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Force and Translational Dynamics

AP Physics C: Mechanics · Topic 2

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2.1

Systems and Center of Mass

Syllabus
Learning ObjectiveEssential Knowledge

2.1.A
Describe the properties and interactions of a system.

  • 2.1.A.1 System properties are determined by the interactions between objects within the system.
  • 2.1.A.2 If the properties or interactions of the constituent objects within a system are not important in modeling the behavior of the macroscopic system, the system can itself be treated as a single object.
  • 2.1.A.3 Systems may allow interactions between constituent parts of the system and the environment, which may result in the transfer of energy or mass.
  • 2.1.A.4 Individual objects within a chosen system may behave differently from each other as well as from the system as a whole.
  • 2.1.A.5 The internal structure of a system affects the analysis of that system.
  • 2.1.A.6 As variables external to a system are changed, the system's substructure may change.

2.1.B
Describe the location of a system's center of mass with respect to the system's constituent parts.

  • 2.1.B.1 For objects or systems with symmetrical mass distributions, the center of mass is located on lines of symmetry.
  • 2.1.B.2 The location of a system's center of mass along a given axis can be calculated using the equation
    • Equation: $\vec{x}_{\text{cm}} = \dfrac{\sum m_i \vec{x}_i}{\sum m_i}$
  • 2.1.B.3 For a nonuniform solid that can be considered as a collection of differential masses, $dm$, the solid's center of mass can be calculated using the equation
    • Equation: $\vec{r}_{\text{cm}} = \dfrac{\int \vec{r}\, dm}{\int dm}$
    • 2.1.B.3.i The linear mass density of a rod or other linear rigid body is the derivative of the rod's mass with respect to the position of the differential mass element on the rigid body.
      • Equation: $\lambda = \dfrac{d}{d\ell} m(\ell)$
    • 2.1.B.3.ii If a function of mass density is given for a solid, the total mass can be determined by integrating the mass density over the length (one dimension), area (two dimensions), or volume (three dimensions) of the solid. For example:
      • Equation: $M_{\text{total}} = \int \rho(r)\, dV$
  • 2.1.B.4 A system can be modeled as a singular object that is located at the system's center of mass.

Source: College Board AP Course and Exam Description

A system 系统 is the object or objects you analyze, treated as a point at its center of mass 质心. For a set of masses, $x_{\text{cm}}=\dfrac{\sum m_i x_i}{\sum m_i}$; for a continuous body, $x_{\text{cm}}=\dfrac{1}{M}\int x\,dm$. Only external forces move the center of mass.

Vocabulary Train
English Chinese Pinyin
system 系统 xì tǒng
center of mass 质心 zhì xīn
2.2

Forces and Free-Body Diagrams

Syllabus
Learning ObjectiveEssential Knowledge

2.2.A
Describe a force as an interaction between two objects or systems

  • 2.2.A.1 Forces are vector quantities that describe the interactions between objects or systems.
    • 2.2.A.1.i A force exerted on an object or system is always due to the interaction of that object or system with another object or system.
    • 2.2.A.1.ii An object or system cannot exert a net force on itself.
  • 2.2.A.2 Contact forces describe the interaction of an object or system touching another object or system and are macroscopic effects of interatomic electric forces.

2.2.B
Describe the forces exerted on an object or system using a free-body diagram.

  • 2.2.B.1 Free-body diagrams are useful tools for visualizing forces being exerted on a single object or system and for determining the equations that represent a physical situation.
  • 2.2.B.2 The free-body diagram of an object or system shows each of the forces exerted on the object or system by the environment.
  • 2.2.B.3 Forces exerted on an object or system are represented as vectors originating from the representation of the center of mass, such as a dot. A system is treated as though all of its mass is located at the center of mass.
  • 2.2.B.4 A coordinate system with one axis parallel to the direction of acceleration of the object or system simplifies the translation from free-body diagram to algebraic representation. For example, in a free-body diagram of an object on an inclined plane, it is useful to set one axis parallel to the surface of the incline.

Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism only expect students to depict the forces exerted on objects, not the force components on free-body diagrams. On the AP Physics exams, individual forces represented on a free-body diagram must be drawn as individual straight arrows, originating on the dot and pointing in the direction of the force. Individual forces that are in the same direction must be drawn side by side, not overlapping.

Source: College Board AP Course and Exam Description

A force is a push or pull (a vector, in newtons). A free-body diagram 受力图 draws one object with an arrow for every force on it – weight, normal, tension, friction, applied, drag. Draw it first; it sets up every dynamics equation.

A free-body diagram shows every force acting on one object A free-body diagram shows every force acting on one object

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Balance the forces on a free-body diagram

A free-body diagram shows every force on one object as an arrow. The object accelerates only if the forces don't cancel — the net force sets $a=F/m$.

Vocabulary Train
English Chinese Pinyin
force
free-body diagram 受力图 shòu lì tú
2.3

Newton's Third Law

Syllabus
Learning ObjectiveEssential Knowledge

2.3.A
Describe the interaction of two objects or systems using Newton's third law and a representation of paired forces exerted on each object or system.

  • 2.3.A.1 Newton's third law describes the interaction of two objects or systems in terms of the paired forces that each exerts on the other.
    • Equation: $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$
  • 2.3.A.2 Interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass.
  • 2.3.A.3 Tension is the macroscopic net result of forces that infinitesimal segments of a string, cable, chain, or similar system exert on each other in response to an external force.
    • 2.3.A.3.i An ideal string has negligible mass and does not stretch when under tension.
    • 2.3.A.3.ii The tension in an ideal string is the same at all points within the string.
    • 2.3.A.3.iii In a string with nonnegligible mass, tension may not be the same at all points within the string.
    • 2.3.A.3.iv An ideal pulley is a pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction.

Source: College Board AP Course and Exam Description

Newton's third law 牛顿第三定律: A pushes B, B pushes back equally and oppositely. The pair acts on different objects, so it never cancels within one free-body diagram.

A Newton's third-law pair: equal and opposite forces on two different objects A Newton's third-law pair: equal and opposite forces on two different objects

Vocabulary Train
English Chinese Pinyin
Newton's third law 牛顿第三定律 niú dùn dì sān dìng lǜ
2.4

Newton's First Law

Syllabus
Learning ObjectiveEssential Knowledge

2.4.A
Describe the conditions under which a system's velocity remains constant.

  • 2.4.A.1 The net force on a system is the vector sum of all forces exerted on the system.
  • 2.4.A.2 Translational equilibrium is the configuration of forces such that the net force exerted on a system is zero.
    • Derived equation: $\sum \vec{F}_i = 0$
  • 2.4.A.3 Newton's first law states that if the net force exerted on a system is zero, the velocity of that system will remain constant.
  • 2.4.A.4 Forces may be balanced in one dimension but unbalanced in another. The system's velocity will change only in the direction of the unbalanced force.
  • 2.4.A.5 An inertial reference frame is one from which an observer would verify Newton's first law of motion.

Source: College Board AP Course and Exam Description

Newton's first law (inertia 惯性): with zero net force 合力, velocity stays constant – the object is in translational equilibrium 平动平衡.

Vocabulary Train
English Chinese Pinyin
inertia 惯性 guàn xìng
net force 合力 hé lì
translational equilibrium 平动平衡 píng dòng píng héng
2.5

Newton's Second Law

Syllabus
Learning ObjectiveEssential Knowledge

2.5.A
Describe the conditions under which a system's velocity changes.

  • 2.5.A.1 Unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero.
  • 2.5.A.2 Newton's second law of motion states that the acceleration of a system's center of mass has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force.
    • Equation: $\vec{a}_{\text{sys}} = \dfrac{\sum \vec{F}}{m_{\text{sys}}} = \dfrac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$
  • 2.5.A.3 The velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.

