šŸŽÆā­ INTERACTIVE LESSON

Uniform Circular Motion

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Uniform Circular Motion - Complete Interactive Lesson

Part 1: Circular Motion Basics

šŸ”„ Circular Motion Basics

Part 1 of 7 — Uniform Circular Motion

An object moving in a circle at constant speed is in uniform circular motion. While the speed stays the same, the direction of motion is constantly changing — which means the velocity is changing, and there must be an acceleration!

In this lesson you will learn:

  • Period, frequency, and angular speed
  • The relationship v=2Ļ€r/Tv = 2\pi r / T
  • How to convert between period and frequency
  • Calculating speed for objects in circular paths

Key Definitions

Period (TT)

The time for one complete revolution (one full circle).

  • Units: seconds (s)
  • Example: A Ferris wheel completes one rotation in 60 s → T=60T = 60 s

Frequency (ff)

The number of revolutions per second.

  • Units: hertz (Hz) = 1/s = rev/s
  • f=1/Tf = 1/T and T=1/fT = 1/f

Speed in Circular Motion

The distance traveled in one revolution is the circumference: C=2Ļ€rC = 2\pi r.

v=distancetime=2Ļ€rT=2Ļ€rfv = \frac{\text{distance}}{\text{time}} = \frac{2\pi r}{T} = 2\pi r f

QuantitySymbolUnitsFormula
PeriodTTsT=1/f=2Ļ€r/vT = 1/f = 2\pi r/v
FrequencyffHzf=1/T=v/(2Ļ€r)f = 1/T = v/(2\pi r)
Speedvvm/sv=2Ļ€r/T=2Ļ€rfv = 2\pi r/T = 2\pi rf

Important Note

In uniform circular motion, speed is constant but velocity is not (because the direction changes continuously).

Circular Motion Concepts šŸŽÆ

Speed and Period Calculations 🧮

  1. A merry-go-round has radius 4 m and completes one revolution in 8 s. What is the speed of a rider on the edge (in m/s, use Ļ€ā‰ˆ3.14\pi \approx 3.14, round to 3 significant figures)?

  2. A satellite orbits Earth at v=7800v = 7800 m/s in a circular orbit of radius 6.7Ɨ1066.7 \times 10^6 m. What is its orbital period (in seconds, round to nearest 100)?

  3. A fan blade tip is 0.3 m from the center and spins at 20 Hz. What is its speed (in m/s, round to 3 significant figures)?

Concept Connections šŸ”

Exit Quiz — Circular Motion Basics āœ…

Part 2: Period, Frequency & Speed

šŸŽÆ Centripetal Acceleration

Part 2 of 7 — Uniform Circular Motion

Even though an object in uniform circular motion moves at constant speed, it is always accelerating. This acceleration — called centripetal acceleration — points toward the center of the circle and changes the direction of velocity.

In this lesson you will learn:

  • Why constant speed still requires acceleration
  • The formula ac=v2/ra_c = v^2/r
  • Equivalent forms using period and frequency
  • How to calculate centripetal acceleration in real scenarios

Why Is There Acceleration?

Acceleration = rate of change of velocity (a vector).

Even if speed (magnitude) is constant, the direction of velocity is continuously changing as the object moves around the circle. A change in velocity — in any way — requires acceleration.

The Centripetal Acceleration Formula

ac=v2ra_c = \frac{v^2}{r}

Equivalent Forms

Since v=2Ļ€r/Tv = 2\pi r/T:

ac=v2r=(2Ļ€r/T)2r=4Ļ€2rT2a_c = \frac{v^2}{r} = \frac{(2\pi r/T)^2}{r} = \frac{4\pi^2 r}{T^2}

Since f=1/Tf = 1/T:

ac=4Ļ€2rf2a_c = 4\pi^2 r f^2

FormulaWhen to Use
ac=v2/ra_c = v^2/rWhen speed and radius are known
ac=4Ļ€2r/T2a_c = 4\pi^2 r/T^2When period and radius are known
ac=4Ļ€2rf2a_c = 4\pi^2 rf^2When frequency and radius are known

Key Relationships

  • acāˆv2a_c \propto v^2: doubling speed → 4Ɨ the acceleration
  • acāˆ1/ra_c \propto 1/r: at constant speed, larger radius → less acceleration
  • acāˆra_c \propto r (at constant TT): at constant period, larger radius → more acceleration

Centripetal Acceleration Concepts šŸŽÆ

Centripetal Acceleration Calculations 🧮

  1. A car rounds a curve of radius 50 m at 20 m/s. What is the centripetal acceleration (in m/s²)?

  2. A record player rotates at 45 RPM. A coin is placed 10 cm from the center. What is the centripetal acceleration of the coin (in m/s², round to 3 significant figures)?

