AP Physics C: Mechanics · Topic 2.10

Topic 2.10: Circular Motion

Q: Does the AP Physics C CED ever say centripetal force?

No. Every occurrence of the word centripetal in the AP Physics C: Mechanics course and exam description is followed by the word acceleration, and the phrase centripetal force appears nowhere in the document. Where a force is meant, the framework writes it out: essential knowledge 2.10.A.2.ii refers to the net force producing centripetal acceleration of an object traveling in a circle on a banked surface, and 2.10.A.2.i says the gravitational force is the only force that causes the centripetal acceleration at the top of a loop at minimum speed. The AP Physics 1 framework uses the same convention. The reason is that centripetal describes a role played by ordinary forces, so an extra arrow labelled centripetal force on a free-body diagram double counts something already drawn.

Q: Can AP Physics C ask about a banked curve with friction?

Yes, quantitatively. AP Physics 1 prints a boundary statement on its circular motion topic saying it only expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion, and that analysis of situations where friction is required is limited to qualitative descriptions. AP Physics C: Mechanics prints no such boundary statement; its Topic 2.10 has only the one about Kepler's first and second laws. Essential knowledge 2.10.A.2.ii explicitly names components of the static friction force and the normal force as contributors to the net force producing centripetal acceleration on a banked surface.

Q: What is the difference between AP Physics 1 Topic 2.9 and AP Physics C Topic 2.10?

They are the same topic with different numbers, because AP Physics C inserts Resistive Forces at 2.9 and pushes Circular Motion to 2.10. The nine essential knowledge statements under the first objective are word for word the same in both frameworks, as is the Kepler's third law statement under the second. Two things differ. AP Physics 1 carries an extra boundary statement limiting banked curves with friction to qualitative descriptions, which AP Physics C does not. And the AP Physics C equation sheet prints centripetal acceleration in two forms, v squared over r and r omega squared, where the AP Physics 1 sheet prints only the first.

Q: Is r omega squared on the AP Physics C equation sheet?

Yes. The AP Physics C: Mechanics Table of Information prints centripetal acceleration as v squared over r equals r omega squared, giving both forms on one line. The AP Physics 1 sheet prints only v squared over r. The second form is useful whenever a question specifies a rotation rate rather than a linear speed, since it saves converting through v equals r omega, which is also printed on the C sheet. Both sheets define omega as angular frequency or angular speed in their symbol lists.

Q: Does an object in circular motion have to move at constant speed?

No. Essential knowledge 2.10.A.1 defines centripetal acceleration as the component of an object's acceleration directed toward the centre of the circular path, and a component implies there can be another. Statement 2.10.A.3 names it: tangential acceleration is the rate at which an object's speed changes and is directed tangent to the path. Statement 2.10.A.4 says the net acceleration of an object moving in a circle is the vector sum of the two. Uniform circular motion, defined in 2.10.A.5 as travelling at constant speed, is the special case where the tangential component is zero, and only that case is described using period and frequency.

Q: Which of Kepler's laws does AP Physics C Mechanics require?

Only the third. The boundary statement under Topic 2.10 reads that AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion. Essential knowledge 2.10.B.1 covers the third law, stating 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. Its derived equation is that the period squared equals four pi squared divided by G times the central mass, all multiplied by the orbital radius cubed. That equation is not printed on the equation sheet, so you derive it.

Q: Why do you keep the axes horizontal for a banked curve but tilt them for a ramp?

Because essential knowledge 2.2.B.4 says to align one axis with the direction of the acceleration, and the two situations have their acceleration in different directions. A block sliding on a ramp accelerates along the ramp surface, so tilting the axes puts one along the acceleration and leaves zero acceleration on the other. A car on a banked curve accelerates horizontally, toward the centre of the circle, even though the surface is tilted, so the axes stay horizontal and vertical. Tilting them with the bank is a standing error here, and it comes from applying the ramp habit without checking where the acceleration points.

