AP Physics 1 · Topic 5.5

Topic 5.5: Rotational Equilibrium and Newton's First Law in Rotational Form

Unit 5: Torque and Rotational Dynamics10-15% of the multiple-choice section

A system is in rotational equilibrium when the net torque on it is zero, and its angular velocity then stays constant. That is Newton's first law in rotational form. Rotational equilibrium is independent of translational equilibrium: net torque can be zero while the center of mass accelerates.

AP Physics: Unit 5 (topics 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form). AP Physics 1 Unit 5, Topic 5.5, covering learning objective 5.5.A, describe the conditions under which a system's angular velocity remains constant, and essential knowledge 5.5.A.1 through 5.5.A.2. The CED lists four suggested skills for this topic: 1.C, 2.A, 2.B, and 3.B. One boundary statement applies: AP Physics 1 does not expect students to simultaneously analyze rotation in multiple planes. Unit 5 is weighted at 10 to 15 percent of the multiple-choice section and about 15 to 20 class periods.

What the CED asks of Topic 5.5

Topic 5.5 carries one learning objective. Objective 5.5.A asks you to describe the conditions under which a system's angular velocity remains constant. It is written as the deliberate twin of objective 2.4.A, which asks the same question about linear velocity, and of objective 5.6.A, which asks when angular velocity changes instead.

Four essential knowledge statements hang off it.

5.5.A.1. A system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa.
5.5.A.1.i. Free-body and force diagrams describe the nature of the forces and torques exerted on an object or rigid system.
5.5.A.1.ii. Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero. The relevant equation is $\sum_i \tau_i = 0$ .
5.5.A.1.iii. The rotational analog of Newton's first law is that a system will have a constant angular velocity only if the net torque exerted on the system is zero.
5.5.A.2. A rotational corollary to Newton's second law states that if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing.

One boundary statement limits the topic: AP Physics 1 does not expect students to simultaneously analyze rotation in multiple planes. Every question is a single plane of rotation, so the torques you add carry one sign or the other and never point off at an angle to each other.

The CED lists four suggested skills for Topic 5.5: 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. In the CED's own grouping, 1.C sits under Practice 1, Creating Representations; 2.A and 2.B under Practice 2, Mathematical Routines; and 3.B under Practice 3, Scientific Questioning and Argumentation.

Notice the pairing of 2.A with 2.B: this topic wants both the symbolic setup and the number at the end.

Rotational equilibrium means constant angular velocity, not zero

The parenthesis in EK 5.5.A.1 does the important work. Rotational equilibrium is defined as constant angular velocity, and zero is only one of the constant values available.

\sum_i \tau_i = 0 \quad \Longleftrightarrow \quad \omega \text{ is constant}

A grindstone held at a steady 15 rad/s by a motor that exactly cancels bearing friction is in rotational equilibrium. So is a bolt that has not moved since it was tightened. Both satisfy the same equation and both are solved with the same algebra, exactly as static and dynamic translational equilibrium are in Topic 2.4.

Read 5.5.A.1.ii and 5.5.A.1.iii together, because they say different things. Statement ii defines rotational equilibrium as a configuration of torques whose sum is zero. Statement iii adds the implication that makes it useful: a system will have a constant angular velocity only if the net torque exerted on the system is zero, so zero net torque is a necessary condition. EK 5.5.A.2 closes the loop from the other side, since unbalanced torques force the angular velocity to change.

That gives you a two-way test you can run on any exam question.

Told that the angular velocity is constant, you may write $\sum \tau = 0$ and solve for an unknown force or distance.
Told the torques do not balance, you may state without further work that the angular velocity is changing, and hand the problem to Topic 5.6 to find how fast.

The equation $\sum \tau_i = 0$ is not printed on the AP Physics 1 equation sheet. What the sheet gives you is the rotational second law solved for angular acceleration, $\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}$ , and setting $\alpha_{\text{sys}} = 0$ drives the numerator to zero. Produce the equilibrium condition from the printed line in one step rather than trying to recall it; the full AP Physics 1 formula sheet is shorter than most students assume.