  3. A ball on a string moves in a circle at v=6v = 6 m/s with ac=18a_c = 18 m/s². What is the radius of the circle (in m)?

Proportional Reasoning šŸ”

Exit Quiz — Centripetal Acceleration āœ…

Part 3: Centripetal Acceleration

āž”ļø Direction of Centripetal Acceleration

Part 3 of 7 — Uniform Circular Motion

The centripetal acceleration always points toward the center of the circular path. This is what "centripetal" literally means — "center-seeking." Understanding direction is essential for solving force and motion problems.

In this lesson you will learn:

  • Why acceleration points toward the center
  • How to identify the direction of velocity and acceleration at any point
  • The perpendicular relationship between vāƒ—\vec{v} and aāƒ—c\vec{a}_c
  • Common misconceptions about "centrifugal force"

Direction Analysis

Velocity Direction

At any point on the circle, velocity is tangent to the circle — perpendicular to the radius at that point.

Acceleration Direction

Centripetal acceleration always points radially inward — from the object toward the center of the circle.

The vāƒ—\vec{v} and aāƒ—\vec{a} Relationship

PropertyVelocity (vāƒ—\vec{v})Acceleration (aāƒ—c\vec{a}_c)
DirectionTangent to circleToward center
MagnitudeConstant (vv)Constant (v2/rv^2/r)
Angle between them—Always 90°90°

Why 90°90°?

The acceleration is perpendicular to the velocity. This is exactly what's needed to change direction without changing speed:

  • If aāƒ—\vec{a} were parallel to vāƒ—\vec{v}: speed would change (speeding up or slowing down)
  • If aāƒ—\vec{a} is perpendicular to vāƒ—\vec{v}: only direction changes, speed stays constant

This is the hallmark of uniform circular motion!

At Specific Points (counterclockwise motion)

PositionVelocity DirectionAcceleration Direction
Top of circleLeft (←)Down (↓) toward center
Bottom of circleRight (→)Up (↑) toward center
Right sideUp (↑)Left (←) toward center
Left sideDown (↓)Right (→) toward center

Direction Identification šŸŽÆ

The "Centrifugal Force" Misconception

What Students Often Think

"When I go around a curve, I feel pushed outward — there must be an outward (centrifugal) force!"

The Reality

  • There is no outward force on you (in an inertial reference frame)
  • What you feel is your body's inertia — wanting to continue in a straight line
  • The seat/door pushes you inward (toward the center), and your body resists this change in direction
  • This inward push IS the centripetal force

The Correct Explanation

When a car turns left:

  1. Your body wants to keep going straight (Newton's 1st Law)
  2. The car seat pushes you to the left (toward the center of the turn)
  3. You feel "pushed" against the right door, but actually the car is turning under you
  4. The contact force from the door provides the centripetal force

On the AP Exam

  • Never refer to "centrifugal force" — it doesn't exist in an inertial frame
  • Always identify the real force providing centripetal acceleration (tension, friction, gravity, normal force)

Misconception Busters šŸ”

Exit Quiz — Direction of Acceleration āœ…

Part 4: Direction of Velocity & Acceleration

šŸ“ Describing Circular Motion with Vectors

Part 4 of 7 — Uniform Circular Motion

To fully describe circular motion, we need to track how the position, velocity, and acceleration vectors change over time. This vector description is essential for AP Physics problems involving circular motion in two dimensions.

