01Circular Motion and Rotation
Analyze centripetal acceleration and force, angular velocity, period, and frequency for objects moving in circular paths.
Circular motion appears everywhere — from car turns to planetary orbits. Understanding centripetal force is essential before studying gravity and orbital mechanics.
Why does a car need a larger centripetal force to round a tight curve at high speed — and what happens when that force isn't enough?
Lesson Overview
When an object moves in a circle at constant speed, its velocity direction changes continuously — meaning it accelerates even though its speed stays the same. This centripetal acceleration always points toward the center of the circle and requires a net inward force called the centripetal force. We describe circular motion using period (T), frequency (f), and angular velocity (ω), and connect them to tangential speed through v = ωr. Extending these ideas to rotation, we introduce rotational kinematics (θ, ω, α), torque (τ = rF sin θ), moment of inertia (I), and angular momentum (L = Iω). The conservation of angular momentum explains everything from spinning figure skaters to the formation of solar systems.
Key Equations
Worked Examples
A car rounds a curve of radius r = 50 m at a speed of v = 20 m/s. (a) Find the centripetal acceleration. (b) Find the centripetal force if the car's mass is m = 1200 kg.
Part (a): Use aₒ = v²/r
aₒ = (20)² / 50 = 400 / 50 = 8 m/s²
Part (b): Use Fₒ = maₒ
Fₒ = 1200 × 8 = 9600 N
This force is provided by friction between the tires and the road.
An object completes one full revolution in T = 4 s. (a) Find the angular velocity ω. (b) Find the tangential speed if the radius is r = 3 m.
Part (a): ω = 2π / T
ω = 2π / 4 = π/2 ≈ 1.571 rad/s
Part (b): v = ωr
v = (π/2) × 3 = 3π/2 ≈ 4.71 m/s
A satellite orbits Earth at a radius of r = 7 × 10⁶ m with an orbital speed of v = 7546 m/s. Find the orbital period T.
For circular orbit: circumference = 2πr, so T = 2πr / v
T = 2π × (7 × 10⁶) / 7546
T = (2 × 3.14159 × 7 000 000) / 7546
T = 43 982 297 / 7546 ≈ 5828 s
Convert: 5828 s ÷ 60 ≈ 97.1 minutes
A wrench applies a force of F = 40 N at a distance r = 0.3 m from the pivot. The force is applied at θ = 90° to the lever arm. Find the torque.
Use τ = rF sinθ
τ = 0.3 × 40 × sin(90°)
sin(90°) = 1
τ = 0.3 × 40 × 1 = 12 N·m
Maximum torque occurs when force is perpendicular to the lever arm (θ = 90°).
A figure skater spins with moment of inertia I₁ = 4 kg·m² at ω₁ = 2 rad/s. She pulls her arms in, reducing her moment of inertia to I₂ = 1 kg·m². Find her new angular velocity ω₂.
Apply conservation of angular momentum: L₁ = L₂
I₁ω₁ = I₂ω₂
4 × 2 = 1 × ω₂
8 = ω₂
Pulling arms in decreases I, so ω must increase to conserve L.
Guided Problems
A ball on a string moves in a horizontal circle of radius r = 0.8 m at v = 4 m/s. Find the centripetal acceleration.
Hint: Use aₒ = v²/r. Square the speed first, then divide by the radius.
A merry-go-round completes one revolution every T = 6 s. Find its angular velocity ω and the tangential speed of a child sitting r = 2 m from the center.
Hint: First find ω = 2π/T, then use v = ωr to get tangential speed.
A bolt is tightened by applying F = 25 N at r = 0.2 m from the center. The force makes an angle of θ = 60° with the wrench handle. Find the torque.
Hint: Use τ = rF sinθ. Be careful to use the angle between the force vector and the lever arm direction.
A spinning top has moment of inertia I = 0.005 kg·m² and angular velocity ω = 50 rad/s. Find its angular momentum L.
Hint: Angular momentum is simply L = Iω. Multiply moment of inertia by angular velocity.
A diver tucks into a ball, reducing her moment of inertia from I₁ = 12 kg·m² to I₂ = 3 kg·m². If her initial angular velocity is ω₁ = 1.5 rad/s, find her final angular velocity.
Hint: Use conservation of angular momentum: I₁ω₁ = I₂ω₂. Solve for ω₂.
Key Vocabulary
Uniform Circular Motion
Motion in a circular path at constant speed. Although speed is constant, velocity changes direction continuously, so the object is always accelerating.
Example: A satellite in a circular orbit moves at constant speed but constantly changes direction.
