Unit 5 · Topic 5.5 · about 30 minutes · 7 parts · 12 scored checks, 1 of them multiple choice, plus 2 from earlier lessons

AP Physics C: Mechanics 5.5: Rotational Equilibrium and Newton’s First Law in Rotational Form

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 5: 10 to 15% of the multiple-choice section

What you'll be able to do

Describe the conditions under which a system's angular velocity stays constant, using force diagrams and the sum of the torques exerted on it.

Key vocabulary

Term	What it means	The exam trap
Rigid system	A system whose shape and size do not change as it moves, so every part of it turns through the same angle together.	Treating an object as rigid is a model choice; it stops fitting when the object bends or stretches enough to matter.
Free-body diagram	A drawing of every force exerted on an object, each as an arrow from a single dot that stands for the object.	A dot hides where each force is exerted, so a free-body diagram alone cannot tell you any lever arm.
Force diagram	A drawing of a rigid system with each force exerted on it as an arrow starting at the point where that force is exerted, used to analyze torques.	Place each arrow at its true point of application; an arrow drawn at the wrong point gives the wrong lever arm.
Net torque	The sum of all the torques exerted on a system about one axis, $\sum τ_{i}$ , each with its sign for clockwise or counterclockwise.	Every torque in the sum must be about the same axis, and each keeps its sign; adding sizes alone overstates it.
Rotational equilibrium	A configuration of torques in which the net torque exerted on the system is zero, $\sum τ_{i} = 0$ .	Two large torques can cancel and one small torque can leave the sum unbalanced, so decide from the signed sum, never from the biggest torque.
Translational equilibrium	The condition in which the net force exerted on a system is zero, so its center of mass moves with constant velocity.	Constant velocity is not the same as at rest: a zero net force allows a moving center of mass, as long as its velocity does not change.
Angular velocity	The rate at which a system's angular position changes, $ω = \frac{d θ}{d t}$ , described as clockwise or counterclockwise about the axis.	State which sense is positive before you compare an angular velocity with a torque; a sign is meaningless without it.

A graph of net torque about the pivot in newton meters, counterclockwise positive, against the position x in meters of a sliding block, from negative 1.2 m to 1.2 m, for a plank pivoted at its center with a 30 N load fixed 0.6 m left of the pivot. A solid line for a 20 N block falls from 42 N m to negative 6 N m and crosses zero at 0.9 m. A dashed line for a 30 N block falls from 54 N m to negative 18 N m and crosses zero at 0.6 m. Each zero crossing is marked with a dot, and two labels at the lower left name each line and the position of its zero. Text at the upper right states the setup: plank pivoted at its center; 30 N fixed 0.6 m left of the pivot; block at x, right positive, and counterclockwise torque positive.

Part 1 Think first The potter's wheel

Commit to a guess first. Being wrong now makes the right idea stick better.

Predict first

A potter's wheel spins at a steady two turns per second. Its motor is running, and the bearing and the potter's wet hands rub against it. Which describes the torques on the wheel?

In short. A system is in rotational equilibrium when the net torque exerted on it is zero, $\sum τ_{i} = 0$ , and the rotational form of Newton's first law says its angular velocity stays constant only under that condition.

Draw a force diagram that shows where each force is exerted, then add the torques about the axis with their signs.
If the torques are not balanced, the angular velocity must be changing: speeding up when the net torque is in the sense of rotation, slowing down when it is opposite.
Rotational and translational equilibrium are separate conditions: a system can have zero net torque and a nonzero net force, or the other way around.

Part 2 Learn it Four ideas, with a check after each

5a. Draw the forces where they act

A free-body diagram draws every force on an object as an arrow from one dot. That is enough for the net force, but it hides where each force is exerted, and torque depends on exactly that.

A force diagram draws the rigid system itself, a beam or a board, and starts each force's arrow at the point where it is exerted. Then you can read off each lever arm: the perpendicular distance from the axis to the force's line of action.

Take a diving board bolted down at its back end and resting on a support 1.0 m in front of the bolt, with a diver standing at the tip. Its force diagram has four arrows: the bolt pulling down at the back end, the support pushing up at the support, the board's weight down at its center, and the diver pushing down at the tip.

About an axis through the support, the support's force has a lever arm of zero, so it exerts no torque about that axis. It still counts in the net force. A force whose line of action passes through the axis, wherever along that line it is exerted, also has zero lever arm.