In this lesson you will learn:

  • Position vectors for circular motion
  • Velocity and acceleration vector components
  • How to decompose circular motion into x and y components
  • The relationship between angular position and linear quantities

Position Vector

For an object moving counterclockwise starting from the positive x-axis:

rāƒ—(t)=rcos⁔θx^+rsin⁔θy^\vec{r}(t) = r\cos\theta \hat{x} + r\sin\theta \hat{y}

where Īø=ωt\theta = \omega t and ω=2Ļ€/T=2Ļ€f\omega = 2\pi/T = 2\pi f is the angular velocity (rad/s).

rāƒ—(t)=rcos⁔(ωt)x^+rsin⁔(ωt)y^\vec{r}(t) = r\cos(\omega t) \hat{x} + r\sin(\omega t) \hat{y}

Velocity Vector

Taking the derivative:

vāƒ—(t)=āˆ’rωsin⁔(ωt)x^+rωcos⁔(ωt)y^\vec{v}(t) = -r\omega\sin(\omega t) \hat{x} + r\omega\cos(\omega t) \hat{y}

The magnitude: ∣vāƒ—āˆ£=rω=v|\vec{v}| = r\omega = v āœ“

Acceleration Vector

Taking another derivative:

aāƒ—(t)=āˆ’rω2cos⁔(ωt)x^āˆ’rω2sin⁔(ωt)y^\vec{a}(t) = -r\omega^2\cos(\omega t) \hat{x} - r\omega^2\sin(\omega t) \hat{y}

This can be rewritten as:

aāƒ—(t)=āˆ’Ļ‰2rāƒ—(t)\vec{a}(t) = -\omega^2 \vec{r}(t)

Key Insight

aāƒ—=āˆ’Ļ‰2rāƒ—\vec{a} = -\omega^2 \vec{r} means the acceleration is opposite to the position vector — it points toward the center (since rāƒ—\vec{r} points from center to object).

The magnitude: ∣aāƒ—āˆ£=rω2=v2/r|\vec{a}| = r\omega^2 = v^2/r āœ“

Angular Velocity (ω\omega)

Angular velocity measures how fast the angle changes:

ω=ΔθΔt=2Ļ€T=2Ļ€f\omega = \frac{\Delta\theta}{\Delta t} = \frac{2\pi}{T} = 2\pi f

QuantitySymbolUnitsRelationship
Angular velocityω\omegarad/sω=2Ļ€f=2Ļ€/T\omega = 2\pi f = 2\pi/T
Linear speedvvm/sv=rωv = r\omega
Centripetal accelerationaca_cm/s²ac=rω2=vωa_c = r\omega^2 = v\omega

Converting Between Linear and Angular

v=rωac=rω2=v2rv = r\omega \quad \quad a_c = r\omega^2 = \frac{v^2}{r}

All Points on a Rigid Body

For a rotating solid object (like a wheel):

  • All points have the same ω\omega (same angular velocity)
  • Points farther from center have greater vv (since v=rωv = r\omega)
  • Points farther from center have greater aca_c (since ac=rω2a_c = r\omega^2)

Vector Description Questions šŸŽÆ

Angular Velocity Calculations 🧮

  1. A bicycle wheel has radius 0.35 m and rotates at 3 rev/s. What is the angular velocity ω\omega (in rad/s, round to 3 significant figures)?

  2. A point on the rim of the wheel in problem 1 has what linear speed (in m/s, round to 3 significant figures)?

  3. What is the centripetal acceleration of that point (in m/s², round to nearest whole number)?

Rotational Relationships šŸ”

Exit Quiz — Vectors in Circular Motion āœ…

Part 5: Vertical & Horizontal Circles

šŸŒ€ Non-Uniform Circular Motion Intro

Part 5 of 7 — Uniform Circular Motion

What happens when an object moves in a circle but its speed is changing? This is non-uniform circular motion, and it requires an additional component of acceleration beyond centripetal.

In this lesson you will learn:

  • The difference between uniform and non-uniform circular motion
  • Tangential acceleration vs. centripetal acceleration
  • How to find total acceleration in non-uniform circular motion
  • Real-world examples of non-uniform circular motion

Two Components of Acceleration

In non-uniform circular motion, acceleration has two perpendicular components:

1. Centripetal (Radial) Acceleration — aca_c

  • Direction: toward the center
  • Magnitude: ac=v2/ra_c = v^2/r
  • Changes the direction of velocity

2. Tangential Acceleration — ata_t

  • Direction: tangent to the circle (along velocity direction)
  • Magnitude: at=Ī”v/Ī”ta_t = \Delta v / \Delta t (rate of speed change)
  • Changes the speed (magnitude of velocity)

