A banked curve and its design speed
You will be able to: Resolve real contact forces into vertical and inward directions and assess the role of static friction.
How can the road’s normal force help a car turn?
A tilted road can push a car both upward and toward the turn’s center. At one design speed the normal force alone supplies both the vertical support and inward acceleration.
A useful starting point: Forces at the top and bottom of a loop →
Words and symbols before equations
- Bank angle θ
- Tilt of the road above horizontal.
- Design speed
- Speed needing no sideways static friction in the ideal model.
- Inward force component
- Component toward the horizontal circle’s center.
- Signed friction f
- Here positive when acting down the bank, toward the lower inner edge.
What this picture assumes
Ideal 1000 kg car, horizontal circular path, g = 10 m/s². Positive signed friction acts down the bank. Above the static limit, the force diagram shows required forces for an impossible no-slip candidate, not actual sustained motion.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Design speed 10.75 m/s. At 10 m/s: N = 11160 N; signed down-bank f = -669.9 N (uphill). Limit 4464 N: no slip possible. Inward is left in the diagram.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For a frictionless bank, N cos θ = mg vertically and N sin θ = mv²/r inward. Dividing eliminates N and m: tan θ = v²/(rg), so v = √(rg tan θ). The normal force is one real tilted force; its components are used in separate equations.
At a different speed, static friction may be needed. With f positive down the bank, N cos θ − f sin θ = mg and N sin θ + f cos θ = mv²/r. Solving gives N = mg cos θ + (mv²/r)sin θ and f = (mv²/r)cos θ − mg sin θ.
A positive required f acts down the bank at speeds above the frictionless design value. A negative value acts uphill at lower speeds. No slip requires |f| ≤ μₛN. These are ideal lateral balances; tire behavior, braking and real road conditions are outside this model.
A worked example, step by step
A frictionless 20 m radius curve is banked at θ with tan θ = 0.5. Find the design speed using g = 10 m/s².
- Vertical balance: N cos θ = mg.
- Inward acceleration: N sin θ = mv²/r.
- Divide to get v² = rg tan θ = 20(10)(0.5) = 100 m²/s².
- The design speed is 10 m/s and does not depend on mass in this ideal model.
Normal force is perpendicular to the road, not generally vertical.
Can friction point uphill on a banked turn?
Compare with an explanation
Yes. At speeds below the design value it can prevent the car from slipping down the bank.
Predict. Change one thing. Explain.
Keep radius and bank angle fixed. Move speed above and below the frictionless design value. Predict the sign of the required static friction and compare it with the limit.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Design speed 10.75 m/s. At 10 m/s: N = 11160 N; signed down-bank f = -669.9 N (uphill). Limit 4464 N: no slip possible. Inward is left in the diagram.
Ideal 1000 kg car, horizontal circular path, g = 10 m/s². Positive signed friction acts down the bank. Above the static limit, the force diagram shows required forces for an impossible no-slip candidate, not actual sustained motion.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 10 m radius curve has tan θ = 0.4. Derive the frictionless speed, calculate it and explain the direction of needed friction at a lower speed.
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Compare with the answer and four-point rubric
- 1 point: Write N cos θ = mg and N sin θ = mv²/r.
- 1 point: Divide to obtain v² = rg tan θ.
- 1 point: v = √(10 × 10 × 0.4) = √40 ≈ 6.32 m/s.
- 1 point: Below that speed the required signed f is negative, so static friction acts uphill.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What supplies centripetal acceleration at design speed?
The inward component of the normal force.
RECALL 2Does the design speed depend on car mass?
Not in this ideal frictionless model.
RECALL 3How is a non-design speed checked?
Solve for required static friction and compare its magnitude with μₛN.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A banked curve and its design speed
- Frictionless design: v² = rg tan θ.
- With signed down-bank f: N cos θ − f sin θ = mg.
- N sin θ + f cos θ = mv²/r; check |f| ≤ μₛN.
Remember: Normal force is perpendicular to the road, not generally vertical.
Conditions: Ideal 1000 kg car, horizontal circular path, g = 10 m/s². Positive signed friction acts down the bank. Above the static limit, the force diagram shows required forces for an impossible no-slip candidate, not actual sustained motion.
Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.10.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.10, objectives 2.10.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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