AgentSkillsCN

05 Circular Motion

分析热传递、温度变化以及热力学系统中的能量转换。 当学生需要解决温度、热量、比热、相变、气体定律，或热机相关问题时，此方法尤为适用。

SKILL.md
--- frontmatter
id: circular-motion
subject: physics
display_name: Circular Motion
description: |
  Analyzes objects moving in circular paths including centripetal acceleration and force.
  Use when students need to solve problems involving circular motion, centripetal force, or orbital mechanics.

grade_band: 9-12
khan_tags: [physics, circular-motion, centripetal-force, orbits]
standards:
  - NGSS.HS-PS2-1

objectives:
  - Define centripetal acceleration and its direction
  - Calculate centripetal force using F = mv²/r
  - Solve problems involving horizontal and vertical circular motion
  - Apply circular motion concepts to orbital mechanics
  - Analyze banked curves and conical pendulums

prerequisites:
  - newtons-laws
  - kinematics-2d

estimated_time_minutes: 60

validator:
  type: numeric_solver
  config:
    unit_library: physics
    default_tolerance: 0.02
    require_units: true

sources:
  - name: OpenStax College Physics
    chapter: 6
    url: https://openstax.org/books/college-physics/pages/6-introduction-to-uniform-circular-motion-and-gravitation
    license: CC-BY-4.0

Circular Motion

Misconceptions

MisconceptionCorrection
"There's a centrifugal force pushing objects outward"Centrifugal force is fictitious; objects feel pushed outward due to inertia
"Centripetal force is a new type of force"Centripetal force is any force that causes circular motion (tension, gravity, friction, normal force)
"Objects in circular motion are in equilibrium"Objects in circular motion are accelerating toward the center
"Faster motion means larger radius"For constant centripetal force, faster motion requires smaller radius (or more force)
"Satellites are not falling"Satellites are continuously falling but moving forward fast enough to miss Earth

Key Concepts

Uniform Circular Motion

Motion in a circle at constant speed. Even though speed is constant, velocity is changing (direction changes), so there is acceleration.

Key insight: The acceleration is always directed toward the center of the circle.

Centripetal Acceleration

  • •Magnitude: a_c = v²/r = ω²r = 4π²r/T²
  • •Direction: Always toward the center of the circle
  • •v = linear (tangential) speed
  • •r = radius of circular path
  • •ω = angular velocity (rad/s)
  • •T = period (time for one revolution)

Centripetal Force

  • •The net force required to keep an object moving in a circle
  • •F_c = ma_c = mv²/r
  • •This is NOT a new force—it's the name for whatever force provides the centripetal acceleration
  • •Examples: tension (ball on string), friction (car turning), gravity (orbits), normal force (loop)

Period and Frequency

  • •Period (T): Time for one complete revolution
  • •Frequency (f): Number of revolutions per second, f = 1/T
  • •Angular velocity: ω = 2π/T = 2πf
  • •Linear speed: v = 2πr/T = ωr

Equations

code
[1] a_c = v²/r (centripetal acceleration)
[2] a_c = ω²r (using angular velocity)
[3] F_c = mv²/r (centripetal force)
[4] v = 2πr/T (speed from period)
[5] ω = 2π/T = 2πf (angular velocity)
[6] v = ωr (relating linear and angular speed)
[7] F_gravity = GMm/r² (gravitational force)
[8] v_orbit = √(GM/r) (orbital velocity)

Worked Examples

Example 1: Car on a Curve

Problem: A 1500 kg car travels around a flat curve of radius 50 m at 15 m/s. What friction force is required?

Solution:

  1. •The friction force provides the centripetal force
  2. •F_c = mv²/r = 1500 × (15)² / 50
  3. •F_c = 1500 × 225 / 50 = 6750 N
  4. •Check: This is the friction needed; compare to μmg to see if it's possible

Example 2: Ball on a String (Vertical Circle)

Problem: A 0.5 kg ball on a 1.2 m string swings in a vertical circle. What minimum speed is needed at the top of the circle?

Solution:

  1. •At the top, both gravity and tension point toward center
  2. •At minimum speed, tension = 0, so gravity alone provides centripetal force
  3. •mg = mv²/r → v² = gr
  4. •v = √(gr) = √(9.8 × 1.2) = √11.76 = 3.43 m/s

Example 3: Orbital Speed

Problem: A satellite orbits Earth at altitude 400 km. Find its orbital speed. (Earth radius = 6.37 × 10⁶ m, Earth mass = 5.97 × 10²⁴ kg)

Solution:

  1. •Orbital radius: r = R_Earth + altitude = 6.37 × 10⁶ + 4 × 10⁵ = 6.77 × 10⁶ m
  2. •Gravity provides centripetal force: GMm/r² = mv²/r
  3. •v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.77 × 10⁶)
  4. •v = √(5.88 × 10⁷) = 7670 m/s ≈ 7.67 km/s

Explanation Patterns

  1. •Identify the source of centripetal force - What force (or forces) point toward the center?
  2. •Draw a free body diagram at the position of interest (top, bottom, side of circle)
  3. •Set up F_net = ma_c with forces toward center as positive
  4. •Be careful at top vs. bottom of vertical circles - weight and normal/tension may add or subtract
  5. •For orbits, equate gravitational force to centripetal force
  6. •Check if the required friction is reasonable - compare to μmg

Common Problem Types

  1. •Horizontal circles: Friction or tension provides centripetal force
  2. •Vertical circles: Analyze at top and bottom separately
  3. •Banked curves: Normal force has a horizontal component
  4. •Conical pendulum: Horizontal component of tension provides centripetal force
  5. •Satellites/orbits: Gravity provides centripetal force
  6. •Loop-the-loop: Find minimum speed at top where normal force = 0