Source: College Board AP Course and Exam Description

The general form uses momentum:

$$\sum\vec{F}=\frac{d\vec{p}}{dt}=m\vec{a}\ \ (\text{for constant mass}).$$
Apply it one axis at a time. When forces depend on velocity or position, this becomes a differential equation 微分方程 to solve.

Worked example. A $2.0\ \text{kg}$ block slides down a frictionless incline 斜面 at $30^{\circ}$. Only the along-ramp component of gravity drives it, so $a=g\sin30^{\circ}=9.8(0.5)=4.9\ \text{m/s}^2$, independent of the mass.

Worked example (two bodies). A $6.0\ \text{kg}$ cart on a frictionless table is pulled by a string over a light pulley to a hanging $2.0\ \text{kg}$ mass. Treat the pair as one system: only the hanging weight drives it, so $a=\dfrac{2.0(9.8)}{8.0}=2.45\ \text{m/s}^2$. Then isolate the cart alone to find the string tension 张力: $T=6.0(2.45)\approx15\ \text{N}$. System first for $a$, single body for internal forces – that two-step is the standard pattern.

Vocabulary Train
English Chinese Pinyin
differential equation 微分方程 wēi fēn fāng chéng
incline 斜面 xié miàn
tension 张力 zhāng lì
2.6

Gravitational Force

Syllabus
Learning ObjectiveEssential Knowledge

2.6.A
Describe the gravitational interaction between two objects or systems with mass.

  • 2.6.A.1 Newton's law of universal gravitation describes the gravitational force between two objects or systems as directly proportional to each of their masses and inversely proportional to the square of the distance between the systems' centers of mass.
    • Equation: $\left| \vec{F}_g \right| = G \dfrac{m_1 m_2}{r^2}$
    • 2.6.A.1.i The gravitational force is attractive.
    • 2.6.A.1.ii The gravitational force is always exerted along the line connecting the center of mass of the two interacting systems.
    • 2.6.A.1.iii The gravitational force on a system can be considered to be exerted on the system's center of mass.
  • 2.6.A.2 A field models the effects of a noncontact force exerted on an object at various positions in space.
    • 2.6.A.2.i The magnitude of the gravitational field created by a system of mass $M$ at a point in space is equal to the ratio of the gravitational force exerted by the system on a test object of mass $m$ to the mass of the test object.
      • Derived equation: $\left| \vec{g} \right| = \dfrac{\left| \vec{F}_g \right|}{m} = G \dfrac{M}{r^2}$
    • 2.6.A.2.ii If the gravitational force is the only force exerted on an object, the observed acceleration of the object (in $\text{m/s}^2$) is numerically equal to the magnitude of the gravitational field strength (in $\text{N/kg}$) at that location.
  • 2.6.A.3 The gravitational force exerted by an astronomical body on a relatively small nearby object is called weight.
    • Derived equation: $\text{Weight} = F_g = mg$

2.6.B
Describe situations in which the gravitational force can be considered constant.

  • 2.6.B.1 If the gravitational force between two systems' centers of mass has a negligible change as the relative position of the two systems changes, the gravitational force can be considered constant at all points between the initial and final positions of the systems.
  • 2.6.B.2 Near the surface of Earth, the strength of the gravitational field is
    • Equation: $g \approx 10\ \text{N/kg}$

2.6.C
Describe the conditions under which the magnitude of a system's apparent weight is different from the magnitude of the gravitational force exerted on that system.

  • 2.6.C.1 The magnitude of the apparent weight of a system is the magnitude of the normal force exerted on the system.
  • 2.6.C.2 If the system is accelerating, the apparent weight of the system is not equal to the magnitude of the gravitational force exerted on the system.
  • 2.6.C.3 A system appears weightless when there are no forces exerted on the system or when the force of gravity is the only force exerted on the system.
  • 2.6.C.4 The equivalence principle states that an observer in a noninertial reference frame is unable to distinguish between an object's apparent weight and the gravitational force exerted on the object by a gravitational field.