Quick check · Sort it

A uniform beam is hinged to a wall at its left end and held level by a vertical cable tied to its right end. A can of paint sits on the middle of the beam. Tap a card, then say whether that force exerts a torque on the beam about the hinge.

Exerts a torque about the hinge

Exerts no torque about the hinge

5b. Rotational equilibrium: the torques add to zero

Rotational equilibrium is a configuration of torques in which the net torque exerted on the system is zero:

\sum τ_{i} = 0

The sum keeps signs. Pick which sense of rotation is positive, clockwise or counterclockwise about the axis as seen from a stated side, and give every torque its sign. A large torque one way and a large torque the other way can cancel exactly.

A light rod hangs from a string tied at one point, with a 6.0 N ornament hanging 0.30 m to the left of the string. Counterclockwise is positive, seen from the front. The ornament's torque about the string's point is $+ (6.0) (0.30) = + 1.8$ N·m. A 4.0 N ornament on the right gives $- (4.0) r$ , and the sum is zero when $r = 0.45$ m. The lighter ornament balances the heavier one by hanging farther out, because torque is force times lever arm, not force alone.

The graph at the top of this page applies the same sum to a plank: as a block slides along it, the net torque about the pivot falls in a straight line, and it is zero at exactly one position for each block.

When the system is also in translational equilibrium, the net torque is zero about every axis, so you are free to pick one. Put it where an unknown force acts, and that force drops out of the torque sum.

Quick check · Calculate

A seesaw 4.0 m long is pivoted at its center, which is also its center of mass. A child weighing 250 N sits 1.6 m to the left of the pivot. Counterclockwise is positive, seen from the front with left on your left. How far to the right of the pivot must a child weighing 320 N sit for the seesaw to stay level?

Your answer m

5c. Newton's first law, in rotational form

The rotational analog of Newton's first law: a system has a constant angular velocity only if the net torque exerted on it is zero. Constant includes zero, so a level seesaw at rest and a fan turning at a steady rate obey the same rule.

A ceiling fan whose motor exerts 0.30 N·m one way while the air drags with 0.30 N·m the other way has a net torque of zero, and its $ω = \frac{d θ}{d t}$ stays put. Steady spinning does not need a net torque. It needs the torques to cancel.

The corollary runs the other way: if the torques on a rigid system are not balanced, its angular velocity must be changing. Switch the fan's motor off and only the drag's 0.30 N·m remains, opposite to the rotation, so the fan slows. A net torque in the same sense as the rotation would speed it up.

So to decide whether $ω$ is changing, find the net torque, with signs. The size of $ω$ tells you nothing about it, and how fast $ω$ changes for a given net torque is Topic 5.6.

Quick check · Decide and justify

A merry-go-round turns counterclockwise as seen from above. A parent pushes along its rim, exerting a torque of 120 N·m counterclockwise, while friction at the axle exerts 80 N·m clockwise. Counterclockwise is positive, seen from above. Is the merry-go-round's angular velocity constant?

Which reason supports your answer?

5d. One equilibrium without the other

Translational equilibrium means the net force on the system is zero, so its center of mass moves with constant velocity. Rotational equilibrium means the net torque is zero, so its angular velocity is constant. Neither condition guarantees the other.

A book dropped flat, with air resistance ignored, has one force on it: its weight, exerted at its center of mass. About an axis through the center of mass that force has no lever arm, so the net torque is zero and the book does not start to spin. The net force is not zero, though, so the book speeds up as it falls. Rotational equilibrium without translational equilibrium.

A bicycle wheel on a fixed axle, spun up by a hand pulling on its tire, is the reverse. The axle holds its center in place, so the forces on it add to zero, but the pull exerts a net torque and the wheel spins faster.

A system can be in both, like a level seesaw, or in neither. Check the net force and the net torque separately.

Quick check · Match

Match each situation with the equilibrium it is in. Ignore air resistance.