Unit 2: Force and Translational Dynamics20-25% of the multiple-choice section

Circular motion is a Newton's second law problem with the acceleration pointed at the centre. AP Physics C states the content exactly as AP Physics 1 does but removes the boundary that kept banked curves with friction qualitative, so a C question can ask you to derive a maximum cornering speed.

AP Physics: Unit 2 (topics 2.10 Circular Motion). AP Physics C: Mechanics Unit 2, Topic 2.10. Numbered 2.10 here and 2.9 in AP Physics 1, because the C course inserts Resistive Forces at 2.9. Two learning objectives. 2.10.A, describe the motion of an object traveling in a circular path, supported by 2.10.A.1 (centripetal acceleration is the COMPONENT of an object's acceleration directed toward the centre), 2.10.A.1.i (a_c = v^2/r), 2.10.A.1.ii, 2.10.A.2 (centripetal acceleration can result from a single force, more than one force, or components of forces), 2.10.A.2.i (at the top of a vertical circular loop an object requires a minimum speed; at that point the gravitational force is the only force that causes the centripetal acceleration; DERIVED v = sqrt(gr)), 2.10.A.2.ii (components of the static friction force AND the normal force can contribute to the net force producing centripetal acceleration on a banked surface), 2.10.A.2.iii (a component of tension contributes to the net force producing centripetal acceleration for a conical pendulum), 2.10.A.3 (tangential acceleration is the rate at which speed changes, directed tangent to the path), 2.10.A.4 (the net acceleration is the vector sum of the centripetal and tangential accelerations) and 2.10.A.5 with 2.10.A.5.i, 2.10.A.5.ii (T = 1/f) and 2.10.A.5.iii (called a derived equation in its own sentence: T = 2 pi r / v). 2.10.B, describe circular orbits using Kepler's third law, with 2.10.B.1 (DERIVED: T^2 = (4 pi^2 / GM) R^3). Boundary statement, the only one on this topic: AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion. CRITICAL DIFFERENCE from AP Physics 1: that course prints a SECOND boundary statement here, limiting quantitative analysis to banked curves in which no friction is required and restricting the with-friction case to qualitative descriptions. AP Physics C prints no such statement, so a quantitative banked-curve-with-friction derivation is in scope. The phrase 'centripetal force' appears NOWHERE in the AP Physics C: Mechanics CED; every occurrence of 'centripetal' is followed by 'acceleration', and where a force is meant the framework writes 'the net force producing centripetal acceleration'. The AP Physics 1 CED uses the same convention. Verified against the Table of Information appendix: the C sheet prints a_c = v^2/r = r omega^2 where the AP Physics 1 sheet prints only a_c = v^2/r; both also print T = 1/f, and the C sheet prints v = r omega. None of this topic's three derived equations is printed. Suggested skills as badged are 1.A, 2.A, 2.D, 3.C; NOTE the CED badges 1.A while giving the WORDING of 1.C. Sample free-response question 2 aligns to 2.10.A; its scenario is two race cars on a FLAT, HORIZONTAL road (not banked) with an aerodynamic downward force F_down = bv^2, which is not a resistive force in the sense of Topic 2.9. One of the unit's six optional sample instructional activities is on Topic 2.10, the penny on a rotating metre stick.

What changes: one missing boundary statement and one extra sheet line

The nine essential-knowledge statements of AP Physics C Topic 2.10.A are word for word those of AP Physics 1 Topic 2.9.A, and 2.10.B matches 2.9.B. The physics is shared. Two things around it are not.

AP Physics 1 prints a boundary statement that AP Physics C does not. It reads: "AP Physics 1 only expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion. Analysis of situations in which friction is required on a banked curve is limited to qualitative descriptions." AP Physics C: Mechanics prints only one boundary statement on this topic, the Kepler one, and nothing about banked curves at all.

So the banked-curve-with-friction derivation, which the algebra-based course can only describe in words, is a fair quantitative question here. The first worked example below does it, and the answer, $v_{\max}^2 = gr\dfrac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}$ , is not a result you can look up on the sheet.

