AP Physics 1 · Topic 2.9

Topic 2.9: Circular Motion

Unit 2: Force and Translational Dynamics18-23% of the multiple-choice section

Topic 2.9 treats circular motion as a Newton's second law problem. An object on a circular path has a centripetal acceleration pointing at the center, equal to its tangential speed squared over the radius, and real forces supply it. If the speed also changes, a tangential acceleration adds on.

AP Physics: Unit 2 (topics 2.9 Circular Motion). Topic 2.9 carries two CED learning objectives: 2.9.A, describe the motion of an object traveling in a circular path, and 2.9.B, describe circular orbits using Kepler's third law. Two boundary statements limit quantitative banked-curve work to the case where no friction is required, and exclude Kepler's first and second laws. The CED's suggested skills for this topic are 1.B, 2.A, 2.D, 3.A, and 3.C. Topic 2.9 closes Unit 2, weighted at 18 to 23 percent of the multiple-choice section.

What Topic 2.9 requires

Two learning objectives close out Unit 2. LO 2.9.A asks you to describe the motion of an object traveling in a circular path, and it carries nearly all the content: centripetal acceleration, where the inward force comes from, tangential acceleration, the net acceleration, and period and frequency. LO 2.9.B asks you to describe circular orbits using Kepler's third law.

Two boundary statements then cut work out of your revision, and both are worth reading before you study anything else. The CED states that AP Physics 1 only expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion, and that analysis of situations in which friction is required on a banked curve is limited to qualitative descriptions. It also states that AP Physics 1 does not expect students to know Kepler's first or second laws of planetary motion.

The suggested skills are 1.B (create quantitative graphs), 2.A (derive a symbolic expression), 2.D (predict new values or factors of change using functional dependence between variables), 3.A (create experimental procedures), and 3.C (justify or support a claim using evidence). Skill 2.D is the ranking-and-scaling skill, and skill 3.C is argumentation, so the topic is set up for reasoning questions as much as for arithmetic.

Centripetal acceleration is a component of the acceleration

EK 2.9.A.1 defines it with unusual care: centripetal acceleration is the component of an object's acceleration directed toward the center of the object's circular path. A component. Not a force, and not necessarily the whole acceleration.

Its magnitude is the ratio of the object's tangential speed squared to the radius of the circular path:

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

That line is printed on the AP Physics 1 equation sheet. What the sheet does not print is any expression for a centripetal force. You build that yourself by putting $a_c$ into $\vec{F}_{net} = m\vec{a}$ , which the sheet does print. EK 2.9.A.2 backs that up: centripetal acceleration can result from a single force, more than one force, or components of forces exerted on an object in circular motion.

The frame question settles itself the same way. The exam's convention list, printed alongside the Table of Information, states that the frame of reference of any problem is assumed to be inertial unless otherwise stated. In an inertial frame there is no outward force to draw, so the arrows on your diagram sum to something pointing inward rather than to zero. The outward sensation a passenger reports belongs to the rotating frame of the vehicle, which the exam does not use. Topic 1.4 covers frames, and the centripetal force guide works the free-body diagrams case by case.

Tangential acceleration and the acceleration that is actually there

Uniform circular motion is the special case where the speed does not change. When it does, EK 2.9.A.3 adds a second piece: tangential acceleration is the rate at which an object's speed changes, and it is directed tangent to the object's circular path. EK 2.9.A.4 then states that the net acceleration of an object moving in a circle is the vector sum of the centripetal acceleration and the tangential acceleration.

The two are perpendicular by construction, one along the radius and one along the tangent, so they combine the way any two perpendicular components do:

|\vec{a}| = \sqrt{a_c^2 + a_T^2}

Notice which result survives the loss of constant speed and which does not. The expression $a_c = v^2/r$ still holds at every instant, because the CED writes it in terms of tangential speed rather than constant speed. You simply use the speed at the instant in question. What does not survive is the constant-acceleration kinematic set from Topic 1.3. Even in uniform circular motion the acceleration is continually changing direction, so no axis carries a constant acceleration and none of those three equations applies to a circular path.