Common mistakes

The mistake	Why it is wrong	Do this instead
Deciding that an object spinning at a steady rate must have a net torque in the sense it spins.	A net torque changes the angular velocity. If the torque driving a fan were larger than the drag, the fan would keep speeding up.	For a constant angular velocity, set the net torque to zero, and let the driving torque equal the opposing torques in size.
Comparing forces instead of torques to decide whether something balances.	Torque depends on the lever arm as well as the force. A 4.0 N weight 0.45 m out balances a 6.0 N weight 0.30 m out on the other side.	Multiply each force by its lever arm about the axis, give each torque its sign, and add.
Measuring a lever arm from the end of a beam instead of from the axis.	Torque is about a chosen axis, and the lever arm is the perpendicular distance from that axis to the force's line of action.	Mark the axis on your force diagram first, then measure every lever arm from it.

Part 3 Watch it done A painter on a plank

Watch each step, then do the last one yourself.

A uniform plank 4.0 m long and weighing 200 N rests level on two sawhorses, one under each end. A painter weighing 600 N stands 1.0 m from the left end. Take up as positive for forces, and counterclockwise as positive for torques, seen from the front with the left end on your left. What force does each sawhorse exert on the plank?

Step 1 · Draw the force diagram
Four forces act on the plank: $F_{L}$ up at the left end, $F_{R}$ up at the right end, the plank's weight, 200 N, down at its center, 2.0 m from the left end, and the painter's push, 600 N, down at 1.0 m from the left end.
Step 2 · Take torques about the left end
With the axis at the left end, $F_{L}$ has no lever arm and drops out. $\sum τ_{i} = F_{R} (4.0) - (200) (2.0) - (600) (1.0) = 0$ , so $F_{R} = \frac{1000}{4.0} = 250$ N, upward.
Why this step?
Why put the axis at the left sawhorse and not at the plank's center?
The plank is at rest, so it is in translational and rotational equilibrium, and the net torque is zero about any axis. An axis at the left end removes one unknown, $F_{L}$ , so the torque equation has only $F_{R}$ in it. About the center, both sawhorse forces would appear and you would need a second equation before solving.
Your step · Calculate

Finish it. What force does the left sawhorse exert on the plank? Up is positive.

Your answer N

Part 4 Try it New situations

Try it · Rank it

Four identical uniform rods are each pivoted at their centers, with two weights hung on each, held level and then released from rest. Rod A: 10 N at 0.40 m left, 8 N at 0.50 m right. Rod B: 6 N at 0.30 m left, 4 N at 0.30 m right. Rod C: 5 N at 0.20 m left, 2 N at 0.50 m right. Rod D: 3 N at 0.50 m left, 6 N at 0.40 m right. Distances are from the pivot. Rank the rods by the size of the net torque about the pivot just after release.

Largest first

Give each one a place, 1 for first. Give two the same place if they are equal.

Rod A
Rod B
Rod C
Rod D

Your ranking: none yet

Try it · Calculate

A flywheel turns counterclockwise as seen from above, and keeps turning counterclockwise for the whole time described; counterclockwise is positive. Its motor exerts a torque $τ_{m} (t) = 6.0 - 2.0 t$ , with $τ_{m}$ in N·m and $t$ in seconds, for $0 \leq t \leq 5.0$ s (a negative value means the motor brakes). Friction at the axle exerts a steady torque of 2.0 N·m clockwise. At what time is the flywheel's angular velocity greatest?

Your answer s

On paper. A uniform shelf board 1.2 m long and weighing 30 N is screwed to two brackets, one 0.2 m from each end, and a 50 N stack of books sits at the board's right end. Draw a force diagram of the board, with each force starting at the point where it is exerted, and label each force's lever arm about the left bracket. Then write the torque sum about the left bracket with counterclockwise positive, seen from the front, and set it to zero. You do not need to solve it. There is nothing to submit here. To compare: measured from the left bracket, the left bracket's force is at 0, the board's 30 N weight at 0.4 m, the right bracket's force at 0.8 m and the books' 50 N at 1.0 m, and the sum is $F_{R} (0.8) - (30) (0.4) - (50) (1.0) = 0$ .

Part 5 Explain it Diagram, equations, graph, prediction

Nothing you type leaves this page. Check your answer against the model and each scoring point.

Free response is 50% of the exam score. This one is shaped like Question 2, Translation Between Representations.

Free-response practice · Question 2 style

A uniform beam of length $L$ and weight $W_{b}$ is attached to a wall by a hinge at its left end and held level by a vertical rope tied to its right end. A box of weight $W_{k}$ sits on the beam a distance $x$ from the hinge, where $0 \leq x \leq L$ . Take up as positive for forces, and counterclockwise as positive for torques, seen from the front with the wall on your left.