Rotational and translational equilibrium are independent conditions

EK 5.5.A.1 says a system may be in rotational equilibrium without being in translational equilibrium, and the reverse. Hold a concrete case of each in your head.

Rotational equilibrium without translational equilibrium. Drop a textbook flat. In a uniform gravitational field the gravitational force may be treated as acting at the center of mass, so it exerts no torque about an axis through that point. The net torque is zero, the book's angular velocity stays constant at zero, and it lands without tumbling, while its center of mass accelerates downward at $9.8\ \mathrm{m/s^2}$ . Any object in free fall with air resistance negligible is in this state, which is one reason Topic 2.1 works so hard on locating the center of mass.

Translational equilibrium without rotational equilibrium. Push the left rim of a steering wheel up while pulling the right rim down with equal force. The forces are equal and opposite, so the net force is zero and the wheel's center goes nowhere. Their torques about the center both turn the wheel the same way, so they add rather than cancel and the wheel angularly accelerates. That is why a net force of zero is never evidence that the torques balance.

All four combinations occur.

Net force	Net torque	Example	What stays constant
Zero	Zero	A bolt in a wall, a bicycle wheel coasting on a fixed axle at steady speed	Velocity and angular velocity both
Nonzero	Zero	A dropped book, a block sliding down a frictionless ramp without spinning	Angular velocity only
Zero	Nonzero	Two equal opposite forces on opposite rims of a steering wheel	Velocity only
Nonzero	Nonzero	A ball rolling down a ramp, a falling wrench thrown with spin	Neither

Rows two and three each defeat a shortcut. Row two defeats "it is accelerating, so the torques must be unbalanced." Row three defeats "the forces cancel, so nothing is happening."

One subtlety in row two. Rotational equilibrium is always claimed about a specified axis, and when the center of mass is itself accelerating, the useful axis is the one through it. The dropped book has zero net torque about that axis, which is why it never begins to spin. About some arbitrary point fixed in the room it does not, and that contradicts nothing.

Declare a sign convention before you write the first torque

Torque adds as a signed quantity in AP Physics 1, and the sign is yours to assign. EK 5.1.A.1.ii sets the rule in the CED's own words: one direction of angular displacement about an axis of rotation, either clockwise or counterclockwise, is typically indicated as mathematically positive, with the other direction becoming mathematically negative.

The convention used throughout this page, and a safe default for the exam, is counterclockwise positive.

That means, for every torque you write down:

Look at the force and the point where it is applied.
Ask which way it alone would turn the system about the chosen axis.
Counterclockwise gets a plus sign, clockwise gets a minus sign.
Add the signed values. Rotational equilibrium is the statement that the sum is zero, which is the same as saying the counterclockwise torques and the clockwise torques have equal magnitudes.

Two rules keep this honest. The convention must be stated before the first torque is written, and it must survive to the last line. A convention that silently flips halfway through is the most common way to lose an otherwise correct equilibrium problem.

The Topic 5.3 boundary statement explains why a plus or minus sign is enough here. While AP Physics 1 expects students to mathematically manipulate the magnitude of torque using vector conventions, the direction of torque is beyond the scope of the course. You will never be asked for a torque vector pointing along the rotation axis. One plane, two signs.

Magnitudes come from Topic 5.3 and the printed equation

\tau = r_{\perp} F = r F \sin\theta

where $\theta$ is the angle between the force vector and the position vector from the axis of rotation to the point of application of the force. A force pointing straight at or straight away from the axis has $\theta = 0^\circ$ or $180^\circ$ , so it contributes nothing to either side of the balance. If you need the arithmetic drilled, the torque calculator and the guide to calculating torque cover the procedure step by step.

The pivot is yours to choose, so choose it to delete an unknown

Here is the technique that pays for the whole topic. When a system is in rotational equilibrium, the net torque is zero about every axis, not just about the physical hinge or pivot. You are free to compute torques about any point you like, and the equation still reads zero.

That freedom is worth points, because a force whose line of action passes through your chosen axis has $r_{\perp} = 0$ and contributes no torque. Putting the axis on a force you do not know and do not want removes it from the equation entirely.