The equation sheet gives you a second form of centripetal acceleration. The AP Physics C: Mechanics Table of Information prints

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

while the AP Physics 1 sheet prints only $a_c = v^2/r$ . The $r\omega^2$ half is useful whenever the question gives you a rotation rate rather than a speed, and it saves converting through $v = r\omega$ , which is also printed here.

Everything else on this topic is the same in both courses, and it is worth saying so rather than manufacturing a difference. The relevant equation in 2.10.A.1.i is the same, the derived $v = \sqrt{gr}$ at the top of a loop is the same, $T = 1/f$ is the same, $T = 2\pi r/v$ is the same, and Kepler's third law is the same.

The CED never says centripetal force, and that is deliberate

Search the AP Physics C: Mechanics course and exam description for the word "centripetal" and every occurrence is followed by "acceleration". The phrase "centripetal force" does not appear anywhere in the document. Where the framework needs to talk about the force, it writes the force out.

2.10.A.2: centripetal acceleration can result from a single force, more than one force, or components of forces exerted on an object in circular motion.
2.10.A.2.i: at the top of a vertical, circular loop, an object requires a minimum speed to maintain circular motion. At this point, and with this minimum velocity, the gravitational force is the only force that causes the centripetal acceleration.
2.10.A.2.ii: components of the static friction force and the normal force can contribute to the net force producing centripetal acceleration of an object traveling in a circle on a banked surface.
2.10.A.2.iii: a component of tension contributes to the net force producing centripetal acceleration experienced by a conical pendulum.
2.10.B.1: for a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction.

The same is true of the AP Physics 1 framework, where all nine occurrences of "centripetal" are also followed by "acceleration".

That wording is not fussiness. It heads off a persistent error. A centripetal force is not a new kind of force to be added to a free-body diagram; it is a job description for whatever real forces happen to be doing the pointing-inward. Adding an extra arrow labelled centripetal force double counts the tension, friction or gravity that is already on your diagram, and it fails the test of statement 2.2.A.1.i, that a force on an object is always due to the interaction of that object with another object or system.

So the working rule is: never write "centripetal force" on a diagram. Write the real force, then say what its inward component sums to.

\sum F_{\text{toward centre}} = \frac{mv^2}{r}

The right-hand side is $ma_c$ , and it belongs on the right-hand side, where the mass times acceleration goes. Putting $mv^2/r$ on the left as though it were another force is the algebraic version of the same mistake. The centripetal versus centrifugal force comparison handles the other half of the confusion.

What the CED requires of Topic 2.10

Two learning objectives. Suggested skills 1.A, 2.A, 2.D and 3.C.

Objective 2.10.A: describe the motion of an object traveling in a circular path.

2.10.A.1: centripetal acceleration is the component of an object's acceleration directed toward the centre of the object's circular path.
2.10.A.1.i: its magnitude is the ratio of the object's tangential speed squared to the radius, $a_c = v^2/r$ .
2.10.A.1.ii: centripetal acceleration is directed toward the centre of an object's circular path.
2.10.A.2: it can result from a single force, more than one force, or components of forces.
2.10.A.2.i: at the top of a vertical circular loop an object requires a minimum speed; at that point gravity alone causes the centripetal acceleration. Derived equation $v = \sqrt{gr}$ .
2.10.A.2.ii: components of static friction and the normal force can contribute to the net force producing centripetal acceleration on a banked surface.
2.10.A.2.iii: a component of tension contributes to the net force producing centripetal acceleration for a conical pendulum.
2.10.A.3: tangential acceleration is the rate at which an object's speed changes and is directed tangent to the circular path.
2.10.A.4: the net acceleration of an object moving in a circle is the vector sum of the centripetal acceleration and tangential acceleration.
2.10.A.5: the revolution of an object traveling in a circular path at a constant speed (uniform circular motion) can be described using period and frequency, with 2.10.A.5.i defining period $T$ , 2.10.A.5.ii defining frequency $f$ with $T = 1/f$ , and 2.10.A.5.iii giving the derived equation $T = 2\pi r/v$ .

