Unit 2 · Topic 2.10 · about 30 minutes · 7 parts · 13 scored checks, 2 of them multiple choice, plus 2 from earlier lessons

AP Physics C: Mechanics 2.10: Circular Motion

Q: Does a heavier satellite take longer to orbit?

No. In G M mR2 = mv2R the satellite's mass m appears on both sides and cancels, so the period at a given radius depends only on the mass of the central body.

Q: What changes when the speed on the circle is not constant?

The acceleration gains a tangential component, dvdt, along the path. The centripetal part, v2r, still uses the speed at that instant, and the net acceleration is the vector sum of the two.

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

Unit 2: 20 to 25% of the multiple-choice section

What you'll be able to do

Describe the motion of an object moving in a circle: its acceleration toward the center and along the path, the forces that supply the acceleration toward the center, and its period and frequency.
Describe a circular orbit with Kepler's third law, relating its period and radius to the mass of the central body.

Key vocabulary

Term	What it means	The exam trap
Centripetal acceleration	The component of an object's acceleration directed toward the center of its circular path, with magnitude $a_{c} = \frac{v^{2}}{r}$ .	It is there even when the speed never changes; a steady speedometer reading on a curve does not mean zero acceleration.
Tangential acceleration	The rate at which an object's speed changes, $a_{t} = \frac{d v}{d t}$ , directed tangent to its circular path.	It changes only how fast the object goes, so an object moving at a constant speed on a circle has none, although it is still accelerating.
Uniform circular motion	Motion in a circular path at a constant speed.	Uniform describes the speed, not the velocity; the velocity's direction turns the whole time.
Period	The time $T$ to complete one full circular path, one full rotation or one full cycle of an oscillation.	On a graph of a repeating motion, measure a period from one point to the next point where the motion repeats, not from the start of the record to its end.
Frequency	The rate $f$ at which an object completes revolutions, in hertz (one per second), with $T = \frac{1}{f}$ .	A rate given per minute has to be changed to per second before it goes into $T = \frac{1}{f}$ with SI units.
Kepler's third law	For a satellite in a circular orbit of radius $R$ around a central body of mass $M$ , $T^{2} = \frac{4 π^{2}}{G M} R^{3}$ .	$R$ is measured from the center of the central body, not from its surface, so add the body's radius to an altitude before you use it.

A cross section of a road banked at an angle theta, where sin theta is 0.60 and cos theta is 0.80, about 37 degrees. The road rises from its inner edge on the left to its outer edge on the right. A car sits on the slope, moving into the page. A dashed vertical line on the left marks the axis through the center of the curve, and a dashed horizontal line from the car to that axis is labeled r = 50 m. No forces are drawn.

Part 1 Think first Pressed against the door

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

Predict first

You ride in the back seat of a car that turns sharply left at a steady speed. You slide across the seat until your right shoulder presses against the door, and you stay there through the turn. While the car turns, which way does the door push on you?

In short. An object moving in a circle has an acceleration component toward the center, the centripetal acceleration $a_{c} = \frac{v^{2}}{r}$ , and a component along the path, $\frac{d v}{d t}$ , whenever its speed changes.

The net acceleration is the vector sum of the centripetal and tangential accelerations, which are perpendicular to each other.
Real forces supply the net force toward the center: gravity, tension, a normal force or friction, acting alone, together or through their components.
Uniform circular motion repeats with period $T = \frac{2 π r}{v}$ and frequency $f = \frac{1}{T}$ , and a circular orbit obeys $T^{2} = \frac{4 π^{2}}{G M} R^{3}$ .