2.6.D
Describe inertial and gravitational mass.

  • 2.6.D.1 Objects have inertial mass, or inertia, a property that determines how much an object's motion resists changes when interacting with another object.
  • 2.6.D.2 Gravitational mass is related to the force of attraction between two systems with mass.
  • 2.6.D.3 Inertial mass and gravitational mass have been experimentally verified to be equivalent.

2.6.E
Describe the gravitational force exerted on an object by a uniform spherical distribution of mass.

  • 2.6.E.1 The net gravitational force exerted on an object by a uniform spherical distribution of mass is the sum of the individual forces from small differential masses that comprise the distribution.
  • 2.6.E.2 Newton's shell theorem describes the net gravitational force exerted on an object by a uniform spherical shell of mass.
    • 2.6.E.2.i The net gravitational force exerted on an object inside a thin spherical shell is zero.
    • 2.6.E.2.ii The net gravitational force exerted on an object outside a thin spherical shell can be determined by treating the shell as a single massive object located at the center of the shell.
    • 2.6.E.2.iii An object inside a sphere of uniform density experiences a net gravitational force from only a partial mass of the sphere.
    • 2.6.E.2.iv The partial mass of a sphere that contributes to the net gravitational force exerted on an object within that sphere is the portion of the sphere's mass located a distance less than or equal to the object's distance from the center of the sphere and can be calculated using the density of the sphere.
      • Derived equation: $m_{\text{partial}} = \rho \dfrac{4}{3} \pi \left( r_{\text{partial}} \right)^3$
  • 2.6.E.3 The gravitational force exerted on an object within a uniform sphere can be shown to be proportional to the object's distance from the sphere's center.
    • Derived equation: $F_{g,\text{partial}} = -k r_{\text{partial}}$

Boundary statement: AP Physics C: Mechanics does not expect students to mathematically prove or derive Newton's shell theorem.

Source: College Board AP Course and Exam Description

Orbital motion (Kepler's 2nd law)

Near a surface, weight is $F_g=mg$. In general, Newton's law of gravitation 万有引力定律:

$$F_g=\frac{Gm_1m_2}{r^2},$$
attractive and inverse-square. The gravitational field is $g=\dfrac{GM}{r^2}$.

Two masses attract each other with equal, opposite, inverse-square forces along the line joining them Two masses attract each other with equal, opposite, inverse-square forces along the line joining them

Vocabulary Train
English Chinese Pinyin
Newton's law of gravitation 万有引力定律 wàn yǒu yǐn lì dìng lǜ
Exercise sheet
2.7

Kinetic and Static Friction

Syllabus
Learning ObjectiveEssential Knowledge

2.7.A
Describe kinetic friction between two surfaces.

  • 2.7.A.1 Kinetic friction occurs when two surfaces in contact move relative to each other.
    • 2.7.A.1.i The kinetic friction force is exerted in a direction opposite the motion of each surface relative to the other surface.
    • 2.7.A.1.ii The force of friction between two surfaces does not depend on the size of the surface area of contact.
  • 2.7.A.2 The magnitude of the kinetic friction force exerted on an object is the product of the normal force the surface exerts on the object and the coefficient of kinetic friction.
    • Equation: $\left| \vec{F}_{f,k} \right| = \left| \mu_k \vec{F}_N \right|$
    • 2.7.A.2.i The coefficient of kinetic friction depends on the material properties of the surfaces that are in contact.
    • 2.7.A.2.ii Normal force is the perpendicular component of the force exerted on an object by the surface with which it is in contact; it is directed away from the surface.

2.7.B
Describe static friction between two surfaces.