(a) On paper, draw a force diagram of the beam. Draw each force as an arrow that starts at the point where it is exerted, and label it.

(b) Derive an expression for the tension $T$ in the rope in terms of $x$ , $L$ , $W_{b}$ and $W_{k}$ .

(d) The beam is 2.0 m long, $W_{b} = 80$ N and $W_{k} = 120$ N. On paper, on one set of axes, sketch $T$ and $F_{H}$ as functions of $x$ from 0 to 2.0 m. Label the value of each at $x = 0$ and at $x = 2.0$ m.

(e) Use your graph to predict where the box must sit for the rope and the hinge to exert forces of the same size. Justify your prediction using your force diagram from part (a).

Your answer stays on this page. It is never sent or saved.

Model answer

(a) Four vertical forces on the beam: $T$ up at the right end, $W_{b}$ down at the center ( $\frac{L}{2}$ from the hinge), the box's push $W_{k}$ down at $x$ , and $F_{H}$ up at the hinge. Nothing pulls sideways, so the hinge exerts no horizontal force. Compare: each arrow should start at its own point on the beam, not at one dot.

(b) About the hinge, $F_{H}$ has no lever arm: $\sum τ_{i} = T L - W_{b} \frac{L}{2} - W_{k} x = 0$ , so $T = \frac{W_{b}}{2} + \frac{W_{k} x}{L}$ .

(c) The beam is also in translational equilibrium: $F_{H} + T - W_{b} - W_{k} = 0$ , so $F_{H} = \frac{W_{b}}{2} + W_{k} (1 - \frac{x}{L})$ .

(d) Both graphs are straight lines. $T$ rises from 40 N at $x = 0$ to 160 N at $x = 2.0$ m, and $F_{H}$ falls from 160 N at $x = 0$ to 40 N at $x = 2.0$ m. Compare your sketch with that: two straight lines, one rising and one falling, with those four labeled values and labeled axes.

(e) The lines cross at $x = 1.0$ m, the middle of the beam, where both forces are 100 N. With the box at the center, both downward forces act at $\frac{L}{2}$ , the same distance from the hinge as from the rope, so the diagram is symmetric and the two supports share the 200 N load equally. Setting $T = F_{H}$ in (b) and (c) gives $\frac{x}{L} = 1 - \frac{x}{L}$ , so $x = \frac{L}{2}$ , which agrees.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Draws $T$ up at the right end, $W_{b}$ down at the center, the box's force $W_{k}$ down at $x$ and $F_{H}$ up at the hinge, each starting at its own point.
Writes a signed torque sum about a stated axis, sets it to zero, and derives $T = \frac{W_{b}}{2} + \frac{W_{k} x}{L}$ .
Derives $F_{H} = \frac{W_{b}}{2} + W_{k} (1 - \frac{x}{L})$ from the force sum or a second torque sum.
Sketches $T$ as a straight line rising from 40 N to 160 N and $F_{H}$ as a straight line falling from 160 N to 40 N, with labeled axes.
Predicts $x = 1.0$ m, the middle of the beam, and justifies it with the crossing of the lines and the symmetry of the force diagram.

Part 6 Check yourself A wheel on a motor

The case questions use the lab group's table. Answer them from it.

A lab group mounts a bicycle wheel on a fixed horizontal axle. A small motor with a torque sensor drives the wheel, and a second sensor records its angular velocity. Counterclockwise is positive, seen from the motor's side, and the wheel turns counterclockwise throughout. The group changes the motor's torque twice:

Torques on the wheel and its angular velocity, by time interval
Interval	Motor torque (N·m)	Axle friction torque (N·m)	$ω$ at start (rad/s)	$ω$ at end (rad/s)
0 to 2 s	+1.2	−0.4	2.0	10.0
2 s to 5 s	+0.4	−0.4	10.0	10.0
5 s to 7 s	0	−0.4	10.0	6.0

Case question 1 · Multiple choice

During which interval is the wheel in rotational equilibrium?

Case question 2 · Calculate

After 7 s the group wants the wheel to keep turning counterclockwise at a steady 6.0 rad/s. Assume the axle friction torque stays at its value in the table. What motor torque must they set? Counterclockwise is positive.

Your answer N·m

Case question 3 · Decide and justify

During the interval from 0 to 2 s, is the wheel in translational equilibrium?

