Circular Motion

AP Physics C: Mechanics· Unit 2: Newton's Laws of Motion, Topic 2.E Circular Motion· 12 min read

1. Uniform Circular Motion Core Principles★★☆☆☆⏱ 3 min

📘 Definition

Uniform Circular Motion

UCM

Motion along a fixed-radius circular path where tangential speed is constant, so only the direction of velocity changes over time

Example:

A car driving around a perfectly circular race track at a steady 15 m/s

a_c = \frac{v_t^2}{r} = \omega^2 r

📐 Worked Example

A 0.4 kg remote controlled car travels at a constant 2.2 m/s on a circular track of radius 1.2 m. Calculate the magnitude of its centripetal acceleration.

1
Identify given values: tangential speed $v_{t} = 2.2$ m/s, radius $r = 1.2$ m
2
Substitute directly into the centripetal acceleration formula
$a_c = \frac{(2.2)^2}{1.2}$
3
Calculate final value, rounding to 2 significant figures
$a_c = 4.0 \text{ m/s}^2$

✓ Quick check

Test your understanding of UCM basics:

What is the direction of centripetal acceleration for an object in UCM?
- Tangent to the path
- Pointing towards the center of the circle
- Pointing away from the center of the circle
- Opposite to the direction of motion
Reveal answer
Pointing towards the center of the circle —
Centripetal acceleration is always radial and points inward to change the direction of velocity.

2. Derivation of Centripetal Acceleration★★★★☆⏱ 4 min

Exam tip:

College Board explicitly awards points for full vector derivation of centripetal acceleration on FRQs, so memorize every step.

3. Force Analysis for Circular Paths★★★☆☆⏱ 4 min

📐 Worked Example

A 0.3 kg mass attached to a 0.8 m string is swung in a perfectly horizontal circle at 3 m/s. Calculate the tension in the string.

1
Note that tension is the only force providing the full centripetal net force
2
Set tension equal to mass multiplied by centripetal acceleration
$F_T = m \frac{v^2}{r}$
3
Substitute all given values
$F_T = 0.3 \times \frac{(3)^2}{0.8} = 3.4 \text{ N}$

4. Frictionless Banked Curve Scenarios★★★★☆⏱ 4 min

📐 Worked Example

A highway curve of radius 70 m is designed for traffic moving at 25 m/s with no friction. Calculate the required bank angle.

1
Use the derived formula for frictionless banked curves
$\tan\theta = \frac{v^2}{rg}$
2
Substitute values $v = 25$ m/s, $r = 70$ m, $g = 9.8$ m/s²
$\tan\theta = \frac{625}{70 \times 9.8} \approx 0.91$
3
Take inverse tangent to find the angle
$\theta \approx 42^\circ$

5. Common Pitfalls

Wrong move:

Adding 'centripetal force' as a separate independent force in free body diagrams

Why:

Centripetal force is the net radial force, not a new force distinct from tension, gravity, or friction

Correct move:

Sum all existing radial forces and set the total equal to $m v^{2} / r$

Wrong move:

Ignoring tangential acceleration for non-uniform circular motion

Why:

Students only use centripetal acceleration when speed is changing, leading to incorrect total acceleration values

Correct move:

Calculate total acceleration as the vector sum of tangential and centripetal components

Wrong move:

Assume $a_{c} = g$ at the top of every vertical circular loop

Why:

This only holds true at the minimum possible speed where normal force equals zero

Correct move:

Include normal force in your net force calculation unless minimum speed is explicitly requested

Wrong move:

Using degrees per second for angular velocity in $a_{c} = ω^{2} r$

Why:

The formula only returns correct values when angular velocity is measured in radians per second

Correct move:

Convert all angular quantities to rad/s before plugging into kinematic equations

Wrong move:

Pointing friction away from the center on banked curve problems

Why:

Friction points inward to prevent the car from sliding up the bank at high speeds

Correct move:

Resolve friction parallel to the bank, with its radial component pointing towards the path center

6. Quick Reference Cheatsheet

Scenario	Net Radial Force	Key Formula	Common Constraint
Uniform Horizontal Circle	Tension / Static Friction	$F_{n e t} = m v^{2} / r$	Speed is constant
Top of Vertical Loop	Gravity + Normal Force	$m g + N = m v^{2} / r$	$N \geq 0$ for no fall
Bottom of Vertical Loop	Normal Force - Gravity	$N - m g = m v^{2} / r$	$N > m g$ always
Frictionless Banked Curve	Horizontal Normal Component	$tan θ = v^{2} / (r g)$	No sideways friction

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · Multiple Choice 1
Banked curve friction scenario
2023 · Free Response 2
Roller coaster vertical loop analysis
2021 · Free Response 2
Rotating platform tension force problem

What's Next

Mastering circular motion is a critical bridge between Newtonian linear kinematics and the upcoming unit on rotational motion, where you will extend these concepts to angular momentum, torque, and rigid body rotation. You will frequently combine circular motion principles with the work-energy theorem to solve multi-step FRQ problems that appear on nearly every AP Physics C exam. This foundation will also help you analyze gravitational orbital motion, a high-yield topic that makes regular appearances on both multiple choice and free response sections. Practice applying these skills to the linked sub-topics to build fluency before your full unit assessment.