The routine:

Identify every force and where each is applied. A force diagram, per EK 5.3.B.1, differs from a free-body diagram in exactly this respect: it shows where each force acts relative to the axis of rotation, not just its direction.
Find the force you know least about, usually the contact force from a hinge, pin, axle, or support, which has both an unknown magnitude and an unknown direction.
Put the axis right there.
Write $\sum \tau = 0$ about that axis. A two-unknown problem collapses to one equation in one unknown.
If you still need that contact force, get it from $\sum F_x = 0$ and $\sum F_y = 0$ , or repeat step 4 with the axis on a different unknown.

Step 5 is the part students skip. A hinge force cannot be recovered from torques about the hinge, since it was deliberately eliminated, so it has to come from the force equations or from a second pivot choice. Worked example 1 does it both ways, which is also a free way to check your work.

A caution: the free choice holds because the system is in rotational equilibrium. If the net torque is not zero, the value of $\sum \tau$ depends on which axis you picked and you no longer get to move it for convenience. In that case, covered in Topic 5.6, use the physical axis the system actually rotates about, or the axis through the center of mass if it is also translating.

Static equilibrium needs both conditions at once

Most Topic 5.5 questions are static equilibrium questions: a beam, a sign, a ladder, a bridge, a plank. Static equilibrium is the conjunction of two separate conditions, and each supplies its own equations.

\sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum \tau = 0

Three equations, so three unknowns are solvable. That ceiling tells you when you have enough relationships and can stop hunting for another one.

Neither condition implies the other, which is why both get written. The table above is the proof: the steering wheel satisfies the first two equations and fails the third, and the dropped book satisfies the third and fails the second.

An ordering that saves time under exam conditions:

Draw the diagram first, with each force at the place it actually acts. The free-body diagram builder is a fast way to check that a set of arrows really sums to zero before you commit to algebra.
Write the torque equation before the force equations, with the axis chosen to delete the worst unknown. It usually has the fewest unknowns in it.
Use the force equations to mop up what the torque equation left behind.

Two habits prevent most of the lost marks. Put the weight of a uniform beam at its geometric center and give it the full lever arm from your chosen axis. And check that every force in your diagram appears in the equations with the right sign, since a force that vanished between the picture and the algebra is the most common single error here.

Degrees for the geometry, radians for the motion

Two kinds of angle appear in this unit and they use different units, so decide which you are holding before you touch the calculator.

Quantity	Unit
$\theta$ inside $\tau = r F \sin\theta$ , and the angle of any cable, rope, or incline	Degrees
Angular displacement $\Delta\theta$	Radians
Angular velocity $\omega$	Radians per second
Angular acceleration $\alpha$	Radians per second squared
Torque $\tau$	Newton meters, never joules

The geometric angles are in degrees. The AP Physics 1 Table of Information prints trigonometric values for common angles listed as $0^\circ$ , $30^\circ$ , $37^\circ$ , $45^\circ$ , $53^\circ$ , $60^\circ$ , and $90^\circ$ , so degree mode is the expected setting when you evaluate a sine.

The motion quantities are in radians. EK 5.1.A.1 states it directly: angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis. The radian is required, not preferred, by the equations converting between linear and rotational quantities in Topic 5.2. Feed degrees into $v = r\omega$ and the speed is too small by a factor of $180/\pi$ .

Torque takes neither. Its unit is the newton meter, and the radian never appears in it, because the radian is a ratio of two lengths and carries no dimension. A newton meter is not a joule even though the two are dimensionally identical.

What the graphs look like, and how the topic is tested

Suggested skill 1.C is sketching qualitative graphs, so be ready to draw rotational equilibrium three ways.

Graph	Shape when the net torque is zero
Angular position $\theta$ versus time	A straight line, sloping if the system is turning, horizontal if it is at rest
Angular velocity $\omega$ versus time	A horizontal line at the constant angular velocity, at any height including zero
Angular acceleration $\alpha$ versus time	A horizontal line lying on the time axis

The angular velocity graph is the test to run. Flat at any height means the net torque is zero. Any slope at all means it is not, because the slope of an angular velocity versus time graph is the angular acceleration, exactly as in Topic 5.1.