Objective 2.10.B: describe circular orbits using Kepler's third law.

2.10.B.1: for a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction, and the period and radius of the circular orbit are related to the mass of the central body. Derived equation

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

Boundary statement, the only one on this topic: "AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion."

One small printing note. The CED's Unit at a Glance table badges Topic 2.10's first suggested skill as 1.A while giving the wording of skill 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of the physical system. The topic page itself does the same. Both codes are Science Practice 1, and the table above reports the badge as printed.

Uniform circular motion is the special case, not the definition

Three statements together say something students often miss: an object moving in a circle can also be speeding up or slowing down.

Statement 2.10.A.1 defines centripetal acceleration as the component of an object's acceleration directed toward the centre. A component, not the whole thing. Statement 2.10.A.3 names the other component: tangential acceleration is the rate at which an object's speed changes and is directed tangent to the path. Statement 2.10.A.4 then adds them: the net acceleration of an object moving in a circle is the vector sum of the centripetal and tangential accelerations.

So the general picture has two perpendicular components:

a_c = \frac{v^2}{r} \quad \text{(changes the direction)} \qquad a_T = \frac{dv}{dt} \quad \text{(changes the speed)}

and the magnitude of the net acceleration is $\sqrt{a_c^2 + a_T^2}$ , at an angle $\arctan(a_T/a_c)$ from the inward radius.

Uniform circular motion, which 2.10.A.5 defines as travelling at constant speed, is the case $a_T = 0$ . It is a special case, and statement 2.10.A.5 uses period and frequency only for that case. A pendulum bob swinging through the bottom of its arc, a car accelerating out of a bend, a ball on a string in a vertical circle: all of these have both components at once, and only the last is at constant radius with a genuinely varying speed throughout.

Two things follow that are worth stating flatly.

Constant speed is not constant velocity. Velocity is a vector, its direction is changing every instant, so there is an acceleration and therefore a net force. Uniform circular motion is not equilibrium, and Newton's first law does not apply to it.

Centripetal acceleration does no work. It is perpendicular to the velocity at every instant, so in $W = \int \vec{F}\cdot d\vec{r}$ the dot product vanishes. That is why the speed in uniform circular motion never changes on its own, and why the tangential component is the only one that can change it. The sheet prints the work integral in that dot-product form for exactly this reason.

The centripetal acceleration glossary entry and the uniform circular motion glossary entry cover the definitions, and the centripetal force guide walks the standard problem types.

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

Statement 2.10.A.2 says the centripetal acceleration can come from a single force, more than one force, or components of forces, and the framework then names three specific situations. Those three are the ones to have rehearsed.

The top of a vertical loop, 2.10.A.2.i. At the minimum speed, the gravitational force is the only force causing the centripetal acceleration, so the track or string contributes nothing: $mg = mv^2/r$ , giving the derived equation $v = \sqrt{gr}$ . Above that speed there is also a normal force or tension pushing inward; below it the object leaves the track. Note what "minimum" means: it is the slowest you can go and still be on a circular path, not a speed you are trying to achieve.

A banked surface, 2.10.A.2.ii. Components of the static friction force and the normal force can contribute to the net force producing centripetal acceleration. Both are named in the statement, which is the tell that the with-friction case is in scope for this course.

The axes here are the standard trap. Do not tilt them with the bank. The acceleration is horizontal, toward the centre of the circle, so keep the axes horizontal and vertical, exactly as statement 2.2.B.4 instructs: one axis parallel to the direction of acceleration.

The frictionless case is the one the algebra-based course also gets: $\tan\theta = v^2/(gr)$ , a single design speed at which the bank alone does the job. With friction there is a range, from a minimum below which the car slides down the bank to a maximum above which it slides up.

A conical pendulum, 2.10.A.2.iii. A component of tension provides the net force producing centripetal acceleration. The vertical component of the tension balances gravity and the horizontal component points at the centre of the horizontal circle. The tension is always greater than the weight, because it has to do both jobs.