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

2a. Acceleration toward the center

Calculus shows which way an object on a circle accelerates. Put the center at the origin and let an object move counterclockwise at a constant speed $v$ around a circle of radius $R$ :

\vec{r} (t) = R \cos (\frac{v t}{R}) \hat{i} + R \sin (\frac{v t}{R}) \hat{j}

Differentiate once and $\vec{v} (t) = - v \sin (\frac{v t}{R}) \hat{i} + v \cos (\frac{v t}{R}) \hat{j}$ , which has size $v$ and is perpendicular to $\vec{r}$ , tangent to the circle. Differentiate again and $\vec{a} (t) = - \frac{v^{2}}{R^{2}} \vec{r} (t)$ . The minus sign says the acceleration points from the object back toward the center, and its size is $\frac{v^{2}}{R}$ , usually written

a_{c} = \frac{v^{2}}{r}

with $r$ the radius of the path. This is the centripetal acceleration: the component of the acceleration directed toward the center of the path.

If the speed also changes, the acceleration has a second component along the path, the tangential acceleration $a_{t} = \frac{d v}{d t}$ . The two are perpendicular, and the net acceleration is their vector sum. A runner on a curve of radius 25 m, moving at 5.0 m/s and speeding up at 0.50 m/s², has $a_{c} = \frac{(5.0)^{2}}{25} = 1.0$ m/s² and a net acceleration of $\sqrt{{1.0}^{2} + {0.50}^{2}} \approx 1.1$ m/s².

Quick check · Calculate

A test car drives counterclockwise, seen from above, around a flat circular track of radius 40 m. The distance it has covered along the track is $s (t) = 4 t + t^{2}$ , with $s$ in meters and $t$ in seconds, so it is speeding up. What is the magnitude of the car's net acceleration at $t = 3$ s?

Your answer m/s²

2b. Real forces supply it

Newton's second law says the net force points the same way as the acceleration. So for an object moving in a circle, the force components toward the center must add up to

Σ F_{toward center} = m \frac{v^{2}}{r}

"Centripetal" names a direction, not a new kind of force. On a free-body diagram you draw only the forces something actually exerts: gravity, tension, a normal force, friction. Then you ask which of them point toward the center.

Sometimes it is a single force. A puck on a string sliding in a circle on frictionless ice has only the string's tension pointing toward the center. Sometimes it is more than one force. At the top of a vertical loop, a roller coaster car rides upside down with the track above it, so the track's normal force and gravity both point down, toward the center:

F_{N} + m g = m \frac{v^{2}}{r}

The track can only push, so $F_{N}$ can never be negative. The smallest speed that keeps the car on the circle makes $F_{N} = 0$ , leaving gravity as the only force that causes the centripetal acceleration: $m g = m \frac{v^{2}}{r}$ , so

v = \sqrt{g r}

Any slower and gravity pulls the car down more sharply than the loop curves, so it falls away from the track.

Quick check · Draw the forces

A student swings a pail of water in a vertical circle at the end of a rope. At the very top of the circle the pail is upside down, moving to the left, and the rope is taut. Air resistance is negligible. Build the free-body diagram of the pail and the water together at that instant.

The dot is the pail and the water.

Pick each force exerted on the pail and the water, then choose the way it points. The arrows are drawn for you.

2c. Components of forces

Often no force points straight at the center, and only a component of a force does the job. Use one axis toward the center and one along the vertical: the acceleration is horizontal, so it lies along one axis.

A banked curve. A road tilted toward the center of a curve pushes on a car perpendicular to its surface, so the normal force leans toward the center. Its horizontal component, $F_{N} \sin θ$ , points toward the center, and its vertical component, $F_{N} \cos θ$ , holds the car up. Static friction acts along the road's surface, so its horizontal component can add to the normal force's part or work against it, depending on which way along the slope friction points.

A conical pendulum. A ball on a string swings in a horizontal circle, so the string sweeps out a cone at an angle $θ$ from the vertical. Only two forces act: gravity and tension. The tension's vertical component balances gravity, and its horizontal component points toward the center of the circle:

F_{T} \cos θ - m g = 0 F_{T} \sin θ = m \frac{v^{2}}{r}

That is two equations in two unknowns, the tension and the speed. Notice that $r$ is the radius of the horizontal circle, not the length of the string.