The rotational column of the equation sheet writes the tangential piece as $a_T = r\alpha$ , which is the Unit 5 way of saying the same thing once you are describing the motion with angular quantities.

Period and frequency

EK 2.9.A.5 states that the revolution of an object traveling in a circular path at a constant speed can be described using period and frequency. The CED then defines each. Period $T$ is the time to complete one full circular path, one full rotation, or a full cycle of oscillatory motion. Frequency $f$ is the rate at which an object is completing revolutions. The sheet prints the link between them:

T = \frac{1}{f}

For constant speed, one period covers exactly one circumference, which gives the CED's derived equation for this case:

T = \frac{2\pi r}{v}

That second one is not printed on the equation sheet. It is listed in the CED as a derived equation, meaning you are expected to produce it, and the derivation is one line: the distance around is $2\pi r$ , the speed is $v$ , and time is distance over speed.

Put numbers on it. A stone whirled on the end of a 0.80 m string completes 12 revolutions in 6.0 s. Then $f = 12 / 6.0 \, \mathrm{s} = 2.0 \, \mathrm{Hz}$ , so $T = 1/f = 0.50 \, \mathrm{s}$ , and $v = 2\pi r / T = 2\pi (0.80 \, \mathrm{m}) / (0.50 \, \mathrm{s}) = 10 \, \mathrm{m/s}$ . Its centripetal acceleration is then $v^2/r \approx 130 \, \mathrm{m/s^2}$ , about 13 times $g$ , which is why the string has to pull with roughly 13 times the stone's weight.

Where the inward force comes from: the CED's three named cases

After stating that the inward acceleration can come from one force, several forces, or components of forces, the CED names three specific situations.

Top of a vertical loop. At the minimum speed that keeps an object on a circular path there, the gravitational force is the only force causing the centripetal acceleration. Setting $mg = mv^2/r$ and canceling the mass gives the CED's derived equation:

v = \sqrt{gr}

The mass drops out, so the minimum speed depends only on the radius. A loaded and an empty car need the same speed at the top of the same loop.

Banked surface. Components of the static friction force and the normal force can both contribute to the net force producing centripetal acceleration for an object circling on a banked surface. The boundary statement limits your quantitative work to the case where no friction is needed. There the normal force alone does the job, and resolving it gives $F_N\cos\theta = mg$ vertically and $F_N\sin\theta = mv^2/r$ horizontally. Dividing the second by the first removes both the mass and the normal force and leaves $\tan\theta = v^2/(rg)$ , worked below.

Conical pendulum. A component of tension contributes to the net force producing the centripetal acceleration. The geometry is the banked curve turned inside out: the vertical component of the tension carries the weight while the horizontal component turns the bob.

Circular orbits and Kepler's third law

LO 2.9.B is short. EK 2.9.B.1 states that for a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction, and that the period and radius of the circular orbit are related to the mass of the central body. The CED gives the derived equation:

T^2 = \frac{4\pi^2}{GM}R^3

Read the structure before touching arithmetic, since skill 2.D is exactly that. $T^2$ is proportional to $R^3$ , so a satellite moved to four times the orbital radius takes eight times as long to circle. The satellite's own mass appears nowhere, which is why an astronaut on a spacewalk drifts alongside the station rather than falling behind it. And $M$ is the mass of the central body, so one period and one radius together weigh a planet.

Being a derived equation, it is fair game to ask you to build it: set the sheet's $|\vec{F}_g| = G\frac{m_1m_2}{r^2}$ equal to $mv^2/R$ , substitute $v = 2\pi R/T$ , and rearrange. The constant $G = 6.67 \times 10^{-11} \, \mathrm{N \cdot m^2/kg^2}$ is in the CED's Table of Information.

The second boundary statement keeps the scope tight: Kepler's first and second laws are not expected, so ellipses, foci, and equal areas swept in equal times stay off the list. Unit 6 returns to orbits with its own topic on the motion of orbiting satellites.