Statement 2.10.B.1 supplies a fourth: for a satellite in circular orbit, the centripetal acceleration is caused only by gravitational attraction. Setting $GMm/R^2 = mv^2/R$ and substituting $v = 2\pi R/T$ from 2.10.A.5.iii gives Kepler's third law in the form the CED prints. The second worked example runs that derivation.

The boundary statement fences the rest of Kepler off: the course does not expect students to know Kepler's first or second laws of planetary motion. Ellipses and equal areas in equal times are not required, which also means every orbit you meet in this topic is circular.

If you want the algebra-based version of this topic

Watch the numbering. AP Physics 1 calls this topic 2.9 because it has no Resistive Forces topic; AP Physics C calls it 2.10. If you are taking AP Physics 1, the page for you is AP Physics 1 Topic 2.9: Circular Motion and Orbits. If you are taking AP Physics C: Mechanics, this is the page, and the banked-curve-with-friction derivation is the reason.

	AP Physics 1 Topic 2.9	AP Physics C Topic 2.10
Topic number	2.9	2.10
Statements under the first objective	nine, same wording	nine, same wording
Banked curve with friction	qualitative descriptions only	no such limit
Kepler's first and second laws	excluded	excluded
Boundary statements	two	one
Sheet: centripetal acceleration	$a_c = v^2/r$	$a_c = v^2/r = r\omega^2$
Sheet: $v = r\omega$	printed	printed
Suggested skills	1.B, 2.A, 2.D, 3.A, 3.C	1.A, 2.A, 2.D, 3.C
Says "centripetal force"	never	never

The last row is worth carrying between both courses. Neither framework uses the phrase, and neither should you.

Also on this site: the centripetal force guide and centripetal force calculator for the standard routines, Topic 2.6 for the gravitation that drives orbits, Topic 2.7 for the friction on a banked road, and the circular motion and gravitation practice set.

How Topic 2.10 is tested

Unit 2 carries 20 to 25 percent of the multiple-choice section across about 15 to 25 class periods, the highest minimum weighting of any unit in AP Physics C: Mechanics (only Unit 3, at 15 to 25 percent, reaches the same 25 percent ceiling). Topic 2.10 holds two of the unit's nineteen learning objectives and two of its eleven derived equations, plus a third result, $T = 2\pi r/v$ , that 2.10.A.5.iii calls a derived equation in its own sentence. None of the three is printed on the equation sheet.

The suggested skills are 1.A as badged, with the wording of 1.C; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Skill 2.A is the headline. It carries 25 to 30 percent of the multiple-choice section, the largest band the CED lists for any single skill, and the CED lists it for exactly four topics in Unit 2: 2.6, 2.7, 2.9 and 2.10. Those are the four with derived equations to produce, and this topic has three.

The CED's sample free-response question 2, the Translation Between Representations question worth 12 points, aligns to 2.10.A among its seven learning objectives, and the whole question is a circular-motion scenario. Two identical race cars of mass $m$ are driven on a flat, horizontal road through a turn modelled as a circular segment of radius $R_1$ , with a coefficient of static friction $\mu$ between tyres and road. The cars are built so that at speed $v$ relative to the air, the air exerts a downward force of magnitude $F_{\text{down}} = bv^2$ on them. Note that the road in that question is flat, not banked, and that the $bv^2$ force is downward rather than backward, so it is not a resistive force in the sense of Topic 2.9.

The four parts are worth knowing as a template for this topic. Part A is a free-body diagram with the arrow rules restated and a requirement that arrow lengths reflect relative magnitudes. Part B asks students to derive an expression for a critical radius in terms of $m$ , $\mu$ , $b$ and physical constants, and instructs them to begin the derivation by writing a fundamental physics principle or an equation from the reference information. Part C asks for a graph sketch against a given reference curve. Part D asks a comparison of two time intervals with a justification that may reference equations but must include conceptual reasoning beyond algebraic solutions.