Total Acceleration

Since aāƒ—c\vec{a}_c and aāƒ—t\vec{a}_t are perpendicular:

atotal=ac2+at2a_{\text{total}} = \sqrt{a_c^2 + a_t^2}

The angle of total acceleration relative to the radius:

tan⁔ϕ=atac\tan\phi = \frac{a_t}{a_c}

Type of Motionaca_cata_t
Uniform circularv2/rv^2/r (constant)0
Non-uniform circular (speeding up)v2/rv^2/r (changing)Positive (along vāƒ—\vec{v})
Non-uniform circular (slowing down)v2/rv^2/r (changing)Negative (opposite vāƒ—\vec{v})

Real-World Examples

Car on a Circular On-Ramp

A car accelerating from 20 m/s to 30 m/s while going around a curve of radius 100 m:

  • Has centripetal acceleration: ac=v2/ra_c = v^2/r (increasing as vv increases)
  • Has tangential acceleration: at>0a_t > 0 (speeding up)
  • Total acceleration is not directed toward the center

Roller Coaster Loop

At different points of a vertical loop:

  • Speed changes due to gravity (not uniform)
  • ac=v2/ra_c = v^2/r changes because vv changes
  • Gravity provides both centripetal and tangential components depending on position

A Ball on a String in a Vertical Circle

  • At the top: gravity adds to centripetal acceleration, speed is minimum
  • At the bottom: gravity opposes centripetal acceleration, speed is maximum
  • At the sides: gravity is entirely tangential (changing speed)

Key AP Insight

On the AP exam, most circular motion problems assume uniform circular motion. Non-uniform circular motion appears mainly in:

  • Vertical circle problems
  • "Car speeds up around a curve" scenarios
  • Conceptual questions about acceleration components

Non-Uniform Circular Motion šŸŽÆ

Non-Uniform Circular Motion Calculations 🧮

  1. A car moves at 15 m/s around a curve of radius 50 m while accelerating at at=2a_t = 2 m/s². What is the centripetal acceleration (in m/s²)?

  2. For the same car, what is the magnitude of the total acceleration (in m/s², round to 3 significant figures)?

  3. A ball moves in a vertical circle of radius 2 m. At the side of the circle, its speed is 5 m/s and its tangential acceleration is g=10g = 10 m/s² (due to gravity). What is the total acceleration (in m/s², round to 3 significant figures)?

Uniform vs. Non-Uniform šŸ”

Exit Quiz — Non-Uniform Circular Motion āœ…

Part 6: Problem-Solving Workshop

šŸ”§ Problem-Solving Workshop

Part 6 of 7 — Uniform Circular Motion

Time to put it all together! In this workshop, we'll tackle a variety of circular motion problems — from simple calculations to multi-step AP-level scenarios.

In this lesson you will:

  • Apply v=2Ļ€r/Tv = 2\pi r/T and ac=v2/ra_c = v^2/r in context
  • Solve problems with multiple rotational quantities
  • Connect circular motion to real-world applications
  • Practice AP-level problem-solving strategies

Problem-Solving Strategy

Step 1: Identify the Circular Motion

  • What object is moving in a circle?
  • What is the radius?
  • What provides the centripetal force? (Preview of next topic!)

Step 2: Choose the Right Formula

GivenFindUse
vv and rraca_cac=v2/ra_c = v^2/r
TT and rrvvv=2Ļ€r/Tv = 2\pi r/T
TT and rraca_cac=4Ļ€2r/T2a_c = 4\pi^2 r/T^2
ff and rrvvv=2Ļ€rfv = 2\pi rf
ω\omega and rrvvv=rωv = r\omega

Step 3: Convert Units

  • RPM → Hz: divide by 60
  • Hz → rad/s: multiply by 2Ļ€2\pi
  • km/h → m/s: divide by 3.6
  • Diameter → radius: divide by 2

Step 4: Solve and Check

  • Does the answer have correct units?
  • Is the magnitude reasonable?

Warm-Up Calculations 🧮

  1. A 0.5 m radius wheel makes 120 RPM. What is the speed of a point on the rim (in m/s, round to 3 significant figures)?

  2. A satellite in low Earth orbit has period T=90T = 90 min and orbital radius r=6.6Ɨ106r = 6.6 \times 10^6 m. What is its orbital speed (in m/s, round to nearest 100)?