How Topic 2.9 is tested

The CED's sample multiple-choice set includes one item aligned to LO 2.9.A and EK 2.9.A.2, with skill 2.B. It puts an object on a light string, releases it from rest with the string horizontal, and asks for the tension when the string swings to vertical. Two steps are needed and neither is circular motion on its own: energy gives the speed at the lowest point, and only then does the radial equation give the tension. Work it symbolically and the answer is three times the object's weight, not equal to it.

The CED's suggested classroom activities push the same way. One has students walk out a described driving action while holding one arm as the velocity vector and the other as the acceleration vector, which forces the two apart. One asks for a case where the smaller circle carries the greater net force and a case where the larger circle does, which is skill 2.D on two variables at once. A third swings a known weight on a force sensor through a 180 degree arc and asks what the sensor reads at the bottom, where the object is neither speeding up nor slowing down. The CED anticipates that students will answer with the weight, and asks the class to work out why that is wrong.

That activity and the sample question are the same physics. At the bottom of a swing the speed is momentarily unchanging, but the direction of the velocity is not, so there is an inward acceleration and the support must exceed the weight. Once the setup is right, the centripetal force calculator will check the arithmetic.

Errors that cost points on circular motion

Five failures account for most lost marks on this topic, and only one of them is arithmetic.

Drawing an inward force labeled centripetal. The inward force is the sum of the real arrows already on the diagram, so adding another one double counts it. Draw gravity, normal force, tension, and friction, then add them up.
Requiring constant speed before using the ratio of speed squared to radius. The CED writes that expression with tangential speed. It is valid at any instant of a circular path, and only the value of $v$ changes.
Applying the constant-acceleration kinematic equations to a circular path. The acceleration turns continuously, so no axis has a constant acceleration to feed those equations.
Subtracting at the top of a loop. Gravity and the normal force both point downward there, which is toward the center, so they add rather than oppose.
Putting friction into a banked-curve calculation. The boundary statement limits quantitative banked-curve work to the frictionless design speed. If a question mentions friction on a bank, it wants words, not numbers.

Numbers on this page are computed with $g = 9.8 \, \mathrm{m/s^2}$ , the figure in the CED's Table of Information. Do not read that as the figure the exam uses: a Topic 1.3 boundary statement says the exam will take $g \approx 10 \, \mathrm{m/s^2}$ whenever a numerical value is required, while accepting $9.81$ or $9.8$ as correct. The banked-curve example below shows how little the choice usually changes once you round.

Speeding up through a curve: centripetal plus tangential

A car rounds a curve of radius 40.0 m. At one instant its speed is 12.0 m/s and it is gaining speed at $1.5 \, \mathrm{m/s^2}$ . Find the centripetal acceleration, the magnitude of the car's total acceleration, and the angle between the total acceleration and the line pointing to the center of the curve.

Set the convention. Take the radial direction as positive when it points toward the center of the curve, and the tangential direction as positive in the direction the car is traveling. Both stay fixed for the whole calculation.
Centripetal component, using the speed at this instant: $a_c = v^2/r = (12.0 \, \mathrm{m/s})^2 / 40.0 \, \mathrm{m} = 144 / 40.0 = 3.60 \, \mathrm{m/s^2}$ , pointing at the center.
Tangential component. The car is gaining speed, so EK 2.9.A.3 gives $a_T = 1.5 \, \mathrm{m/s^2}$ directed forward along the tangent.
The two are perpendicular, so combine them as components rather than adding them: $|\vec{a}| = \sqrt{a_c^2 + a_T^2} = \sqrt{(3.60)^2 + (1.5)^2} = \sqrt{12.96 + 2.25} = \sqrt{15.21} = 3.90 \, \mathrm{m/s^2}$ .
Angle from the radial line: $\tan\phi = a_T / a_c = 1.5 / 3.60 = 0.4167$ , so $\phi = 22.6^\circ$ .
Round to the inputs. The tangential acceleration is given to two significant figures, so the combined magnitude and the angle stop there as well. Note also that the car's mass never entered: accelerations do not need it, and only a question about force would.