Quick check · Calculate

A 0.20 kg ball hangs from a light string 0.50 m long and swings in a horizontal circle of radius 0.30 m, counterclockwise seen from above, so the ball stays 0.40 m below the point where the string is tied. Air resistance is negligible; use $g = 10$ m/s². What is the ball's speed?

Your answer m/s

2d. Period, frequency and orbits

Motion in a circle at a constant speed is uniform circular motion, and it repeats. The period $T$ is the time to complete one full circular path, and the same word names the time for one full rotation or one full cycle of an oscillation. The frequency $f$ is the rate at which revolutions are completed, in revolutions per second, or hertz:

T = \frac{1}{f}

In one period the object goes once around, a distance $2 π r$ , so $v = \frac{2 π r}{T}$ , or $T = \frac{2 π r}{v}$ .

For a satellite in a circular orbit, the gravitational force from the central body is the only force, so it alone causes the centripetal acceleration. Background from Topic 2.6: two objects of mass $M$ and $m$ whose centers are $R$ apart attract each other with a force of size $\frac{G M m}{R^{2}}$ . Setting that equal to $m \frac{v^{2}}{R}$ with $v = \frac{2 π R}{T}$ gives Kepler's third law for a circular orbit:

T^{2} = \frac{4 π^{2}}{G M} R^{3}

The satellite's own mass cancels. The period and radius depend only on each other and on the mass $M$ of the central body, which is how astronomers weigh a planet by timing its moons.

Quick check · Fill the statement

The tip of a ceiling fan blade moves at a steady speed around a circle of radius 0.60 m, counterclockwise seen from below, and makes 120 revolutions every minute.

The tip's frequency is , its period is , and its speed is .

Common mistakes

The mistake	Why it is wrong	Do this instead
Drawing a centripetal force as one more arrow on a free-body diagram.	Centripetal names the direction of the net force, not a force of its own. Something real, such as tension, gravity, a normal force or friction, has to supply it, and an extra arrow counts that force twice.	Draw only forces that some object exerts, then add their components toward the center and set the sum equal to $m \frac{v^{2}}{r}$ .
Adding the centripetal and tangential accelerations as plain numbers.	One points toward the center and the other along the path, so they are perpendicular and their sizes do not simply add.	Combine them as a vector: $\| \vec{a} \| = \sqrt{a_{c}^{2} + a_{t}^{2}}$ .
Swapping the period and the frequency.	The period is seconds per revolution and the frequency is revolutions per second, so using one in place of the other inverts every result built on it.	Write $T = \frac{1}{f}$ first and check the units, then use $T = \frac{2 π r}{v}$ .

Part 3 Watch it done A car on a banked curve

Watch each step, then do the last one yourself.

A 1200 kg car rounds a curve of radius 50 m at a steady 10 m/s, counterclockwise seen from above. The road is banked at an angle $θ$ with $\sin θ = 0.60$ and $\cos θ = 0.80$ , about 37°, rising toward the outside of the curve. Take $+ x$ horizontal toward the center of the curve and $+ y$ up, and use $g = 10$ m/s². What static friction force does the road exert on the car?