That instruction in part B is the one to internalise: start from something printed. For this topic, that means starting from $\vec{a}_{\text{sys}} = \sum\vec{F}/m_{\text{sys}}$ with $a_c = v^2/r$ , not from a memorised banking formula.

One of the unit's six optional sample instructional activities is on Topic 2.10: a penny is placed on a metre stick that pivots on a pencil point through a hole drilled at its centre, the stick is rotated faster and faster until the penny slips, and students take measurements to find the coefficient of static friction between metre stick and penny. That is 2.10.A.2 and 2.7.B.2 in one apparatus.

A banked curve with friction, which AP Physics 1 cannot ask

A curve of radius $r = 85$ m is banked at $\theta = 18^\circ$ . The coefficient of static friction between tyres and road is $\mu_s = 0.35$ . Derive an expression for the maximum speed at which a car can round the curve without sliding, evaluate it, and compare it with the frictionless design speed. Use $g = 9.8\ \mathrm{m/s^2}$ .

Note why this is a Physics C question. The AP Physics 1 boundary statement on its Topic 2.9 limits quantitative analysis to banked curves where no friction is required, and restricts the with-friction case to qualitative descriptions. AP Physics C: Mechanics prints no such boundary statement, and 2.10.A.2.ii explicitly names both the static friction force and the normal force as contributors.
Draw the free-body diagram: three arrows on a dot, gravity straight down, the normal force perpendicular to the banked surface, and static friction along the surface. At the maximum speed the car tends to slide up and out, so friction points down the bank.
Choose axes. The acceleration is horizontal, pointing at the centre of the circle, so keep the axes horizontal and vertical. Do not tilt them with the bank; that is the standard error here, and it is the case where the Topic 2.2 habit misfires.
Declare the convention: positive $x$ horizontally toward the centre of the circle, positive $y$ vertically upward.
Vertical equation, with no vertical acceleration: $F_N\cos\theta - f\sin\theta - mg = 0$ . Both the normal force and the friction have vertical components, and friction's points downward because friction points down the bank.
Horizontal equation, with the acceleration equal to $v^2/r$ toward the centre: $F_N\sin\theta + f\cos\theta = \dfrac{mv^2}{r}$ .
At the maximum speed friction is at its ceiling, from 2.7.B.2.ii: $f = \mu_s F_N$ . Substitute into both equations.
Vertical becomes $F_N\left(\cos\theta - \mu_s\sin\theta\right) = mg$ , and horizontal becomes $F_N\left(\sin\theta + \mu_s\cos\theta\right) = \dfrac{mv^2}{r}$ .
Divide the second by the first. Both $F_N$ and $m$ cancel, which is the sign the derivation is going right: $\dfrac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta} = \dfrac{v^2}{gr}$ .
So $v_{\max} = \sqrt{gr\,\dfrac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta}}$ , or equivalently $\sqrt{gr\,\dfrac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}}$ after dividing top and bottom by $\cos\theta$ . That symbolic result is the answer a C question wants.
Check the limits. With $\mu_s = 0$ it reduces to $v = \sqrt{gr\tan\theta}$ , the frictionless design speed. With $\theta = 0$ it reduces to $v = \sqrt{\mu_s g r}$ , the flat-road result. Both are right.
Numbers. $\tan 18^\circ = 0.3249$ , so the numerator is $0.3249 + 0.35 = 0.6749$ and the denominator is $1 - 0.35(0.3249) = 1 - 0.1137 = 0.8863$ . Their ratio is $0.7615$ .
$gr = 9.8(85) = 833.0\ \mathrm{m^2/s^2}$ , so $v_{\max}^2 = 833.0(0.7615) = 634.4\ \mathrm{m^2/s^2}$ and $v_{\max} = 25.2$ m/s.
The frictionless design speed is $\sqrt{833.0(0.3249)} = \sqrt{270.7} = 16.5$ m/s. Friction raises the safe maximum by more than half, from 16.5 to 25.2 m/s, which is the practical reason roads are built with both a bank and a rough surface.
One more reading. The minimum speed comes from the same derivation with friction reversed, giving $\tan\theta - \mu_s$ on top. Here that is $0.3249 - 0.35 = -0.025$ , which is negative, so there is no minimum: friction is more than enough to hold the car on this gentle bank even at rest.