How the topic shows up in questions:

Multiple choice. Which of these systems has a constant angular velocity? Rank the tensions in three cables holding differently loaded beams. Predict what happens to a support force when a load slides toward it, which is answered from the lever arms alone.
Free response. A beam, a plank, or a sign with one unknown support force, solved with a chosen pivot. Skill 3.B asks you to justify a claim by naming the condition, so write the sentence and not just the equation: the net torque about the hinge is zero because the beam's angular velocity is constant.
Experimental design. Balance a meter stick on a knife edge, hang known masses, and test whether force times lever arm really is equal on both sides.

From here, Topic 5.3 supplies the torque magnitudes, Topic 5.4 supplies the rotational inertia you need the moment the balance breaks, and Topic 5.6 picks up where the sum does not vanish. Unit 5 is weighted at 10 to 15 percent of the AP Physics 1 multiple-choice section and about 15 to 20 class periods; the Unit 5 overview puts the six topics in order.

A hinged beam and a support cable, solved from two different pivots

A uniform beam of mass 12.0 kg and length 4.00 m is hinged to a wall at its left end and sticks straight out horizontally. A cable runs from the far right end of the beam back up to the wall, making an angle of $30^\circ$ with the beam. A 30.0 kg sign hangs from the beam 3.00 m from the hinge. The beam's angular velocity is constant at zero. Find the tension in the cable and the force the hinge exerts on the beam. Use $g = 9.8\ \mathrm{m/s^2}$ .

Set the conventions first. Take $x$ pointing right, away from the wall, $y$ pointing up, and counterclockwise as the positive sense of rotation. These hold to the last line.
List the forces and where each acts. The beam is uniform, so its weight $W_b = (12.0)(9.8) = 117.6\ \mathrm{N}$ acts at its center, 2.00 m from the hinge. The sign's weight $W_s = (30.0)(9.8) = 294\ \mathrm{N}$ acts 3.00 m out. The tension $T$ acts 4.00 m out, pulling up and back toward the wall at $30^\circ$ above the beam. The hinge force $\vec{H}$ has unknown magnitude and direction, which is two unknowns.
Choose the axis to delete the worst unknown: put it at the hinge, so $\vec{H}$ has $r_{\perp} = 0$ and drops out of the torque equation entirely. The tension's perpendicular component is $T \sin 30^\circ$ acting 4.00 m out, turning the beam counterclockwise: $+ (4.00)(T)(\sin 30^\circ) = +2.00\,T$ . Both weights point down at positive $x$ , so both turn it clockwise: $-(2.00)(117.6) = -235.2\ \mathrm{N \cdot m}$ and $-(3.00)(294) = -882\ \mathrm{N \cdot m}$ .
Apply $\sum \tau = 0$ : $2.00\,T - 235.2 - 882 = 0$ , so $T = \frac{1117.2}{2.00} = 558.6\ \mathrm{N}$ , which is $559\ \mathrm{N}$ to three significant figures.
Now get the hinge force from the force equations. The tension's components are $T_x = -T\cos 30^\circ = -483.8\ \mathrm{N}$ and $T_y = +T\sin 30^\circ = +279.3\ \mathrm{N}$ . Horizontal: $\sum F_x = H_x - 483.8 = 0$ , so $H_x = +484\ \mathrm{N}$ , pointing away from the wall. The cable pulls the beam toward the wall, so the hinge pushes it back out.
Vertical: $\sum F_y = H_y + 279.3 - 117.6 - 294 = 0$ , so $H_y = +132.3\ \mathrm{N}$ , upward. Then $|\vec{H}| = \sqrt{(483.8)^2 + (132.3)^2} = 502\ \mathrm{N}$ , at $\tan^{-1}(132.3 / 483.8) = 15.3^\circ$ above the horizontal.
Check $H_y$ independently by moving the pivot to the far right end. The tension acts there, so now it is the tension that drops out. The hinge force acts 4.00 m to the left, the beam's weight 2.00 m to the left, and the sign's weight 1.00 m to the left: $-(4.00)H_y + (2.00)(117.6) + (1.00)(294) = 0$ , giving $H_y = \frac{529.2}{4.00} = 132.3\ \mathrm{N}$ . The two routes agree.

The cable tension is 559 N, and the hinge pushes on the beam with 502 N directed 15.3 degrees above the horizontal, away from the wall. Putting the axis at the hinge removed the hinge force and left one unknown; putting it at the far end removed the tension and confirmed the vertical hinge component. The net torque is zero about both points, and about every other point, because the beam is in rotational equilibrium.

Rotational equilibrium without translational equilibrium, and the reverse

A uniform rod of mass 2.00 kg and length 1.20 m lies on frictionless horizontal ice, seen from above, so gravity plays no part in the horizontal analysis. Its rotational inertia about a vertical axis through its center is $I = 0.240\ \mathrm{kg \cdot m^2}$ , a value the exam would provide. Two cases are applied to the rod, each starting from rest. Case A: two 5.00 N horizontal forces perpendicular to the rod, one at each end, pointing in opposite directions. Case B: a single 5.00 N horizontal force perpendicular to the rod, applied at its center. Decide which equilibrium each case satisfies.

Keep counterclockwise positive, and use the axis through the rod's center of mass, the correct axis when the rod is free to translate as well as rotate.
Case A, forces. The two forces are equal and opposite, so $\sum \vec{F} = 0$ , the center of mass acceleration is zero, and the rod is in translational equilibrium: its center stays where it started.
Case A, torques. Each force acts 0.600 m from the center, perpendicular to the rod, so each has lever arm 0.600 m and magnitude $\tau = (5.00)(0.600) = 3.00\ \mathrm{N \cdot m}$ . They point opposite ways at opposite ends, which means both turn the rod the same way, so they add: $\sum \tau = +3.00 + 3.00 = +6.00\ \mathrm{N \cdot m}$ . Rotational equilibrium fails, and $\alpha = \frac{\tau_{\text{net}}}{I} = \frac{6.00}{0.240} = 25.0\ \mathrm{rad/s^2}$ . The rod spins up about a center that never moves.
Case B. The single force is applied at the center of mass, so its line of action passes through the axis, $r_{\perp} = 0$ , and $\sum \tau = 0$ . The rod is in rotational equilibrium: it started from rest, so its angular velocity stays at zero and it never begins to spin. But nothing balances the 5.00 N, so $\sum F = 5.00\ \mathrm{N}$ and $a_{\text{cm}} = \frac{5.00}{2.00} = 2.50\ \mathrm{m/s^2}$ . Translational equilibrium fails while rotational equilibrium holds.
Note that the same 5.00 N force did completely different things depending on where it was applied. That is what a force diagram records and a free-body diagram does not, which is why EK 5.3.B.1 asks for the location of each force relative to the axis.

Case A is in translational equilibrium but not rotational equilibrium: the center of mass stays put while the rod angularly accelerates at 25.0 rad/s squared. Case B is in rotational equilibrium but not translational equilibrium: the rod never starts spinning while its center of mass accelerates at 2.50 m/s squared. That is EK 5.5.A.1 made concrete: neither condition can be inferred from the other.

A grindstone in rotational equilibrium at nonzero angular velocity

A grindstone with rotational inertia $0.80\ \mathrm{kg \cdot m^2}$ turns counterclockwise at a constant 15 rad/s while its motor applies a torque of $2.4\ \mathrm{N \cdot m}$ about the axle. Find the frictional torque from the bearings. Then the motor is switched off with the friction unchanged. Find the angular acceleration and how long the grindstone takes to stop.

Take counterclockwise positive, matching the direction of rotation, so the initial angular velocity is $\omega_0 = +15\ \mathrm{rad/s}$ and the motor torque is $+2.4\ \mathrm{N \cdot m}$ .
The angular velocity is constant, so by EK 5.5.A.1.iii the net torque must be zero. This is rotational equilibrium at a nonzero angular velocity: nothing in the analysis needs the stone to be at rest. Write it out: $\sum \tau = +2.4 + \tau_f = 0$ , so $\tau_f = -2.4\ \mathrm{N \cdot m}$ , a torque of magnitude $2.4\ \mathrm{N \cdot m}$ acting clockwise against the rotation.
With the motor off, only friction remains, so $\sum \tau = -2.4\ \mathrm{N \cdot m}$ . The torques are unbalanced, and EK 5.5.A.2 says the angular velocity must be changing. The printed equation says how fast: $\alpha = \frac{\tau_{\text{net}}}{I} = \frac{-2.4}{0.80} = -3.0\ \mathrm{rad/s^2}$ , negative because it opposes the counterclockwise motion, so the stone slows.
The torque is constant, so the angular acceleration is constant and the rotational kinematics equation applies: $\omega = \omega_0 + \alpha t$ gives $0 = 15 + (-3.0)t$ , so $t = 5.0\ \mathrm{s}$ .
As a bonus, the angle turned while stopping is $\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = (15)(5.0) + \frac{1}{2}(-3.0)(5.0)^2 = 75 - 37.5 = 37.5\ \mathrm{rad}$ , which is $37.5 / (2\pi) = 5.97$ revolutions. Radians here, degrees nowhere.

The bearings exert a frictional torque of 2.4 newton meters clockwise. With the motor off the angular acceleration is 3.0 rad/s squared opposing the motion, and the stone stops after 5.0 s, having turned through 37.5 rad or about 5.97 revolutions. While the motor was on, the stone was in rotational equilibrium at 15 rad/s, which is the case that the phrase "in equilibrium" most often hides.

Frequently asked questions

What does rotational equilibrium mean in AP Physics 1?

Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero. The AP Physics 1 CED defines it as constant angular velocity, so it does not mean the system is not spinning. A flywheel turning at a steady 15 rad/s with the motor torque exactly cancelling bearing friction is in rotational equilibrium, and so is a bolt that has not moved in years.

Can an object be in rotational equilibrium while it is accelerating?

Yes, and the CED says so explicitly: a system may exhibit rotational equilibrium without being in translational equilibrium, and vice versa. A textbook dropped flat is the standard case. The gravitational force acts effectively at its center of mass and so exerts no torque about that axis, meaning the net torque is zero and the book does not start to tumble, while its center of mass accelerates downward at 9.8 meters per second squared. Zero net torque and zero net force are independent conditions.

What is the difference between rotational equilibrium and static equilibrium?

Rotational equilibrium is one condition, that the net torque is zero, which keeps the angular velocity constant. Static equilibrium is stronger: it requires the net force to be zero as well, so both the velocity and the angular velocity stay constant. Static equilibrium gives you three equations in a plane, the sum of the horizontal forces, the sum of the vertical forces, and the sum of the torques, each equal to zero. Rotational equilibrium gives you only the third.

Where should I put the pivot in a torque problem?

Put it on the line of action of the force you know least about, usually a hinge, an axle, or a support. A force whose line passes through the chosen axis has a lever arm of zero and contributes no torque, so it disappears from the equation and a two-unknown problem becomes one equation in one unknown. This is legitimate because when a system is in rotational equilibrium the net torque is zero about every axis, not only about the physical pivot, so you may compute torques about any point you like.

Is clockwise positive or negative for torque?

Either, as long as you declare it first and keep it. The AP Physics 1 CED states that one direction of angular displacement about an axis, clockwise or counterclockwise, is typically indicated as mathematically positive, with the other becoming mathematically negative. Counterclockwise positive is the usual default. What loses marks is not the choice but changing it partway through a solution.

Is the equation for the sum of the torques equal to zero on the AP Physics 1 equation sheet?

No. The AP Physics 1 Table of Information prints the rotational second law solved for angular acceleration, alpha equals the sum of the torques divided by the rotational inertia of the system, along with torque equals r perpendicular times F. The equilibrium condition is derived from that in one step by setting the angular acceleration to zero, which forces the numerator to zero. Expect to write it rather than look it up.