Step 1 · List the forces and the acceleration
Three forces act on the car: gravity, $m g = 12 000$ N, straight down; the normal force $F_{N}$ , perpendicular to the road, leaning toward the center; and static friction $f$ along the road's surface. Take $f$ as pointing up the slope; if it comes out negative, it points down the slope. The car's acceleration is $\frac{v^{2}}{r} = \frac{(10)^{2}}{50} = 2.0$ m/s², horizontal, toward the center.
Why this step?
Why horizontal and vertical axes, and not axes along the slope?
The acceleration is horizontal, toward the center of the curve. With a horizontal axis the whole acceleration lies along it, so the vertical forces just add to zero. Axes along the slope would split the acceleration into two parts, and both equations would carry it.
Step 2 · Write one equation per axis
The normal force has parts $F_{N} \sin θ$ toward the center and $F_{N} \cos θ$ up. Friction up the slope has parts $- f \cos θ$ , away from the center, and $+ f \sin θ$ , up. So
$F_{N} \sin θ - f \cos θ = m \frac{v^{2}}{r} = 2400 N F_{N} \cos θ + f \sin θ - m g = 0$
Step 3 · Solve for the normal force
Multiply the first equation by $\sin θ$ and the second by $\cos θ$ and add them; the friction terms cancel and $\sin^{2} θ + \cos^{2} θ = 1$ . So $F_{N} = m g \cos θ + m \frac{v^{2}}{r} \sin θ = (12 000) (0.80) + (2400) (0.60) = 9600 + 1440 = 11 040$ N.
Your step · Calculate

Finish it. How large is the static friction force the road exerts on the car? Up the slope is positive for friction.

Your answer N

Part 4 Try it New situations

Try it · Decide and justify

A puck tied to a string slides in a circle on frictionless, horizontal ice, counterclockwise seen from above; take east as $+ x$ and north as $+ y$ . The string is the only horizontal force on the puck. Just as the puck passes the point due east of the center, moving north, the string snaps. Which way does the puck move just after?

Which reason supports your answer?

Try it · Match

An object moves in a circle at a constant speed. Match each change to what happens to its centripetal acceleration.

Try it · Rank it

Three children ride playground merry-go-rounds that turn at steady rates, counterclockwise seen from above. Ana sits 1.0 m from the center of a ride that turns once every 4.0 s. Ben sits 2.0 m from the center of the same ride. Cara sits 2.0 m from the center of a different ride that turns once every 8.0 s. Rank the children by the size of their centripetal acceleration.

Greatest first

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

Ana
Ben
Cara

Your ranking: none yet

On paper. The car from Watch it done now rounds the same banked curve at 25 m/s, faster than the speed, about 19 m/s, at which the road would need no friction at all. Draw its free-body diagram: each force a straight arrow starting on the dot, labeled with what exerts it, and no centripetal force. Decide which way friction points along the slope at this speed. Then write the horizontal and vertical equations from your diagram and check that the signs agree with your choice. There is nothing to submit here.

Part 5 Explain it Predict it, derive it, then compare

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 4, Qualitative/Quantitative Translation.

Free-response practice · Question 4 style

A car of mass $m$ drives over the top of a hill at speed $v$ , moving to the right. Near the top, the road's cross section is a circular arc of radius $r$ that curves downward on both sides. Take up as positive, and ignore any forces along the road at the top.

(a) On paper, draw the free-body diagram of the car at the top of the hill. Draw each force as a straight arrow starting on the dot, and label what exerts it.

(b) Without writing equations, predict whether the normal force from the road at the top is greater than, less than or equal to $m g$ . Briefly explain your reasoning.

(c) Derive an expression for the normal force exerted by the road on the car at the top, in terms of $m$ , $v$ , $r$ and $g$ .

(d) Explain whether your expression in (c) agrees with your prediction in (b). Then use it to find the greatest speed at which the car can pass over the top without leaving the road, and explain why.

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

Model answer

(a) A dot with two arrows: $F_{g}$ , exerted by Earth, straight down, and $F_{N}$ , exerted by the road, straight up. No centripetal force and no force of motion.

(b) Less than $m g$ . At the top the car moves on a circle whose center is below it, so its acceleration points down, toward the center. The net force must point down too, so the downward gravitational force has to be larger than the upward normal force.

(c) With up as positive, the acceleration is $- \frac{v^{2}}{r}$ , so $F_{N} - m g = - m \frac{v^{2}}{r}$ and $F_{N} = m g - m \frac{v^{2}}{r}$ .

