2.9 Circular Motion

Syllabus
2024
Topic
2.9
Level

Learning objectives

2.9A—Describe the motion of an object traveling in a circular pathDescribe the motion of an object traveling in a circular path.• Centripetal acceleration is the component of an object’s acceleration directed toward the center of the object’s circular path.- 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. Relevant equation: =av rc 2- ii. Centripetal acceleration is directed toward the center of an object’s circular path.• Centripetal acceleration can result from a single force, more than one force, or components of forces exerted on an object in circular motion.- 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 speed, the gravitational force is the only force that causes the centripetal acceleration. Derived equation: =vg r- 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.- iii. A component of tension contributes to the net force producing centripetal acceleration experienced by a conical pendulum.• T angential acceleration is the rate at which an object’s speed changes and is directed tangent to the object’s circular path.• The net acceleration of an object moving in a circle is the vector sum of the centripetal acceleration and tangential acceleration.• The revolution of an object traveling in a circular path at a constant speed (uniform circular motion) can be described using period and frequency.- i. The time to complete one full circular path, one full rotation, or a full cycle of oscillatory motion is defined as period, T.- ii. The rate at which an object is completing revolutions is defined as frequency, f. Relevant equation: =Tf 1- iii. For an object traveling at a constant speed in a circular path, the period is given by the derived equation AP Physics 1: Algebra-Based Course and Exam Description Force and Translational Dynamics UNIT 22.9B—Describe circular orbits using Kepler’s third lawDescribe circular orbits using Kepler’s third law.• 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: BOUNDARY STATEMENT 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. BOUNDARY STATEMENT AP Physics 1 does not expect students to know Kepler’s first or second laws of planetary motion. AP Physics 1: Algebra-Based Course and Exam Description AP PHYSICS 18–23% AP EXAM WEIGHTING ~22–27 CLASS PERIODS 59 | AP Physics 1: Algebra-Based Course and Exam Description Remember to go to AP Classroom to assign students the online Progress Check for this unit. Whether assigned as homework or completed in class, the Progress Check provides each student with immediate feedback related to this unit’s topics and science practices. Progress Check 3 Multiple-choice: ~18 questions Free-response: 4 questions

Circular motion needs an inward acceleration

Separate the directions

At any point on the circle:

  • Velocity v\vec v: tangent to the path; gives the instantaneous motion direction.
  • Centripetal acceleration ac\vec a_c: toward the center; changes velocity direction.
  • Tangential acceleration at\vec a_t: tangent to the path; changes speed.
  • Net acceleration: a=ac+at\vec a=\vec a_c+\vec a_t; changes direction, speed, or both.

Connect speed, radius, and timing

For radius rr and tangential speed vv, ac=v2ra_c=\dfrac{v^2}{r}. In uniform circular motion, at=0a_t=0 even though ac0a_c\ne0. One revolution takes period TT and frequency is revolutions per second: T=1fT=\dfrac{1}{f} and T=2πrvT=\dfrac{2\pi r}{v}.

Identify the real inward forces

There is no new force called 'centripetal force.' The inward net component of real forces satisfies Finward=mac\sum F_{\text{inward}}=ma_c. Gravity can supply it alone; normal force and static friction components can supply it on a banked path; a tension component supplies it for a conical pendulum.

Apply the loop threshold

Top-of-loop threshold: at the minimum speed that maintains contact, the contact force has just fallen to zero, so gravity alone supplies the inward force: mg=mv2rmg=m\dfrac{v^2}{r}. Therefore vmin=grv_{\min}=\sqrt{gr}. Below this speed, the required inward force would exceed what gravity provides at that point, so contact cannot be maintained.

Respect the course boundary

Constant speed does not mean zero acceleration in a circle because velocity direction keeps changing. For banked curves, AP Physics 1 expects quantitative analysis only when no friction is required for uniform circular motion; friction-required banked curves are analyzed qualitatively.

Larger circular orbits take longer

Choose the inward force

For a satellite of mass mm in a circular orbit of radius RR around a central mass MM, gravitational attraction supplies the entire inward force. Orbit radius is measured from the central body's center of mass.

Derive the period relation

GMmR2=mv2RG\dfrac{Mm}{R^2}=m\dfrac{v^2}{R} and v=2πRTv=\dfrac{2\pi R}{T}. Substitution cancels the satellite mass and gives Kepler's third-law form for a circular orbit: T2=4π2GMR3T^2=\dfrac{4\pi^2}{GM}R^3.

Predict proportional changes

Comparison Consequence
Same MM: RR increases by factor qq TT increases by q3/2q^{3/2}
Same RR: MM increases by factor qq TT decreases by q\sqrt q
Satellite mass mm changes Circular-orbit period at the same RR is unchanged

Apply the ratio and boundary

Ratio example: around the same central body, if orbit B has RB=4RAR_B=4R_A, then TBTA=(RBRA)3/2=43/2=8\dfrac{T_B}{T_A}=\left(\dfrac{R_B}{R_A}\right)^{3/2}=4^{3/2}=8. The farther circular orbit takes eight times as long. AP Physics 1 does not require Kepler's first or second laws; keep this derivation to circular orbits.