Centripetal Acceleration
The acceleration directed toward the center of a circular path. Its magnitude is aₒ = v²/r = ω²r. It changes the direction of velocity, not its magnitude.
Example: A car rounding a curve at 20 m/s on a 50 m radius has aₒ = 8 m/s² pointing toward the center of the curve.
Centripetal Force
The net inward force required to keep an object moving in a circle: Fₒ = mv²/r. It is not a new type of force — it is the name for whatever force (friction, tension, gravity) acts centripetally.
Example: Friction from the road provides the centripetal force that keeps a car on a curved road.
Period (T)
The time for one complete revolution or cycle, measured in seconds. Related to frequency by T = 1/f.
Example: A satellite with T = 5828 s takes about 97 minutes to complete one orbit.
Frequency (f)
The number of complete revolutions per second, measured in hertz (Hz). f = 1/T.
Example: A merry-go-round completing one revolution every 4 s has f = 0.25 Hz.
Angular Velocity (ω)
The rate of change of angular position, measured in radians per second (rad/s). ω = 2π/T = 2πf. Relates to tangential speed by v = ωr.
Example: An object with T = 4 s has ω = π/2 ≈ 1.57 rad/s.
Torque (τ)
The rotational equivalent of force — a measure of how effectively a force causes rotation about a pivot. τ = rF sinθ, where θ is the angle between the force and the lever arm.
Example: Applying 40 N perpendicularly at 0.3 m from a pivot produces τ = 12 N·m.
Angular Momentum (L)
The rotational equivalent of linear momentum: L = Iω, where I is the moment of inertia. In the absence of external torques, angular momentum is conserved.
Example: A figure skater pulling in her arms decreases I, so ω increases to keep L = Iω constant.
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
A car travels at v = 15 m/s around a curve of radius r = 45 m. What is the centripetal acceleration?
An object completes one revolution in T = 2 s. What is its angular velocity?
A 2 kg object moves in a circle of radius 4 m at v = 6 m/s. What centripetal force is required?
A force of 30 N is applied perpendicular to a wrench at 0.5 m from the pivot. What is the torque?
A skater with I₁ = 6 kg·m² spinning at ω₁ = 3 rad/s pulls in her arms to I₂ = 2 kg·m². What is her new angular velocity?
Independent Practice
A car travels around a circular track of radius 50 m at a constant speed of 20 m/s. Calculate (a) the centripetal acceleration and (b) the centripetal force if the car has a mass of 1,200 kg.
A wheel rotates at 300 RPM. Convert this to (a) rad/s and (b) find the linear speed of a point 0.40 m from the center.
A 2.0 kg ball on a 1.5 m string is swung in a horizontal circle at 4.0 m/s. Find the tension in the string.
A solid disk (I = ½mr²) of mass 3.0 kg and radius 0.20 m is accelerated from rest to 10 rad/s in 5.0 s. Find the net torque applied.
★ A figure skater spinning at 2.0 rev/s pulls her arms in, reducing her moment of inertia from 4.0 kg·m² to 1.0 kg·m². (a) Find her new angular velocity. (b) Calculate the initial and final rotational kinetic energies. (c) Explain the source of the energy increase.
ChallengeCommon Mistakes
Calling centripetal force a separate 'extra' force on the free-body diagram
Centripetal force is the NET inward force provided by real forces (tension, gravity, normal force) — never draw it as a separate arrow on an FBD
Confusing angular velocity ω (rad/s) with linear speed v (m/s)
They are related by v = ωr. Angular velocity describes how fast the angle changes; linear speed depends on the radius
Forgetting to convert RPM to rad/s before using rotational equations
Convert: ω (rad/s) = RPM × 2π/60. Always use radians (not degrees or revolutions) in rotational formulas
Applying τ = rF without accounting for the angle between r and F
Torque is τ = rF sinθ where θ is the angle between the position vector and the force. Maximum torque occurs at θ = 90°
Math Tips
Centripetal acceleration: a_c = v²/r = ω²r. The net inward force equals ma_c = mv²/r. Identify which real force(s) provide this inward push
Rotational–linear analogies: θ↔x, ω↔v, α↔a, I↔m, τ↔F. The rotational kinematic equations mirror the linear ones exactly
Moment of inertia depends on both mass AND how it is distributed: I = mr² for a point mass, ½mr² for a solid disk, ⅔mr² for a hollow sphere
Angular momentum L = Iω is conserved when net torque = 0. A spinning skater pulling in their arms decreases r, so I decreases and ω must increase