tau = I\alpha.

This also explains a deep physical point: torque alone doesn’t determine how fast something speeds up rotationally. Two objects with the same torque can have different angular accelerations if their moments of inertia differ.

Notation and concept connections (translation vs rotation)

Seeing the analogies helps you avoid memorizing in isolation.

A common misconception is to treat I like a constant independent of axis. Always tie I to a specific axis.

How to apply \sum \tau = I\alpha (a reliable process)

1. Choose the axis of rotation (often given: a fixed axle; sometimes you choose).
2. Draw a free-body diagram and identify forces that create torque about that axis.
3. Compute each torque using \tau = rF\sin(\theta) or \tau = Fr_\perp with a sign.
4. Sum torques to get \sum \tau.
5. Relate to rotation using \sum \tau = I\alpha and solve for \alpha or another quantity.

If the object is also translating, you may need Newton’s Second Law for translation at the same time. In AP Physics 1, this happens often with a string wrapped around a pulley or a block causing a disk to rotate.

Worked example 1: a solid disk spun by a tangential pull

A solid disk rotates about its center. A string wrapped around the rim is pulled with a constant tangential force F = 10\ \text{N}. The disk radius is R = 0.20\ \text{m}. The disk’s moment of inertia about its center is given as I = 0.080\ \text{kg·m}^2. Find the angular acceleration.

Step 1: Torque from the force.
The force is tangential, so \theta = 90^\circ and \sin(\theta) = 1.

\tau = rF\sin(\theta) = RF = (0.20)(10) = 2.0\ \text{N·m}

Step 2: Apply rotational Newton’s Second Law.

\sum \tau = I\alpha

2.0 = 0.080\alpha

\alpha = 25\ \text{rad/s}^2

Interpretation: The same pull would produce a smaller \alpha if the disk had a larger I (more mass or mass farther from center).

What could go wrong: Using \tau = RF is only valid because the pull is perpendicular to the radius. If the string pulled at an angle, you’d need the sine factor or r_\perp.

Worked example 2: comparing angular accelerations with different moments of inertia

Two objects experience the same net torque \tau_{net} about their centers. Object A has moment of inertia I_A and object B has I_B = 2I_A.

Using:

\alpha = \frac{\sum \tau}{I}

Then:

\alpha_B = \frac{\tau_{net}}{2I_A} = \frac{1}{2}\alpha_A

Meaning: doubling moment of inertia halves the angular acceleration for the same applied torque.

When you must combine translation and rotation (a key AP skill)

Many AP Physics 1 problems involve a force applied to a rotating object via a string, and that same string’s tension is also accelerating a mass linearly. The tension produces torque on the rotating object, while also appearing in \sum F = ma for the translating mass.

Even without going into every variant, the strategic idea is:

• Use \sum F = ma for the translating object.
• Use \sum \tau = I\alpha for the rotating object.
• Add a kinematic link if the string does not slip:

a = \alpha r

Where r is the pulley or disk radius.

A frequent error is to assume the tension is the same on both sides of a massive pulley. For a pulley with rotational inertia, different tensions can exist because the difference in tensions provides the net torque.

Direction, sign, and consistency

In rotational dynamics, students often lose points not because the concept is hard, but because sign conventions are handled inconsistently.

A practical approach:

• Choose counterclockwise as positive for torque and angular acceleration.
• Compute each torque’s sign by imagining the rotation it would cause.
• Your final \alpha sign tells you the direction of angular acceleration relative to your positive choice.

If your result is negative, it doesn’t mean “wrong” automatically; it means the angular acceleration is opposite your assumed positive direction.

Common misconceptions to watch for (within dynamics)

• “Any force on a rotating object causes torque.” A force whose line of action passes through the axis produces zero torque about that axis.
• “Bigger radius always means bigger torque.” Only if the force is applied so that r_\perp increases; if the force becomes more aligned with the radius, \sin(\theta) can shrink and reduce torque.
• “Moment of inertia depends only on mass.” It strongly depends on how mass is distributed relative to the axis.

Exam Focus

• Typical question patterns
  • Given forces on a rotating object (disk, rod, pulley), compute net torque and use \sum \tau = I\alpha to find \alpha.
  • Compare angular accelerations for different shapes or different axes using provided or known I values.
  • Coupled translation-rotation setups using \sum F = ma, \sum \tau = I\alpha, and a = \alpha r.
• Common mistakes
  • Forgetting that torque must be computed about a specified axis and using the wrong lever arm distance.
  • Treating I as the same for the object regardless of axis (it is axis-dependent).
  • Mixing sign conventions: computing some torques as positive “because they’re upward” rather than because of rotation direction.