A conical pendulum in space
You will be able to: Connect a spatial path to vertical balance and the inward component of tension.
How does a tilted string keep a mass moving in a horizontal circle?
A small bob traces a horizontal circle while its string sweeps out a cone. The bob stays at one height even though it has a nonzero acceleration toward the circle center.
A useful starting point: A banked curve and its design speed →
Words and symbols before equations
- Cone angle θ
- Angle between the string and vertical.
- String length L
- Distance from pivot to bob.
- Circle radius r
- Horizontal radius L sin θ, not the string length.
- Tension T
- Single force along the string toward the pivot.
What this picture assumes
Steady conical pendulum with ideal string and g = 10 m/s². The projected spatial path and flat force diagram describe the same state. Camera and phase do not change the calculated radius, tension or speed.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- r = 1.732 m; tension 20 N; speed 5.477 m/s; period 1.987 s. Vertical forces balance; net horizontal force 17.32 N points inward.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Draw two real forces on the bob: tension along the string and weight down. Vertical acceleration is zero, so T cos θ = mg. The inward horizontal component gives T sin θ = mv²/r.
Divide the equations: v² = rg tan θ, with r = L sin θ. For 0 < θ < 90°, angular speed obeys ω² = g/(L cos θ), so period P = 2π√(L cos θ/g). P denotes period here to avoid confusing it with tension T.
The spatial view shows the circular path and string geometry; rotating the camera changes only the projection. Use the flat force diagram and labels to calculate. The normal-force banked-curve problem uses similar component logic but different physical interactions.
A worked example, step by step
A 1 kg bob uses L = 2 m at θ = 60° from vertical. Find radius, tension and speed with g = 10 m/s².
- Radius r = L sin 60° = √3 m ≈ 1.732 m.
- Vertical balance gives T = mg/cos 60° = 20 N.
- v² = rg tan 60° = √3(10)√3 = 30 m²/s².
- v = √30 ≈ 5.48 m/s; its acceleration is horizontal inward, not along the string.
The circular radius is L sin θ. Using L directly changes the required acceleration.
Why is acceleration not along the string?
Compare with an explanation
The downward weight cancels tension’s vertical component, leaving a horizontal net force.
Predict. Change one thing. Explain.
Change the viewing angle without changing θ or L. Then change θ and predict radius and tension. Compare the spatial path with the flat actual-force diagram.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
r = 1.732 m; tension 20 N; speed 5.477 m/s; period 1.987 s. Vertical forces balance; net horizontal force 17.32 N points inward.
Steady conical pendulum with ideal string and g = 10 m/s². The projected spatial path and flat force diagram describe the same state. Camera and phase do not change the calculated radius, tension or speed.
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 2 kg bob has L = 1 m and θ = 60° from vertical. Find radius, tension, speed and period using g = 10 m/s².
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Compare with the answer and four-point rubric
- 1 point: r = sin 60° = √3/2 m ≈ 0.866 m.
- 1 point: T = 20/0.5 = 40 N.
- 1 point: v² = (√3/2)(10)√3 = 15, so v ≈ 3.873 m/s.
- 1 point: P = 2π√(0.5/10) ≈ 1.405 s.
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 are the actual forces?
Tension along the string and gravitational weight downward.
RECALL 2What does camera rotation change?
Only the screen projection, not physical values.
RECALL 3Why is tension greater than weight for nonzero θ?
Only its vertical component balances weight.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A conical pendulum in space
- r = L sin θ; T cos θ = mg.
- T sin θ = mv²/r; v² = rg tan θ.
- P = 2π√(L cos θ/g) for an ideal steady conical pendulum.
Remember: The circular radius is L sin θ. Using L directly changes the required acceleration.
Conditions: Steady conical pendulum with ideal string and g = 10 m/s². The projected spatial path and flat force diagram describe the same state. Camera and phase do not change the calculated radius, tension or speed.
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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