$v_{\max} = \sqrt{gr\dfrac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}} = 25.2$ m/s, against a frictionless design speed of $\sqrt{gr\tan\theta} = 16.5$ m/s. Because $\tan\theta < \mu_s$ here, there is no minimum speed: the car will not slide down the bank even when stationary.

Kepler's third law derived, then used to weigh a planet

Derive the relation between the period and radius of a circular orbit, then use it to find the mass of a planet given that a moon orbits it at $R = 4.22 \times 10^8$ m with a period of $T = 1.53 \times 10^5$ s. Take $G = 6.67\times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ , as printed in the constants box.

Start from something printed, as the CED's sample free-response question instructs. Two printed lines are enough: $\lvert\vec{F}_g\rvert = Gm_1m_2/r^2$ and $a_c = v^2/r$ .
Statement 2.10.B.1 supplies the physical claim: for a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction. So gravity is the entire net force, and it points at the centre.
Write Newton's second law along the inward radius for a moon of mass $m$ orbiting a body of mass $M$ : $\dfrac{GMm}{R^2} = \dfrac{mv^2}{R}$ .
The orbiting mass cancels immediately, giving $v^2 = \dfrac{GM}{R}$ . That cancellation is why every satellite at a given radius has the same orbital speed regardless of its own mass.
Bring in the period with 2.10.A.5.iii's derived equation $T = 2\pi R/v$ , rearranged as $v = 2\pi R/T$ . Substituting, $\dfrac{4\pi^2 R^2}{T^2} = \dfrac{GM}{R}$ .
Rearrange for the period squared: $T^2 = \dfrac{4\pi^2}{GM}R^3$ . That is the derived equation printed at 2.10.B.1, and it is Kepler's third law for a circular orbit. The whole derivation is four lines from two printed equations, which is exactly what the exam means by deriving from a fundamental principle.
Rearrange again for the central mass, which is what the numbers ask for: $M = \dfrac{4\pi^2 R^3}{G T^2}$ .
Compute the pieces. $R^3 = (4.22\times 10^8)^3 = 7.515\times 10^{25}\ \mathrm{m^3}$ , $T^2 = (1.53\times 10^5)^2 = 2.341\times 10^{10}\ \mathrm{s^2}$ , and $4\pi^2 = 39.478$ .
Numerator: $39.478 \times 7.515\times 10^{25} = 2.967\times 10^{27}$ . Denominator: $(6.67\times 10^{-11})(2.341\times 10^{10}) = 1.561$ .
$M = 2.967\times 10^{27} / 1.561 = 1.90\times 10^{27}$ kg.
Cross-check with the orbital speed, which is an independent route through the same physics. $v = 2\pi R/T = 2\pi(4.22\times 10^8)/(1.53\times 10^5) = 1.73\times 10^4$ m/s. Then $M = v^2 R/G = (1.73\times 10^4)^2(4.22\times 10^8)/(6.67\times 10^{-11})$ , which also gives $1.90\times 10^{27}$ kg to three significant figures.
Check the units on the final expression: $\mathrm{m^3}/\left(\mathrm{N\,m^2\,kg^{-2}}\cdot\mathrm{s^2}\right)$ . Writing a newton as $\mathrm{kg\,m\,s^{-2}}$ turns the denominator into $\mathrm{m^3\,kg^{-1}}$ , so the whole thing is in kilograms.
A scope reminder from the boundary statement: this is the only one of Kepler's laws the course requires. AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion, so every orbit in this topic is circular and no ellipse or equal-area argument is needed.

