Force Analysis for Circular Motion — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Centripetal net force definition, free-body diagram construction for circular motion, vertical circular motion force analysis, unbanked/banked curve analysis, and resolving common misconceptions about centripetal vs centrifugal force.

You should already know: Newton's second law of motion, free-body diagram construction, vector resolution of forces into components.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Force Analysis for Circular Motion?

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: Circular Motion and Gravitation, which counts for 6–8% of the total AP Physics 1 exam score. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as part of a multi-concept question that connects to energy, gravitation, or forces. The core convention for this topic 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. For uniform circular motion (constant speed), acceleration is entirely radial, so tangential net force is always zero. Synonyms for centripetal net force include radial net force, and this topic builds all foundational skills for orbital motion analysis later in the unit.

2. The Centripetal Net Force Rule

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 $\omega$ = angular speed. Applying Newton's second law ($F_{net}=ma$), 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: 1) draw a free-body diagram of only real forces acting on the object; 2) align one axis with the radial (center-pointing) direction; 3) resolve all forces into radial and tangential components; 4) 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: $F_{net,c}=T=\frac{m{v}^2}{r}$.
4. Substitute values: $T=\frac{(0.5\text{kg})(4\text{m/s}{)}^2}{2\text{m}}=4\text{N}$.
5. 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

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: at the top, the center is below the object, so positive radial direction is downward; at the bottom, the center is above the object, so positive radial direction is upward. Common AP exam questions for this topic ask for normal force or tension at the top/bottom, or the minimum speed needed to keep the object moving in a 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 $mg$ (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: $F_{net,c}=N+mg=\frac{m{v}^2}{r}$.
4. Rearrange to solve for $N$: $N=m\left(\frac{v^2}{r}-g\right)$.
5. Substitute values: $N=70\text{kg}\left(\frac{225{\text{m}}^2/{\text{s}}^2}{12\text{m}}-9.8{\text{m/s}}^2\right)\approx 630\text{N}$.

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 $v_{min}=\sqrt{gr}$. Memorize this to save time on MCQs.

4. Unbanked and Banked Horizontal Curve Analysis

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 $\theta$ 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 $v$), we derive the relationship by balancing the vertical component of normal force with gravity, and setting the horizontal component equal to centripetal net force: $\tan \theta =\frac{v^2}{rg}$.

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 $\theta$ 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: vertical component $N\cos \theta =mg$, horizontal radial component $N\sin \theta =\frac{m{v}^2}{r}$.
4. Divide the two equations to eliminate $N$ and $m$: $\tan \theta =\frac{v^2}{rg}$.
5. Substitute values: $\tan \theta =\frac{(20\text{m/s}{)}^2}{(50\text{m})(9.8{\text{m/s}}^2)}\approx 0.816$, so $\theta \approx 39^{\circ }$.

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. Common Pitfalls (and how to avoid them)

• 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 $N-mg=\frac{m{v}^2}{r}$ instead of $N+mg=\frac{m{v}^2}{r}$. 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 is what 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 $\sin \theta$ and $\cos \theta$ in banked curve problems, leading to $\cot \theta =\frac{v^2}{rg}$ instead of $\tan \theta$. Why: Students misidentify how the bank angle relates to the normal force's components. Correct move: Test your trig with the edge case: if $\theta =0$ (flat road), $\tan \theta =0$, which means $v=0$ for no friction, which is logical. If your formula gives $\tan \theta =\frac{rg}{v^2}$, the edge case is wrong, so flip your trig.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

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 $mg$? A) $N>mg$ B) $N=mg$ C) $N<mg$ D) $N=0$ for any non-zero speed

Worked Solution: 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 $mg-N=\frac{m{v}^2}{r}$. Rearranging gives $N=mg-\frac{m{v}^2}{r}$. For any non-zero speed, $\frac{m{v}^2}{r}$ is positive, so $N$ is always less than $mg$. $N$ only equals zero at the specific speed $v=\sqrt{rg}$ where the car loses contact, not for any non-zero speed. The correct answer is C.

Question 2 (Free Response)

A student swings a 0.25 kg stone in a vertical circle on the end of a 1.0 m long massless string. The stone moves at enough speed to keep the string taut throughout the motion. (a) Draw a free-body diagram for the stone when it is at the bottom of the circle, label all forces. (b) The tension in the string is measured to be 15 N when the stone is at the bottom of the circle. Calculate the speed of the stone at this point. (c) If the student swings the stone more slowly, describe what will happen to the tension at the top of the circle as speed decreases, and explain why. If speed is low enough, what happens to the string?

Worked Solution: (a) Two forces are drawn: 1) Weight $mg$, pointing vertically downward; 2) Tension $T$, pointing vertically upward (toward the center of the circle, which is above the stone at the bottom). No additional forces are included. (b) Align positive radial direction upward (toward center). Sum forces: $T-mg=\frac{m{v}^2}{r}$. Solve for $v$: $$v=\sqrt{\frac{r(T-mg)}{m}}=\sqrt{\frac{1.0\text{m}(15\text{N}-(0.25\text{kg})(9.8{\text{m/s}}^2))}{0.25\text{kg}}}\approx 7.1\text{m/s}$$ (c) At the top of the circle, tension follows $T=\frac{m{v}^2}{r}-mg$. As speed decreases, $\frac{m{v}^2}{r}$ decreases, so tension also decreases. When speed drops low enough that $\frac{m{v}^2}{r}<mg$, tension would need to be negative to maintain circular motion, which a string cannot do (strings only pull, they cannot push). The string goes slack, and the stone stops moving in a circular path, falling under gravity.

Question 3 (Application / Real-World Style)

NASA uses a horizontal centrifuge to test the effects of high acceleration on astronauts. The centrifuge has a radius of 8.8 m, and spins an astronaut in a horizontal circle at a constant rate. The centrifuge is designed to produce a centripetal acceleration of 12$g$ (12 times gravitational acceleration) on the astronaut. If the astronaut has a mass of 70 kg, what is the magnitude of the net force acting on the astronaut, and what is the tangential speed of the astronaut?

Worked Solution: The net force on the astronaut equals mass times centripetal acceleration, by Newton's second law. First calculate centripetal acceleration: $a_c=12g=12(9.8{\text{m/s}}^2)=117.6{\text{m/s}}^2$. Net force is: $$F_{net,c}=m{a}_c=(70\text{kg})(117.6{\text{m/s}}^2)\approx 8200\text{N}$$ Use the centripetal acceleration formula to solve for tangential speed: $$v=\sqrt{a_{c}r}=\sqrt{(117.6{\text{m/s}}^2)(8.8\text{m})}\approx 32\text{m/s}$$ In context, this net force is equivalent to 12 times the astronaut's body weight, which mimics the high acceleration the astronaut will experience during rocket launch.

7. Quick Reference Cheatsheet

8. 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.

• Newton's Law of Universal Gravitation
• Orbital Motion and Kepler's Laws
• Energy Conservation for Vertical Circular Motion