  • 2.7.B.1 Static friction may occur between the contacting surfaces of two objects that are not moving relative to each other.
  • 2.7.B.2 Static friction adopts the value and direction required to prevent an object from slipping or sliding on a surface.
    • Equation: $\left| \vec{F}_{f,s} \right| \leq \left| \mu_s \vec{F}_n \right|$
    • 2.7.B.2.i Slipping and sliding refer to situations in which two surfaces are moving relative to each other.
    • 2.7.B.2.ii There exists a maximum value for which static friction will prevent an object from slipping on a given surface.
      • Derived equation: $F_{f,s,\text{max}} = \mu_s F_N$
  • 2.7.B.3 The coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces.

Source: College Board AP Course and Exam Description

Friction 摩擦力 opposes sliding along a surface: kinetic 动摩擦 $f_k=\mu_k N$ while sliding, and static 静摩擦 $f_s\le\mu_s N$ up to a maximum before sliding. $N$ is the normal force 法向力. Note the inequality: static friction is only as large as it needs to be, and $\mu_s N$ is its ceiling, not its value.

Worked example. The same block on the same $30^{\circ}$ incline, now with $\mu_k=0.20$. Perpendicular to the ramp: $N=mg\cos30^{\circ}$. Along the ramp: $ma=mg\sin30^{\circ}-\mu_k mg\cos30^{\circ}$, so $a=g(\sin30^{\circ}-0.20\cos30^{\circ})=9.8(0.500-0.173)=3.2\ \text{m/s}^2$ – the mass cancels again.

Explore

Slide a block down a slope with friction

Friction opposes motion up to a maximum $\mu N$. Tilt the slope until gravity's pull along it beats static friction and the block starts to slide.

Vocabulary Train
English Chinese Pinyin
Friction 摩擦力 mó cā lì
kinetic 动摩擦 dòng mó cā
static 静摩擦 jìng mó cā
normal force 法向力 fǎ xiàng lì
2.8

Spring Forces

Syllabus
Learning ObjectiveEssential Knowledge

2.8.A
Describe the force exerted on an object by an ideal spring.

  • 2.8.A.1 An ideal spring has negligible mass and exerts a force that is proportional to the change in its length as measured from its relaxed length. A nonideal spring either has nonnegligible mass or exerts a force that is not proportional to the change in its length as measured from its relaxed length.
  • 2.8.A.2 The magnitude of the force exerted by an ideal spring on an object is given by Hooke's law:
    • Equation: $\vec{F}_s = -k \Delta \vec{x}$
  • 2.8.A.3 The force exerted on an object by a spring is always directed toward the equilibrium position of the object–spring system.

2.8.B
Describe the equivalent spring constant of a combination of springs exerting forces on an object.

  • 2.8.B.1 A collection of springs that exert forces on an object may behave as though they were a single spring with an equivalent spring constant $k_{\text{eq}}$.
    • 2.8.B.1.i The inverse of the equivalent spring constant of a set of springs in series is equal to the sum of the inverses of the individual spring constants.
      • Derived equation: $\dfrac{1}{k_{\text{eq, series}}} = \sum_i \dfrac{1}{k_i} = \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dots$
    • 2.8.B.1.ii The equivalent spring constant of a set of springs arranged in series is smaller than the smallest constituent spring constant.
    • 2.8.B.1.iii The equivalent spring constant of a set of springs arranged in parallel is the sum of the individual spring constants.
      • Derived equation: $k_{\text{eq, parallel}} = \sum_i k_i = k_1 + k_2 + \dots$

Boundary statement: AP Physics C: Mechanics only expects students to find the effective spring constant of systems of springs that are arranged either in series or in parallel and does not expect students to find the effective spring constant of a system in which springs are arranged in both series and parallel.

Source: College Board AP Course and Exam Description

Hooke's law & the elastic limit

An ideal spring obeys Hooke's law 胡克定律 $F_s=-kx$, a restoring force set by the spring constant 弹簧劲度系数 $k$. Its stored energy is $U=\tfrac12 kx^2$.

