Forces at the top and bottom of a loop
You will be able to: Write the inward force equation at each location and enforce the physical tension condition.
Why can a string become slack at the top of a loop?
A small mass moves in a vertical circle on a string. At the top both gravity and tension can point inward. At the bottom tension points inward while gravity points outward.
A useful starting point: Radial and tangential acceleration →
Words and symbols before equations
- Local inward axis
- Positive direction toward the center, chosen separately at each point.
- Taut-string condition
- Tension must be nonnegative; a flexible string cannot push.
- Minimum top speed
- Speed at which gravity alone supplies the required inward acceleration.
What this picture assumes
Separate instantaneous top/bottom cases on a flexible string, g = 10 m/s². Speeds at these locations are not assumed equal in a natural orbit. Negative required tension means the prescribed circle fails.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Required tension 8 N. Taut-string condition is possible at this instant. Minimum top speed 4.472 m/s.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
At the top the inward direction is downward, so T + mg = mv_top²/r. Thus T = mv_top²/r − mg. At the bottom inward is upward: T − mg = mv_bottom²/r, giving T = mv_bottom²/r + mg.
At the limiting top speed, T = 0 and v_top = √(gr). Slower than this would require a negative tension to maintain the specified circle, so the string goes slack and the assumed circular path fails. The threshold does not apply to a rigid rod that can push as well as pull.
The top and bottom speeds generally differ. This lesson compares forces at specified local speeds; it does not assume speed stays constant throughout a gravity-driven loop. A later energy analysis can relate the speeds along the path.
A worked example, step by step
A 0.5 kg mass at the top of a 2 m radius loop has speed 5 m/s. Find tension and decide if a taut circle is possible with g = 10 m/s².
- At the top choose downward as inward.
- Required inward force is mv²/r = 0.5(25)/2 = 6.25 N.
- Gravity supplies 5 N, leaving T = 1.25 N.
- T is positive, so a taut string is possible at this instant; the minimum speed is √20 ≈ 4.47 m/s.
A negative calculated tension means the assumed string-constrained path is impossible, not that the string pushes.
Is tension necessarily zero at the top?
Compare with an explanation
No. It is zero only at the limiting speed in the stated ideal model.
Predict. Change one thing. Explain.
Use the top position and decrease speed through √(gr). Watch the required tension. Then switch to the bottom at the same local speed and explain the difference.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Required tension 8 N. Taut-string condition is possible at this instant. Minimum top speed 4.472 m/s.
Separate instantaneous top/bottom cases on a flexible string, g = 10 m/s². Speeds at these locations are not assumed equal in a natural orbit. Negative required tension means the prescribed circle fails.
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 1 kg object moves at 6 m/s at a point on a 2 m radius vertical loop. Compare required tension if that point is the top or bottom, then find the minimum top speed.
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Compare with the answer and four-point rubric
- 1 point: Required inward net force is 36/2 = 18 N.
- 1 point: Top tension = 18 − 10 = 8 N.
- 1 point: Bottom tension = 18 + 10 = 28 N.
- 1 point: Minimum top speed is √20 ≈ 4.47 m/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 1Which way is inward at the top?
Downward, toward the center.
RECALL 2Why check T ≥ 0?
The string can pull but cannot push.
RECALL 3Does this comparison say top and bottom speeds naturally match?
No. They are separate instantaneous cases with specified speeds.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Forces at the top and bottom of a loop
- Top: T = mv²/r − mg; require T ≥ 0.
- Bottom: T = mv²/r + mg.
- Minimum top speed for a taut string: √(gr).
Remember: A negative calculated tension means the assumed string-constrained path is impossible, not that the string pushes.
Conditions: Separate instantaneous top/bottom cases on a flexible string, g = 10 m/s². Speeds at these locations are not assumed equal in a natural orbit. Negative required tension means the prescribed circle fails.
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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