$T^2 = \dfrac{4\pi^2}{GM}R^3$ , so $M = \dfrac{4\pi^2 R^3}{GT^2} = 1.90\times 10^{27}$ kg. The orbiting body's own mass cancels out of the derivation, and the orbital speed check gives $1.73\times 10^4$ m/s and the same central mass.

Frequently asked questions

Does the AP Physics C CED ever say centripetal force?

No. Every occurrence of the word centripetal in the AP Physics C: Mechanics course and exam description is followed by the word acceleration, and the phrase centripetal force appears nowhere in the document. Where a force is meant, the framework writes it out: essential knowledge 2.10.A.2.ii refers to the net force producing centripetal acceleration of an object traveling in a circle on a banked surface, and 2.10.A.2.i says the gravitational force is the only force that causes the centripetal acceleration at the top of a loop at minimum speed. The AP Physics 1 framework uses the same convention. The reason is that centripetal describes a role played by ordinary forces, so an extra arrow labelled centripetal force on a free-body diagram double counts something already drawn.

Can AP Physics C ask about a banked curve with friction?

Yes, quantitatively. AP Physics 1 prints a boundary statement on its circular motion topic saying it only expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion, and that analysis of situations where friction is required is limited to qualitative descriptions. AP Physics C: Mechanics prints no such boundary statement; its Topic 2.10 has only the one about Kepler's first and second laws. Essential knowledge 2.10.A.2.ii explicitly names components of the static friction force and the normal force as contributors to the net force producing centripetal acceleration on a banked surface.

What is the difference between AP Physics 1 Topic 2.9 and AP Physics C Topic 2.10?

They are the same topic with different numbers, because AP Physics C inserts Resistive Forces at 2.9 and pushes Circular Motion to 2.10. The nine essential knowledge statements under the first objective are word for word the same in both frameworks, as is the Kepler's third law statement under the second. Two things differ. AP Physics 1 carries an extra boundary statement limiting banked curves with friction to qualitative descriptions, which AP Physics C does not. And the AP Physics C equation sheet prints centripetal acceleration in two forms, v squared over r and r omega squared, where the AP Physics 1 sheet prints only the first.

Is r omega squared on the AP Physics C equation sheet?

Yes. The AP Physics C: Mechanics Table of Information prints centripetal acceleration as v squared over r equals r omega squared, giving both forms on one line. The AP Physics 1 sheet prints only v squared over r. The second form is useful whenever a question specifies a rotation rate rather than a linear speed, since it saves converting through v equals r omega, which is also printed on the C sheet. Both sheets define omega as angular frequency or angular speed in their symbol lists.

Does an object in circular motion have to move at constant speed?

No. Essential knowledge 2.10.A.1 defines centripetal acceleration as the component of an object's acceleration directed toward the centre of the circular path, and a component implies there can be another. Statement 2.10.A.3 names it: tangential acceleration is the rate at which an object's speed changes and is directed tangent to the path. Statement 2.10.A.4 says the net acceleration of an object moving in a circle is the vector sum of the two. Uniform circular motion, defined in 2.10.A.5 as travelling at constant speed, is the special case where the tangential component is zero, and only that case is described using period and frequency.

Which of Kepler's laws does AP Physics C Mechanics require?

Only the third. The boundary statement under Topic 2.10 reads that AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion. Essential knowledge 2.10.B.1 covers the third law, stating 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. Its derived equation is that the period squared equals four pi squared divided by G times the central mass, all multiplied by the orbital radius cubed. That equation is not printed on the equation sheet, so you derive it.

Why do you keep the axes horizontal for a banked curve but tilt them for a ramp?

Because essential knowledge 2.2.B.4 says to align one axis with the direction of the acceleration, and the two situations have their acceleration in different directions. A block sliding on a ramp accelerates along the ramp surface, so tilting the axes puts one along the acceleration and leaves zero acceleration on the other. A car on a banked curve accelerates horizontally, toward the centre of the circle, even though the surface is tilted, so the axes stay horizontal and vertical. Tilting them with the bank is a standing error here, and it comes from applying the ramp habit without checking where the acceleration points.