Hooke's law: extension is proportional to load up to the limit of proportionality Hooke's law: extension is proportional to load up to the limit of proportionality

Combined springs act as one equivalent spring of constant $k_{\text{eq}}$. In parallel (side by side, sharing the load) the constants add, $k_{\text{eq}}=k_1+k_2$ – stiffer than either. In series (end to end) they combine as $\dfrac{1}{k_{\text{eq}}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$ – softer than the softest one. Two springs pulling a mass from opposite sides also act in parallel, giving a restoring constant $k_1+k_2$. Whichever you have, the SHM period uses $k_{\text{eq}}$: $T=2\pi\sqrt{m/k_{\text{eq}}}$.

Explore

Stretch a spring (Hooke's law)

A spring's force is proportional to its extension, $F=kx$ (Hooke's law). Pull harder and the extension grows in step — until the spring's limit.

Vocabulary Train
English Chinese Pinyin
Hooke's law 胡克定律 hú kè dìng lǜ
spring constant 弹簧劲度系数 tán huáng jìn dù xì shù
Exercise sheet
2.9

Resistive Forces

Syllabus
Learning ObjectiveEssential Knowledge

2.9.A
Describe the motion of an object subject to a resistive force.

  • 2.9.A.1 A resistive force is defined as a velocity-dependent force in the opposite direction of an object's velocity, for example:
    • Equation: $\vec{F}_r = -k\vec{v}$
  • 2.9.A.2 Applying Newton's second law to an object upon which a resistive force is exerted results in a differential equation for velocity.
    • 2.9.A.2.i Using the method of separation of variables, the velocity can be determined by integrating over the proper limits of integration.
    • 2.9.A.2.ii The acceleration or position of a moving object that is subject to a velocity-dependent force may be determined using initial conditions of the object and methods of calculus, once a function for velocity is determined.
    • 2.9.A.2.iii The position, velocity, and acceleration as functions of time of an object under the influence of a resistive force of the form $\vec{F}_r = -k\vec{v}$ are exponential and have asymptotes that are determined by the initial conditions of the object and the forces exerted on the object.
  • 2.9.A.3 Terminal velocity is defined as the maximum speed achieved by an object moving under the influence of a constant force and a resistive force that are exerted on the object in opposite directions. The terminal condition is reached when the net force exerted on the object is zero.

Source: College Board AP Course and Exam Description

A resistive force 阻力 (drag) opposes motion through a fluid and grows with speed, often modeled as $F=-bv$ or $F=-cv^2$. Newton's second law then gives a differential equation, e.g. $m\dfrac{dv}{dt}=mg-bv$ for a falling object. As speed rises, drag builds until it balances the driving force; the object then stops accelerating and moves at a constant terminal velocity 终极速度, found by setting the net force (and $dv/dt$) to zero.

Worked example. For a falling object with $m\dfrac{dv}{dt}=mg-bv$, the terminal velocity is where $\dfrac{dv}{dt}=0$: $mg=bv_T$, so $v_T=\dfrac{mg}{b}$. With $m=0.10\ \text{kg}$ and $b=0.50\ \text{kg/s}$, $v_T=\dfrac{0.10(9.8)}{0.50}=2.0\ \text{m/s}$.

To get the full motion $v(t)$, separate the variables 分离变量 and integrate from rest:

$$\int_0^{v}\frac{dv'}{mg-bv'}=\int_0^{t}\frac{dt'}{m}\;\Rightarrow\;-\frac{1}{b}\ln\!\frac{mg-bv}{mg}=\frac{t}{m},$$
which rearranges to $v(t)=v_T\left(1-e^{-bt/m}\right)$ – the speed rising exponentially toward $v_T$. Showing this integration is the calculus skill a Physics C free-response rewards, not just quoting the final formula.

