Tilted forces: banks and conical pendulums
You will be able to: Resolve a tilted support force into vertical and inward components.
How can one tilted force both support and turn an object?
A hanging bob moves in a horizontal circle while its string tilts away from vertical. The string pulls partly upward to balance weight and partly inward to turn the bob.
A useful starting point: Resolving forces →
Words and symbols before equations
- Conical pendulum
- A bob moving in a horizontal circle so the string sweeps out a cone.
- Angle θ
- Here the string’s angle from vertical, not from horizontal.
- Bank angle β
- A road’s tilt from horizontal; its normal tilts β from vertical.
What this picture assumes
Frictionless banked circular road, radius 2 m, mass 1 kg, g=10 m/s². Read the design speed, not the maximum speed for a road with friction. At zero angle the no-friction circular design speed is zero.
Connect the picture to the physics
For the bob, vertical acceleration is zero, so T cosθ=mg. The horizontal inward component supplies circular acceleration: T sinθ=mv²/r. Divide these equations to remove tension and mass: tanθ=v²/(rg).
For a frictionless banked road, use the normal force in place of tension: N cosβ=mg and N sinβ=mv²/r. Thus v²=rg tanβ is the design-speed condition for a frictionless turn.
At other speeds, friction may contribute up or down the bank. Decide the tendency to slide before choosing the friction direction; its vertical and inward components also enter the equations. The no-friction formula cannot be assumed valid for every speed.
A worked example, step by step
A 0.5 kg bob travels at 2 m/s around a horizontal circle of radius 0.4 m. Use g=10 m/s². Find the string’s angle from vertical and tension.
- Inward requirement: mv²/r=0.5×4/0.4=5 N.
- Weight is 5 N, so the tension components are 5 N upward and 5 N inward.
- tanθ=5/5=1, so θ=45° from vertical.
- T=√(5²+5²)≈7.07 N. It exceeds weight because it has a horizontal job too.
Tension’s vertical component balances weight; the full tension does not equal mg.
Why must tension exceed mg for a nonzero conical angle?
Compare with an explanation
Only T cosθ supports weight, and cosθ<1, so T must be larger than mg.
Predict. Change one thing. Explain.
Keep circle radius fixed and change bank angle. Predict the frictionless design speed. Compare the upward and inward components of normal force; the upward component stays equal to weight.
Design speed 4.47 m/s at radius 2 m. Normal force 14.14 N; inward component 10 N. Force scale 5 drawing units/N. At 0° the normal points straight upward.
Frictionless banked circular road, radius 2 m, mass 1 kg, g=10 m/s². Read the design speed, not the maximum speed for a road with friction. At zero angle the no-friction circular design speed is zero.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a force or motion relationship to justify your prediction.
Optional 3D: inspect a conical pendulum
The string sweeps a cone above a horizontal circular path. This helps distinguish string length from horizontal radius. Fixed example: string length 2 m, angle 45° from vertical, mass 0.5 kg, g=10 m/s². Radius = 2 sin45° ≈ 1.41 m; vertical drop = 1.41 m; tension ≈ 7.07 N; speed ≈ 3.76 m/s. These numbers differ from the earlier worked example because the string is longer.
The 2D explanations above contain everything needed for the lesson.
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 frictionless circular road of radius 20 m is banked at an angle with tanβ=0.5. Use g=10. (a) Draw weight and normal, (b) write two component equations, (c) find design speed, and (d) explain why the formula cannot cover every speed with zero friction.
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Compare with the answer and four-point rubric
- 1 point: Weight down; normal perpendicular to the road, tilted inward from vertical.
- 1 point: N cosβ=mg and N sinβ=mv²/r.
- 1 point: v=√(20×10×0.5)=10 m/s.
- 1 point: At a different speed the same component ratio cannot simultaneously support and turn the car; another force such as friction is required.
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 1Which angle matters for a component?
The one defined relative to the chosen axis.
RECALL 2Why can one force have two roles?
Its perpendicular components contribute to different axis equations.
RECALL 3When is the bank formula valid?
For a frictionless bank and a horizontal circular path at its design speed.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Tilted forces: banks and conical pendulums
- Conical pendulum: T cosθ=mg; T sinθ=mv²/r.
- Frictionless bank: v²=rg tanβ.
- Horizontal circular path and no vertical acceleration; stated ideal force assumptions.
Remember: Tension’s vertical component balances weight; the full tension does not equal mg.
Conditions: Frictionless banked circular road, radius 2 m, mass 1 kg, g=10 m/s². Read the design speed, not the maximum speed for a road with friction. At zero angle the no-friction circular design speed is zero.
Refresh Kid · Unit 2 · Objectives 2.9.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.9, objectives 2.9.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. 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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