Centripetal acceleration is $3.60 \, \mathrm{m/s^2}$ toward the center, and the total acceleration is 3.9 m/s squared at 23 degrees from the inward radial line, tilted toward the direction of travel. Because the speed is changing, the acceleration is not pointing at the center; it points at the center only in uniform circular motion, where the tangential component is zero.

The banking angle that needs no friction

An engineer is designing a curve of radius 60.0 m so that a car traveling at 20.0 m/s can round it with no friction between the tires and the road. At what angle must the roadway be banked?

Set the convention. Take up as positive vertically, and take the horizontal direction pointing toward the center of the curve as positive. The car moves in a horizontal circle, so its acceleration is horizontal and inward, with no vertical component.
Draw the forces. With no friction, only two act: gravity $mg$ straight down, and the normal force $F_N$ perpendicular to the banked road surface. Tilting the road tilts $F_N$ , which is the whole trick, since a tilted $F_N$ has a horizontal component pointing at the center.
Vertical equation. There is no vertical acceleration, so the vertical component of the normal force carries the full weight: $F_N\cos\theta = mg$ .
Horizontal equation. The horizontal component of the normal force is the entire net inward force: $F_N\sin\theta = mv^2/r$ .
Divide the horizontal equation by the vertical one. Both $F_N$ and $m$ cancel, leaving $\tan\theta = \frac{v^2}{rg}$ . The mass cancelling is the useful part: one banking angle works for a motorcycle and a loaded truck alike.
Substitute: $\tan\theta = (20.0 \, \mathrm{m/s})^2 / [(60.0 \, \mathrm{m})(9.8 \, \mathrm{m/s^2})] = 400 / 588 = 0.680$ , so $\theta = 34.2^\circ$ . Computed instead with the exam's $g \approx 10 \, \mathrm{m/s^2}$ , $\tan\theta = 400/600 = 0.667$ and $\theta = 33.7^\circ$ . Both round to the same two-figure answer.

The curve must be banked at 34 degrees. That angle is the design speed condition and nothing more: at exactly 20.0 m/s no friction is needed, while a car going faster or slower needs friction to make up the difference. The CED's boundary statement keeps that friction case qualitative in AP Physics 1, so a question that adds friction to a bank is asking you to explain, not to compute.

Frequently asked questions

What does AP Physics 1 Topic 2.9 cover?

Two learning objectives. 2.9.A asks you to describe the motion of an object traveling in a circular path, covering centripetal acceleration and its magnitude, the forces that can produce it, tangential acceleration, the net acceleration as a vector sum, and period and frequency. 2.9.B asks you to describe circular orbits using Kepler's third law, which relates the period and radius of a circular orbit to the mass of the central body.

What is the difference between centripetal and tangential acceleration?

Centripetal acceleration is the component pointing at the center of the circle, and it exists whenever the direction of the velocity changes. Tangential acceleration is the rate at which the speed changes, and it points along the tangent. In uniform circular motion the tangential piece is zero. When both are present the net acceleration is their vector sum, and since they are perpendicular the magnitude is the square root of the sum of their squares.

Does AP Physics 1 test banked curves with friction?

Only qualitatively. The CED boundary statement says AP Physics 1 expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion, and limits analysis of the friction case to qualitative descriptions. So learn the frictionless design-speed relationship, tan theta equals v squared over r g, and be ready to explain in words which way friction has to point when the car is faster or slower than that design speed.

Is the equation for the period of circular motion on the AP formula sheet?

The relationship between period and frequency is printed on the sheet. The period of circular motion, two pi r divided by v, is not; the CED lists it as a derived equation. Producing it takes one line, since the object covers one circumference in one period and time equals distance divided by speed. The same is true of Kepler's third law for circular orbits, which the CED also lists as derived.

Do I need Kepler's first and second laws for AP Physics 1?

No. A boundary statement in the CED says AP Physics 1 does not expect students to know Kepler's first or second laws of planetary motion, so elliptical orbits, foci, and equal areas in equal times are outside the course. Only the third law appears, and only for circular orbits, where the satellite's centripetal acceleration is caused entirely by gravitational attraction.