An object falling through a fluid speeds up to a terminal velocity An object falling through a fluid speeds up to a terminal velocity

Vocabulary Train
English Chinese Pinyin
resistive force 阻力 zǔ lì
terminal velocity 终极速度 zhōng jí sù dù
separate the variables 分离变量 fēn lí biàn liàng
2.10

Circular Motion

Syllabus
Learning ObjectiveEssential Knowledge

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 center of the object's circular path.
    • 2.10.A.1.i The magnitude of centripetal acceleration for an object moving in a circular path is the ratio of the object's tangential speed squared to the radius of the circular path.
      • Equation: $a_c = \dfrac{v^2}{r}$
    • 2.10.A.1.ii Centripetal acceleration is directed toward the center of an object's circular path.
  • 2.10.A.2 Centripetal acceleration can result from a single force, more than one force, or components of forces that are 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.
      • Derived equation: $v = \sqrt{gr}$
    • 2.10.A.2.ii Components of the static friction force and the normal force can contribute to the net force producing centripetal acceleration 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.A.3 Tangential acceleration is the rate at which an object's speed changes and is directed tangent to the object's 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.
    • 2.10.A.5.i The time to complete one full circular path, one full rotation, or a full cycle of oscillatory motion is defined as period, $T$.
    • 2.10.A.5.ii The rate at which an object is completing revolutions is defined as frequency, $f$.
      • Equation: $T = \dfrac{1}{f}$
    • 2.10.A.5.iii For an object traveling at a constant speed in a circular path, the period is given by the derived equation
      • Derived equation: $T = \dfrac{2\pi r}{v}$

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. The period and radius of the circular orbit are related to the mass of the central body.
    • Derived equation: $T^2 = \dfrac{4\pi^2}{GM} R^3$

Boundary statement: AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion.

Source: College Board AP Course and Exam Description

Uniform circular motion

Uniform circular motion has a centripetal acceleration 向心加速度 $a_c=\dfrac{v^2}{r}$ pointing to the center, requiring a net inward force $F_c=\dfrac{mv^2}{r}$ supplied by a real force (tension, gravity, friction, normal). There is no outward force. For vertical circles and banked curves, decompose the real forces to find which provides the centripetal requirement.

Worked example. A $0.50\ \text{kg}$ ball is whirled on a $1.0\ \text{m}$ string in a horizontal circle at $3.0\ \text{m/s}$. The string tension supplies the whole centripetal force: $T=\dfrac{mv^2}{r}=\dfrac{0.50(3.0)^2}{1.0}=4.5\ \text{N}$.

Worked example (vertical circle). At the top of a vertical loop of radius $r$, gravity and the normal force both point toward the center: $N+mg=\dfrac{mv^2}{r}$. The slowest possible speed at the top is where the track pushes with nothing at all ($N=0$): $v_{\min}=\sqrt{gr}$. For $r=2.5\ \text{m}$: $v_{\min}=\sqrt{9.8(2.5)}=4.9\ \text{m/s}$. Any slower and the cart leaves the track before the top.

The velocity points along the tangent; the centripetal force points to the centre The velocity points along the tangent; the centripetal force points to the centre

Vocabulary Train
English Chinese Pinyin
centripetal acceleration 向心加速度 xiàng xīn jiā sù dù
2.10

Exam tips

  • Locate the center of mass with $x_{cm}=\tfrac{1}{M}\int x\,dm$ (or $\tfrac{\sum m_i x_i}{\sum m_i}$ for point masses) and use $\lambda,\sigma,\rho$ for the mass element.
  • The net external force moves the center of mass as if all mass sat there: $\vec F_{net}=M\vec a_{cm}$.
  • Define your system clearly — internal forces cancel, so only external forces change its momentum.
  • Exploit symmetry to shortcut a center-of-mass integral.
  • Distinguish center of mass from center of gravity (identical in a uniform field).

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