Acceleration in Uniform Circular Motion

AP Physics 1· AP Physics 1 CED — Circular Motion and Gravitation· 14 min read

1. What is Acceleration in Uniform Circular Motion?★★☆☆☆⏱ 3 min

Uniform circular motion (UCM) is motion of an object along a circular path at constant tangential speed. Because acceleration depends on change in the velocity vector (not just the scalar speed), changing the direction of velocity counts as a change in velocity, so UCM always has non-zero acceleration.

📘 Definition

Centripetal Acceleration

$a_{c}$

The acceleration of an object in circular motion that points toward the center of the circular path, responsible for changing the direction of velocity. For UCM, all acceleration is centripetal.

Example:

A car turning at constant speed has non-zero centripetal acceleration, even though its speed is constant.

This topic is core to AP Physics 1 Unit 3, which counts for 6-8% of your total exam score. Centripetal acceleration is almost always a foundational step for solving centripetal force problems, the most heavily tested skill in the unit. A common early misconception is that acceleration is zero in UCM because speed is constant; this ignores the vector nature of velocity.

2. Direction of Centripetal Acceleration★★☆☆☆⏱ 3 min

A defining property of centripetal acceleration in UCM is that it always points directly toward the center of the circular path, and is always perpendicular to the tangential velocity vector at every point.

To confirm this direction, consider two velocity vectors $v_{1}$ and $v_{2}$ for an object at two nearby points on the circle, with the same magnitude (constant speed) but different directions. The change in velocity $Δ v = v_{2} - v_{1}$ approaches the center direction as $Δ t$ approaches zero. Since acceleration is $a = lim_{Δ t \to 0} Δ v /Δ t$ , acceleration inherits this center-pointing direction.

Because centripetal acceleration is always perpendicular to velocity, it only changes the direction of velocity, not its magnitude, which is why speed stays constant in UCM.

📐 Worked Example

A jogger runs clockwise around a circular track at constant speed. When the jogger is at the topmost point of the track, what is the direction of their centripetal acceleration?

1
Recall the rule for UCM: centripetal acceleration always points toward the center of the circular path.
2
Map the jogger's position: when at the topmost point of the track, the center of the track is directly below the jogger's current position.
3
Confirm tangential acceleration is zero in UCM, so there is no acceleration component parallel or antiparallel to the jogger's direction of motion.
4
Conclusion: The jogger's centripetal acceleration points directly downward, toward the center of the track.

Exam tip:

Always draw the circle and mark the center before answering direction questions — never assume direction based only on the direction the object is moving.

3. Magnitude of Centripetal Acceleration★★★☆☆⏱ 4 min

🔬 Derivation

Goal:

Derive the magnitude of centripetal acceleration

Starting from:

Similar triangles for position and velocity vectors for small $Δ t$ in UCM

1
For a small time interval $Δ t$ , the distance traveled along the circle is $v Δ t$ . The triangle formed by the two position vectors is similar to the triangle formed by the two velocity vectors.
2
This gives the proportionality:
3
$\frac{\Delta v}{v} = \frac{v\Delta t}{r}$
4
Rearranging for $a_{c} = Δ v /Δ t$ gives the core formula.

Result:

The magnitude of centripetal acceleration is given by $a_{c} = v^{2} / r$ .

Using the relationship between tangential speed and angular speed $v = ω r$ , we can substitute to get an equivalent second formula:

a_c = \omega^2 r

Intuitively, centripetal acceleration depends on the square of speed: doubling your speed around a turn quadruples your centripetal acceleration, which explains why high-speed turns require much more force. It is inversely proportional to radius: a tighter turn (smaller $r$ ) gives larger acceleration, matching everyday experience.

📐 Worked Example

A go-kart drives around a circular turn of radius 20 m at a constant speed of 10 m/s. What is the magnitude of the go-kart's centripetal acceleration?

1
Identify given values: $v = 10 m/s$ , $r = 20 m$ , we need to find $a_{c}$ .
2
Select the appropriate formula: we have tangential speed and radius, so use $a_{c} = v^{2} / r$ .
3
Substitute values:
4
$a_c = \frac{(10 \text{ m/s})^2}{20 \text{ m}} = \frac{100}{20} = 5$
5
Add units, confirm direction: $a_{c} = 5 m/s^{2}$ , directed toward the center of the turn.

Exam tip:

If the problem gives you revolutions per minute or second instead of tangential speed, first calculate angular speed $ω = 2 π f$ (where $f$ is frequency in rev/s) then use $a_{c} = ω^{2} r$ to avoid extra calculation steps.

4. Centripetal vs Tangential Acceleration★★★☆☆⏱ 4 min

All circular motion can have two perpendicular acceleration components, and AP Physics 1 questions frequently test the distinction between them:

Centripetal (radial) acceleration: Always present for any circular motion (uniform or non-uniform), responsible for changing the direction of velocity, points toward the center.
Tangential acceleration: Parallel or antiparallel to tangential velocity, responsible for changing the magnitude (speed) of velocity.

In uniform circular motion, speed is constant, so tangential acceleration $a_{t} = 0$ , and all acceleration is centripetal. In non-uniform circular motion (speed changes along the path), both components are non-zero and perpendicular. The magnitude of total acceleration is calculated via the Pythagorean theorem:

a_{total} = \sqrt{a_c^2 + a_t^2}

Crucially, $a_{c} = v^{2} / r$ still holds for non-uniform circular motion at any instant, as long as you use the instantaneous speed $v$ at that moment.

📐 Worked Example

A car speeds up as it exits a circular on-ramp of radius 50 m. At the instant the car's speed is 15 m/s, it has a tangential acceleration of 3 m/s². What is the magnitude of the car's total acceleration at this instant?

1
Calculate the instantaneous centripetal acceleration, which still depends on instantaneous speed:
2
$a_c = v^2/r = (15^2)/50 = 225/50 = 4.5 \text{ m/s}^2$
3
Confirm $a_{c}$ and $a_{t}$ are perpendicular, so we can use the Pythagorean theorem for total acceleration.
4
Substitute values:
5
$a_{total} = \sqrt{(4.5)^2 + (3)^2} = \sqrt{20.25 + 9} = \sqrt{29.25} \approx 5.4 \text{ m/s}^2$
6
If this were uniform circular motion, $a_{t} = 0$ , so total acceleration would just equal $a_{c} = 4.5 m/s^{2}$ , which matches our definition of UCM.

Exam tip:

If a question explicitly says "uniform circular motion", you can immediately set $a_{t} = 0$ without any further calculation — this is a common time-saver on MCQs.

5. AP-Style Practice Problems★★★★☆⏱ 4 min

📐 Worked Example

An object moves in uniform circular motion of radius $r$ with centripetal acceleration $a_{0}$ . If the speed of the object is doubled and the radius is doubled, what is the new centripetal acceleration in terms of $a_{0}$ ?

A) $a_{0} /2$ B) $a_{0}$ C) $2 a_{0}$ D) $4 a_{0}$

1
Start with the original formula for centripetal acceleration: $a_{0} = v^{2} / r$ .
2
The new speed is $v^{'} = 2 v$ and the new radius is $r^{'} = 2 r$ . Substitute into the formula:
3
$a_{new} = \frac{(v')^2}{r'} = \frac{(2v)^2}{2r} = \frac{4v^2}{2r} = 2 \frac{v^2}{r} = 2a_0$
4
Common wrong answers come from forgetting to square the speed (leading to option A) or incorrectly squaring the radius (leading to option D). The correct answer is C.

📐 Worked Example

A student swings a small toy car tied to a 0.8 m long string in a uniform vertical circle at constant speed. The car completes 3 full revolutions every 4 seconds.

(a) Calculate the period of revolution, then calculate the tangential speed of the car. (b) Calculate the magnitude of the centripetal acceleration of the car. (c) State whether tangential acceleration is zero or non-zero at any point in this motion, and explain your reasoning.

1
(a) The period $T$ is time per revolution:
2
$T = 4 \text{ s}/3 \text{ rev} \approx 1.33 \text{ s per revolution}$
3
The circumference of the circle is $C = 2 π r = 2 π (0.8 m) \approx 5.03 m$ . Tangential speed is $v = C / T = 5.03/1.33 \approx 3.8 m/s$ .
4
(b) Use the centripetal acceleration formula:
5
$a_c = v^2/r = (3.8)^2 / 0.8 = 14.44 / 0.8 \approx 18 \text{ m/s}^2$
6
(c) The problem states the motion is uniform circular motion, meaning speed is constant. Tangential acceleration is the component of acceleration that changes speed, so tangential acceleration is zero at all points in this motion.

📐 Worked Example

The Moon orbits the Earth in an approximately uniform circular path with a radius of $3.84 \times 1 0^{8} m$ , and completes one full orbit every 27.3 days. Calculate the centripetal acceleration of the Moon as it orbits the Earth.

1
First, convert the period $T$ to SI units (seconds):
2
$T = 27.3 \text{ days} \times 24 \text{ h/day} \times 3600 \text{ s/h} \approx 2.36 \times 10^6 \text{ s}$
3
Calculate tangential speed:
4
$v = 2\pi r / T = 2\pi (3.84 \times 10^8) / (2.36 \times 10^6) \approx 1020 \text{ m/s}$
5
Calculate centripetal acceleration:
6
$a_c = v^2 / r = (1020)^2 / (3.84 \times 10^8) \approx 0.0027 \text{ m/s}^2$
7
This very small acceleration matches the expected gravitational acceleration of the Moon due to Earth, consistent with gravitational theory.

6. Common Pitfalls

Wrong move:

Claiming acceleration is zero in uniform circular motion because speed is constant.

Why:

Students confuse scalar speed with vector velocity, forgetting acceleration depends on change in velocity, not just speed.

Correct move:

Always check if direction is changing before concluding acceleration is zero — any curved path has non-zero centripetal acceleration, even if speed is constant.

Wrong move:

Stating centripetal acceleration points outward from the center of the circle (the "centrifugal" direction).

Why:

Students incorrectly use the non-inertial frame of the moving object instead of the inertial frame required for AP Physics 1.

Correct move:

Always solve circular motion problems from a stationary inertial frame, where acceleration is always center-seeking for UCM.

Wrong move:

Calculating average acceleration over a full UCM revolution, getting zero, and claiming this means there is no acceleration.

Why:

Students confuse average acceleration over a full cycle with the instantaneous centripetal acceleration the question almost always asks for.

Correct move:

Unless the question explicitly asks for average acceleration over a full revolution, you need to calculate instantaneous centripetal acceleration $a_{c} = v^{2} / r$ .

Wrong move:

Abandoning $a_{c} = v^{2} / r$ for non-uniform circular motion because speed is changing.

Why:

Students associate the formula only with UCM, so they incorrectly assume it does not apply when speed varies.

Correct move:

$a_{c} = v^{2} / r$ always gives instantaneous centripetal acceleration for any circular motion, uniform or non-uniform, as long as $v$ is the instantaneous speed.

Wrong move:

Leaving units mismatched (e.g., radius in kilometers, speed in m/s) when calculating $a_{c}$ .

Why:

Students copy given values without converting to consistent units before plugging into the formula.

Correct move:

Convert all quantities to SI units (meters for radius, m/s for speed, rad/s for angular speed) before substituting into the acceleration formula.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Centripetal acceleration (tangential speed form)	$a_{c} = \frac{v ^{2}}{r}$	Applies to UCM and instantaneous acceleration in any circular motion. Always points toward the center.
Centripetal acceleration (angular speed form)	$a_{c} = ω^{2} r$	$ω$ must be in radians per second. Use when given frequency/revolutions per unit time.
Period of revolution	$T = \frac{2 π r}{v} = \frac{2 π}{ω}$	Time to complete one full revolution, always positive.
Tangential acceleration	$a_{t} = \frac{Δ v}{Δ t}$	Parallel to tangential velocity, changes speed. $a_{t} = 0$ for all uniform circular motion.
Total acceleration magnitude	$a_{t o t a l} = a_{c}^{2} + a_{t}^{2}$	Works because $a_{c}$ and $a_{t}$ are always perpendicular components.
Direction of velocity	Always tangential to path	Perpendicular to centripetal acceleration at all points.
Average acceleration over full UCM revolution	$a_{a v g} = 0$	Only true when velocity returns to original vector. Not equal to instantaneous centripetal acceleration.

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.

2023 · MCQ
Centripetal acceleration proportionality
2022 · FRQ
Total acceleration in non-uniform motion

Going deeper

SyllabusOfficial AP Physics 1 CEDUnit 3 content overview

What's Next

This module lays the foundational kinematic framework for all circular motion problems across the rest of AP Physics 1 Unit 3. Immediately, you will apply what you learned about centripetal acceleration to connect to centripetal force via Newton’s second law, the most heavily tested skill in this unit. Without correctly identifying the magnitude and direction of centripetal acceleration, you cannot correctly set up force equations for circular motion, leading to avoidable errors on both MCQ and FRQ sections. This topic is also a prerequisite for gravitational orbit problems later in this unit, where you will use centripetal acceleration to relate gravitational force to orbital speed and period.