(d) It agrees: $F_{N}$ is $m g$ minus a positive amount, so it is less than $m g$ at any speed, and it gets smaller as the car goes faster. The road can only push, so $F_{N}$ cannot drop below zero. Setting $F_{N} = 0$ gives $v = \sqrt{g r}$ . Faster than that, gravity alone cannot give the car the downward acceleration $\frac{v^{2}}{r}$ that following the arc takes, so the car leaves the road.

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

Draws two forces from the dot, gravity down and the normal force up, each labeled with what exerts it, with no centripetal force.
Predicts that the normal force is less than $m g$ because the acceleration, and so the net force, points down toward the center.
Writes the second law along the vertical with the acceleration toward the center and solves it to $F_{N} = m g - m \frac{v^{2}}{r}$ .
Explains that the expression is $m g$ minus a positive term, so it agrees with the prediction in (b).
Sets $F_{N} = 0$ to get $v = \sqrt{g r}$ and explains that the road can only push, so above that speed the car leaves the road.

Part 6 Check yourself Weighing a planet by its moons

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

A space probe orbiting a distant gas giant times four of the planet's moons. Each moon's orbit is close to a circle centered on the planet, and all four moons orbit counterclockwise as seen from above the planet's north pole. The table gives each orbit's radius $R$ , measured from the planet's center, and each moon's measured period $T$ .

Orbits measured by the probe
Moon	$R$ ( $10^{8}$ m)	$T$ ( $10^{5}$ s)
A	2.00	0.770
B	3.00	1.412
C	5.00	3.046
D	8.00	6.150

Case question 1 · Multiple choice

The probe team wants a graph of the table's data that is a straight line through the origin, so they can find the planet's mass from its slope. What should they plot?

Case question 2 · Calculate

Using moon B's radius and period and $G = 6.67 \times 10^{- 11}$ N·m²/kg², what is the planet's mass? Give your answer in units of $10^{26}$ kg.

Your answer × 10^26 kg

Case question 3 · Multiple choice

A fifth moon is found orbiting the same planet at $R = 16.0 \times 10^{8}$ m, twice moon D's orbital radius. What period does Kepler's third law predict for it?

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

Check yourself · Sort it

Tap a card, then say what supplies the object's centripetal acceleration. Friction and air resistance are negligible unless a card names them.

One force aimed at the center

Two forces aimed at the center

Part of a force aimed elsewhere

Check yourself · Spot the error

A 0.50 kg ball on a 0.80 m string swings in a vertical circle. At the lowest point it moves to the right at 4.0 m/s. Take up as positive and use $g = 10$ m/s². A student finds the tension in the string there. Tap the first line with a mistake.

Part 7 Keep it Bring back earlier ideas

This lesson stands on Newton's second law and on splitting forces into components. Pulling them back up is what makes them last.

On paper. A cart rides on a level aluminum track with negligible friction, and a magnet under it produces a braking force ${\vec{F}}_{r} = - k \vec{v}$ . A string over a pulley at the end of the track ties the cart to a hanging mass, so the hanging weight pulls the cart forward. With hanging weights of 0.10, 0.20, 0.30, 0.40 and 0.50 N, a class measured terminal speeds of 0.41, 0.78, 1.22, 1.59 and 2.02 m/s. Plot terminal speed against hanging weight, with a scale and units on each axis, and draw a best-fit line. Then describe how you would measure one of those terminal speeds in your own lab: what you would record, and how your data would show that the cart had stopped speeding up. There is nothing to submit here.

Check your graph: the points lie close to a line through the origin with slope about 4.0 m/s per newton. Since $v_{T} = \frac{W}{k}$ , the slope is $\frac{1}{k}$ , so $k \approx 0.25$ kg/s. To show a run reached terminal speed, record position against time with a motion sensor and show that the velocity over the last few intervals stays the same within the sensor's precision.