Which reason supports your answer?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

A uniform board 3.0 m long and weighing 90 N rests on a pivot 1.0 m from its left end. A student finds the downward push at the left end that holds the board level, with counterclockwise positive, seen from the front with left on your left. Tap the first line with a mistake.

Check yourself · Put it in order

Put the steps for finding an unknown force on a beam in rotational equilibrium in order.

Part 7 Keep it Bring back earlier ideas

This lesson stands on two earlier ideas: from Topic 1.2, a velocity stays constant only when its rate of change is zero, and from Topic 5.4, rotational inertia, which Topic 5.6 joins with the net torque from this lesson.

From 1.2 Displacement, Velocity, and Acceleration · Decide and justify

A cart moves along a track with its position given by $x (t) = 5 t^{2} - t^{3}$ , with $x$ in meters and $t$ in seconds. Take the direction of increasing $x$ as positive. At $t = 2$ s, is the cart speeding up or slowing down?

Which reason supports your answer?

From 5.4 Rotational Inertia · Calculate

A light rod 0.60 m long lies horizontally and turns about a vertical axis through its left end. Three small objects are fixed to it: 0.50 kg at 0.20 m from the left end, 0.50 kg at 0.40 m from the left end, and 1.0 kg at the right end. What is the rotational inertia of the system about the axis?

Your answer kg·m²

In three lines

A force diagram shows where each force is exerted, so you can read each lever arm about the axis; a force through the axis exerts no torque about it.
Rotational equilibrium is $\sum τ_{i} = 0$ , and a system's angular velocity stays constant only then; unbalanced torques mean it is changing.
Rotational and translational equilibrium are separate: check the net torque and the net force each on its own.

Exam strategy for this topic

Section I, Multiple Choice: 42 questions in 85 minutes.
Section II, Free Response: 4 questions in 95 minutes.
This lesson's Explain it is shaped like free-response question 2, Translation Between Representations.
A four-function, scientific, or graphing calculator is allowed on both sections of the exam.

On a free-response question about a beam or a board, draw the force diagram with each arrow at its point of application before writing any equation, and state your axis and your positive sense. Put the axis where an unknown force acts so it drops out, then use the force sum, or a second torque sum, for the other unknown.

Questions students ask

Can something be spinning and still be in equilibrium?

Yes, in rotational equilibrium. The condition is a zero net torque, which keeps the angular velocity constant, and constant includes a steady nonzero rate, not only zero.

Which axis should I take torques about?

For a system in both translational and rotational equilibrium, any axis gives a zero net torque, so pick the one that makes the algebra easiest: usually where an unknown force acts, since that force then has no lever arm.

Where does a beam's own weight act?

At its center of mass. For a uniform beam that is the middle, so its lever arm is the distance from the axis to the middle of the beam.

Does a zero net torque mean the net force is zero too?

No. The two conditions are separate. A dropped book has a zero net torque about its center of mass but a nonzero net force, and a wheel spun up on a fixed axle has the reverse.

What happens when the torques are not balanced?

The angular velocity must be changing: it grows when the net torque is in the sense of rotation and shrinks when it is opposite. How fast it changes depends on the rotational inertia, which is Topic 5.6.

Beyond the AP Exam

College Board sets this limit on what the AP Exam asks about this topic. Anything past it is not on the exam.

AP Physics C: Mechanics does not expect students to simultaneously analyze rotation in multiple planes.

Course alignment, for teachers

AP Physics C: Mechanics topic 5.5, Unit 5: Torque and Rotational Dynamics.

Skill 1.A: Create diagrams, tables, charts, or schematics to represent physical situations.
Skill 2.A: Derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway.
Skill 2.C: Compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario.
Skill 3.B: Apply an appropriate law, definition, theoretical relationship, or model to make a claim.

This lesson maps all 5 essential knowledge statements of the topic to the part that teaches each one: 5 are checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first The potter's wheel

Part 2 Learn it Four ideas, with a check after each

5a. Draw the forces where they act

5b. Rotational equilibrium: the torques add to zero

5c. Newton's first law, in rotational form

5d. One equilibrium without the other

Part 3 Watch it done A painter on a plank

Part 4 Try it New situations

Part 5 Explain it Diagram, equations, graph, prediction

Part 6 Check yourself A wheel on a motor

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic