Force Analysis for Circular Motion

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

1. Core Framework for Circular Motion Force Analysis★★☆☆☆⏱ 3 min

Force analysis for circular motion is the application of Newton's laws to objects moving along a circular (or nearly circular) path, and it is one of the highest-frequency tested topics in Unit 3, counting for 6–8% of the total AP Physics 1 exam score. The core convention is that we align one coordinate axis along the centripetal (radial, center-pointing) direction, and the other tangent to the circle, rather than defaulting to horizontal/vertical axes used for linear motion.

2. The Centripetal Net Force Rule★★☆☆☆⏱ 3 min

For any uniform circular motion (constant speed along a circular path), acceleration is always directed toward the center of the circle, with magnitude:

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

where $v$ = tangential speed, $r$ = radius of the circular path, and $ω$ = angular speed. Applying Newton's second law ( $F_{n e t} = ma$ ), this means the net force on the object must also point toward the center, with magnitude:

F_{net,c} = m \frac{v^2}{r} = m \omega^2 r

The most important point to remember is that centripetal force is not a new, separate fundamental force like gravity or tension. It is simply the net force resulting from the sum of real forces acting on the object along the radial direction. The step-by-step method for force analysis is:

Draw a free-body diagram of only real forces acting on the object
Align one axis with the radial (center-pointing) direction
Resolve all forces into radial and tangential components
Sum radial components (positive toward center) and set equal to $m v^{2} / r$

📐 Worked Example

A 0.5 kg toy car is tied to a central post with a 2 m long string, moving in a horizontal circle at constant speed 4 m/s. Assuming the string remains horizontal, what is the tension in the string?

1
Identify all real forces: weight (downward), normal force (upward from the ground), tension (horizontal toward the post/center of the circle). Weight and normal cancel vertically, so only tension contributes to radial net force.
2
Align the radial axis to point toward the center of the circle, so tension is positive.
3
Apply Newton's second law:
4
$F_{net,c} = T = \frac{mv^2}{r}$
5
Substitute values:
6
$T = \frac{(0.5\ \text{kg})(4\ \text{m/s})^2}{2\ \text{m}} = 4\ \text{N}$
7
Check: Tension is the only real radial force, so it equals the required centripetal net force, which is consistent.

Exam tip:

Never draw "centripetal force" as a separate force on your free-body diagram. AP Physics 1 FRQ rubrics explicitly award zero points for force diagrams that include this incorrect extra force.

3. Vertical Circular Motion Analysis★★★☆☆⏱ 4 min

Vertical circular motion is typically non-uniform, because gravity changes the object's speed as it moves up and down the circle. However, at the two extreme points (the very top and very bottom of the circle), acceleration is still entirely radial, so force analysis works exactly the same as uniform circular motion at these points. The key difference from horizontal circular motion is that the direction of the center (and thus positive radial direction) changes between the top and bottom of the circle.

📐 Worked Example

A 70 kg student rides a roller coaster through a vertical circular loop of radius 12 m. At the top of the loop, the coaster moves at 15 m/s. What is the normal force the seat exerts on the student at the top of the loop?

1
Draw the free-body diagram: two real forces act on the student, both pointing downward at the top of the loop: gravity $m g$ (downward) and normal force $N$ (downward, because the seat is above the student at the top of the loop).
2
Align positive radial direction downward (toward the center of the loop).
3
Sum radial forces:
4
$F_{net,c} = N + mg = \frac{mv^2}{r}$
5
Rearrange to solve for $N$ :
6
$N = m\left(\frac{v^2}{r} - g\right)$
7
Substitute values:
8
$N = 70\ \text{kg}\left(\frac{225\ \text{m}^2/\text{s}^2}{12\ \text{m}} - 9.8\ \text{m/s}^2\right) ≈ 630\ \text{N}$

Exam tip:

If you are asked for the minimum speed to keep a string taut (or a roller coaster on the track) at the top of a vertical circle, remember tension/normal force equals zero at minimum speed, so $v_{min} = g r$ . Memorize this to save time on MCQs.

4. Unbanked and Banked Horizontal Curve Analysis★★★☆☆⏱ 4 min

Horizontal circular motion for vehicles turning on curved roads is another common AP exam topic, split into unbanked (flat) and banked (tilted) curves. For unbanked (flat) curves, the entire centripetal force is provided by static friction between the vehicle's tires and the road, because friction acts parallel to the road toward the center of the curve. For banked curves, the road is tilted at an angle $θ$ from horizontal, so the horizontal component of the normal force from the road provides some or all of the required centripetal force, reducing reliance on friction. For the ideal case (no friction needed to navigate the curve at speed $v$ ), the relationship is:

\tan\theta = \frac{v^2}{rg}

📐 Worked Example

A highway designer wants to bank a curve of radius 50 m so that a car moving at 20 m/s can navigate the curve without any friction assistance. What angle $θ$ should the curve be banked at?

1
Identify real forces: only gravity (downward) and normal force (perpendicular to the road surface, no friction).
2
Align radial direction horizontally toward the center of the curve, vertical direction upward.
3
Resolve normal force $N$ into components, with vertical component balancing gravity and horizontal component providing centripetal force:
4
$N\cos\theta = mg \\ N\sin\theta = \frac{mv^2}{r}$
5
Divide the two equations to eliminate $N$ and $m$ :
6
$\tan\theta = \frac{v^2}{rg}$
7
Substitute values:
8
$\tan\theta = \frac{(20\ \text{m/s})^2}{(50\ \text{m})(9.8\ \text{m/s}^2)} ≈ 0.816, \text{ so } \theta ≈ 39^\circ$

Exam tip:

For banked curve problems, the mass of the car always cancels out. If your final answer still includes $m$ , you have made an algebra error—go back and check your division step.

5. AP-Style Concept Check Practice★★★★☆⏱ 4 min

✓ Quick check

Test your understanding with these practice problems:

A 1000 kg car drives at constant speed over the top of a circular hill with radius 20 m. When the car is at the top of the hill, which of the following correctly compares the magnitude of the normal force from the road on the car $N$ to the weight of the car $m g$ ?
- A) $N > m g$
- B) $N = m g$
- C) $N < m g$
- D) $N = 0$ for any non-zero speed
Reveal answer
2 —
At the top of the hill, the center of the circular path is below the car, so positive radial direction is downward. Summing radial forces gives $m g - N = \frac{m v ^{2}}{r}$ , so $N = m g - \frac{m v ^{2}}{r}$ . For any non-zero speed, $N < m g$ , so this option is correct.
A student swings a 0.25 kg stone in a vertical circle on a 1.0 m long massless string. The tension at the bottom of the circle is measured to be 15 N. What is the approximate speed of the stone at the bottom?
- A) 2.3 m/s
- B) 5.2 m/s
- C) 7.1 m/s
- D) 15 m/s
Reveal answer
2 —
At the bottom, positive radial direction is upward, so $T - m g = \frac{m v ^{2}}{r}$ . Solving for $v$ gives $v = \frac{r ( T - m g )}{m} \approx 7.1$ m/s.

6. Common Pitfalls

Wrong move:

Drawing "centripetal force" as a separate force on a free-body diagram, in addition to real forces like tension and gravity.

Why:

Students mistake the label "centripetal force" for a fundamental interaction force instead of a net force.

Correct move:

Only draw real forces (tension, gravity, friction, normal) on FBDs, then sum radial components to get the centripetal net force.

Wrong move:

Keeping positive direction upward at the top of a vertical circle, leading to $N - m g = \frac{m v ^{2}}{r}$ instead of $N + m g = \frac{m v ^{2}}{r}$ .

Why:

Students default to the standard horizontal/vertical coordinate system for linear motion, instead of aligning radial positive toward the center.

Correct move:

Always draw the circle, mark the center, and set positive radial direction to point directly at the center before writing Newton's second law.

Wrong move:

Including centrifugal (outward-pointing) force in force analysis for inertial frames of reference, which AP Physics 1 always uses.

Why:

Students feel an outward push when turning in a car, so they assume a real outward force exists.

Correct move:

The outward sensation is due to inertia (your body wants to move straight, so the car pushes inward on you), no outward force exists in an inertial frame. Never include centrifugal force in your analysis.

Wrong move:

Claiming normal force provides centripetal force for an unbanked flat curve.

Why:

Students confuse normal force direction with the required radial direction.

Correct move:

For flat unbanked curves, static friction (parallel to the road, pointing toward the center) is the centripetal force; normal force balances gravity vertically.

Wrong move:

Flipping $sin θ$ and $cos θ$ in banked curve problems, leading to $cot θ = \frac{v ^{2}}{r g}$ instead of $tan θ$ .

Why:

Students misidentify how the bank angle relates to the normal force's components.

Correct move:

Test your trig with the edge case: if $θ = 0$ (flat road), $tan θ = 0$ , which means $v = 0$ for no friction, which is logical. If your formula gives the wrong edge case, flip your trig functions.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Centripetal acceleration	$a_{c} = \frac{v ^{2}}{r} = ω^{2} r$	Always points toward the center, applies to uniform circular motion
Centripetal net force	$F_{n e t, c} = m a_{c} = m \frac{v ^{2}}{r}$	Net force, not a separate real force; positive value toward center
Vertical circle (top)	$N + m g = \frac{m v ^{2}}{r}$	Positive direction downward (toward center); $N$ = normal/tension
Vertical circle (bottom)	$T - m g = \frac{m v ^{2}}{r}$	Positive direction upward (toward center); $T$ = tension/normal
Minimum speed at top (string)	$v_{min} = g r$	Tension = 0 at minimum speed, gravity provides all centripetal force
Unbanked flat curve	$f_{s} = \frac{m v ^{2}}{r}$	Static friction provides centripetal force; $v_{ma x} = μ_{s} g r$
Ideal banked curve (no friction)	$tan θ = \frac{v ^{2}}{r g}$	Mass cancels out; $θ$ = angle of bank from horizontal

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
Vertical loop normal force calculation
2022 · FRQ
Banked curve friction analysis

What's Next

Force analysis for circular motion is the direct prerequisite for the next topic in Unit 3: gravitational orbital motion, where we analyze how gravity provides the centripetal net force for planets, moons, and satellites orbiting larger bodies. Without mastering the core rule that net radial force equals $\frac{m v ^{2}}{r}$ , you will not be able to calculate orbital speeds or derive orbital period relationships, which are common AP exam questions. This topic also connects to energy conservation, as vertical circular motion problems often require relating speed at the top of a circle to speed at the bottom using energy, combining force analysis with earlier dynamics concepts. The core skills of aligned coordinate systems and net force analysis transfer to all other dynamics problems on the exam.