  3. Earth orbits the Sun at vā‰ˆ30,000v \approx 30{,}000 m/s in a roughly circular orbit of radius 1.5Ɨ10111.5 \times 10^{11} m. What is Earth's centripetal acceleration (in m/s², to 3 significant figures)?

Applied Problems šŸŽÆ

Challenge Problems 🧮

  1. A vinyl record (radius 15 cm) plays at 33.3 RPM. How much farther does a point on the outer edge travel in 1 minute compared to a point 5 cm from the center? (in meters, round to 3 significant figures)

  2. An amusement park ride spins riders in a circle of radius 8 m. If the maximum safe centripetal acceleration is 3g3g (g=10g = 10 m/s²), what is the maximum allowed speed (in m/s, round to 3 significant figures)?

  3. A wheel of radius 0.4 m accelerates from rest to 10 rad/s in 5 s. What is the tangential acceleration of a point on the rim (in m/s²)?

Quick Checks šŸ”

Exit Quiz — Problem-Solving Workshop āœ…

Part 7: Synthesis & AP Review

šŸŽ“ Synthesis & AP Review

Part 7 of 7 — Uniform Circular Motion

This final lesson brings together all circular motion concepts for AP exam preparation. We'll cover AP-style questions, common pitfalls, and exam strategies.

In this lesson you will:

  • Tackle AP-style multiple choice questions
  • Identify common exam mistakes
  • Connect circular motion to Newton's Laws (preview of centripetal force)
  • Review the complete circular motion toolkit

Your Circular Motion Toolkit

Essential Formulas

FormulaWhat It Gives
v=2Ļ€r/Tv = 2\pi r/TSpeed from radius and period
ac=v2/ra_c = v^2/rCentripetal acceleration from speed and radius
ac=4Ļ€2r/T2a_c = 4\pi^2 r/T^2Centripetal acceleration from radius and period
v=rωv = r\omegaSpeed from angular velocity
ac=rω2a_c = r\omega^2Centripetal acceleration from angular velocity
T=1/fT = 1/fPeriod-frequency relationship
ω=2Ļ€f\omega = 2\pi fAngular velocity from frequency

Key Concepts Checklist

āœ… Speed is constant; velocity is not (direction changes) āœ… Acceleration points toward the center — always āœ… vāƒ—āŠ„aāƒ—c\vec{v} \perp \vec{a}_c (velocity perpendicular to acceleration) āœ… No outward ("centrifugal") force in an inertial frame āœ… If string breaks, object goes straight (tangent), not outward āœ… Non-uniform: add tangential acceleration, use Pythagorean theorem

Common AP Mistakes

Mistake 1: Confusing acāˆra_c \propto r vs. acāˆ1/ra_c \propto 1/r

  • At constant speed: ac=v2/ra_c = v^2/r → bigger rr means less aca_c
  • At constant period: ac=4Ļ€2r/T2a_c = 4\pi^2r/T^2 → bigger rr means more aca_c
  • Always check what's being held constant!

Mistake 2: Saying acceleration = 0 because speed is constant

  • Acceleration is zero only when velocity (including direction) is constant
  • Circular motion always has centripetal acceleration

Mistake 3: Using diameter instead of radius

  • Common error: forgetting to divide by 2
  • Always double-check: is the problem giving diameter or radius?

Mistake 4: Forgetting unit conversions

  • RPM → Hz: divide by 60
  • Minutes → seconds: multiply by 60
  • km → m: multiply by 1000

AP-Style Multiple Choice šŸŽÆ

AP-Style Calculations 🧮

  1. Mars orbits the Sun at r=2.28Ɨ1011r = 2.28 \times 10^{11} m with period T=687T = 687 days. What is Mars's orbital speed (in m/s, round to nearest 100)?

  2. A coin sits on a turntable 12 cm from the center, spinning at 78 RPM. What is the centripetal acceleration of the coin (in m/s², round to 3 significant figures)?

  3. A space station creates artificial gravity by spinning. If the station has radius 50 m, what angular velocity ω\omega is needed to produce g=10g = 10 m/s² at the rim (in rad/s, round to 3 significant figures)?

Conceptual Review šŸ”

Final Exit Quiz — Uniform Circular Motion āœ