From 2.9 Resistive Forces · Rank it

Four parachute toys are dropped from rest from a high balcony; take down as positive and use $g = 10$ m/s². Model each one's air resistance as ${\vec{F}}_{r} = - k \vec{v}$ . Toy A has a mass of 0.10 kg and $k = 0.20$ kg/s; toy B, 0.20 kg and 0.20 kg/s; toy C, 0.20 kg and 0.40 kg/s; toy D, 0.30 kg and 0.50 kg/s. Rank the toys by terminal speed.

Greatest first

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

Toy A
Toy B
Toy C
Toy D

Your ranking: none yet

A tilted surface with motion along it, then three blocks on level floors. For the skier, picture axes along the slope and across it before you choose directions.

From 2.2 Forces and Free-Body Diagrams · Rank it

Three identical 5.0 kg blocks sit on level floors, and none leaves its floor. A rope pulls block A with 20 N at 30° above the horizontal. A hand pushes block B with 20 N at 30° below the horizontal. A rope pulls block C with 20 N horizontally. Use $g = 10$ m/s² and take up as positive. Rank the blocks by the normal force the floor exerts on each.

Greatest first

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

Block A
Block B
Block C

Your ranking: none yet

In three lines

An object moving in a circle has a centripetal acceleration $\frac{v^{2}}{r}$ toward the center and, when its speed changes, a tangential acceleration $\frac{d v}{d t}$ ; the net acceleration is their vector sum.
Real forces supply the net force toward the center, alone, together or through their components; at the top of a loop the minimum speed is $\sqrt{g r}$ .
Period and frequency are reciprocals, $T = \frac{2 π r}{v}$ for uniform circular motion, and a circular orbit obeys $T^{2} = \frac{4 π^{2}}{G M} R^{3}$ .

Unit 2 check: 10 questions from every lesson in Unit 2

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 4, Qualitative/Quantitative Translation.
A four-function, scientific, or graphing calculator is allowed on both sections of the exam.

On a free-response question about circular motion, draw only real forces, then write the second law along the radius with toward the center as positive: the force components toward the center add up to $m \frac{v^{2}}{r}$ . Write that equation in symbols before you put in any numbers.

Questions students ask

Is there a centrifugal force pushing outward?

Not in a frame attached to the ground, which is where this course does its analysis. The outward feeling in a turning car is your body keeping its motion while the car turns, until the seat or the door pushes you inward. Every real force on an object moving in a circle has to be named by what exerts it, and none of them pushes outward.

Why is the minimum speed at the top of a vertical loop

\sqrt{g r}

At the top, the track's push and gravity both point down toward the center, so $F_{N} + m g = m \frac{v^{2}}{r}$ . The track can only push, so the smallest speed is the one that makes $F_{N} = 0$ , leaving $m g = m \frac{v^{2}}{r}$ and $v = \sqrt{g r}$ .

Does a heavier satellite take longer to orbit?

No. In $\frac{G M m}{R^{2}} = m \frac{v^{2}}{R}$ the satellite's mass $m$ appears on both sides and cancels, so the period at a given radius depends only on the mass of the central body.

How do I know which way friction points on a banked curve?

Assume a direction along the slope, write the horizontal and vertical equations, and solve. A negative answer means friction points the other way. As a check, a car slower than the speed at which the road needs no friction tends to slide down, so friction points up the slope, and a faster car needs friction pointing down the slope.

What changes when the speed on the circle is not constant?

The acceleration gains a tangential component, $\frac{d v}{d t}$ , along the path. The centripetal part, $\frac{v^{2}}{r}$ , still uses the speed at that instant, and the net acceleration is the vector sum of the two.

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 know Kepler's first or second laws of planetary motion.

Course alignment, for teachers

AP Physics C: Mechanics topic 2.10, Unit 2: Force and Translational 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.D: Predict new values or factors of change of physical quantities using functional dependence between variables.
Skill 3.C: Justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

This lesson maps all 14 essential knowledge statements of the topic to the part that teaches each one: 14 are checked by a scored question. The map is in the lesson file.