Identify the real inward force in circular motion

1. Orient: Identify the real inward force in circular motion

Uniform circular motion has constant speed but changing velocity direction. Over a small angular step, the velocity-vector triangle is similar to the position-radius triangle: Δv/v≈Δs/r. Divide by elapsed time and use Δs/Δt≈v to obtain inward acceleration aᵣ=v²/r. The direction changes even though the speed does not.

2. Explain and derive

Newton’s second law in the radial direction requires a net inward force mv²/r. “Centripetal” describes that net force’s role; it is not an extra interaction to add beside tension, gravity or friction. Identify which real forces contribute inward and sum their radial components with a consistent sign convention.

For a horizontal circle supported vertically, the model assigns the inward force to tension. With a hypothetical 20 N connector limit and mass 2 kg, feasibility requires 2 v²/r≤20. Solving gives v≤√(10 r). If the required tension exceeds the limit, the prescribed circular path is not a physically maintainable outcome; the displayed requirement is a rejected design.

In a vertical circle the weight also has a radial component. At the top, inward points downward and T+mg=mv²/r; at the bottom, inward points upward and T−mg=mv²/r. A taut-string solution requires T≥0. If the string goes slack, use unconstrained motion instead of retaining a negative tension. Tangential forces, when present, change speed separately from the radial change in direction.

3. Work an example

For m=2 kg, v=3 m/s and r=1.5 m, inward acceleration is 9/1.5=6 m/s² and required tension is 12 N. Doubling speed at fixed radius multiplies tension by four, giving 48 N, above the assigned 20 N limit. Doubling radius at fixed speed halves the requirement.

Period, frequency and circular orbits

One revolution covers circumference 2πr. With constant speed v, period P=2πr/v and frequency f=1/P. Here P denotes the revolution period to avoid confusing it with string tension; many references call the period T. For r=1.5 m and v=3 m/s, P=π s and f=1/π Hz. Doubling speed at fixed radius halves the period while quadrupling the radial-force requirement.

For a circular orbit dominated by a spherical central mass M, equate gravity with the required radial net force: GMm/r²=mv²/r. Cancel the orbiting mass to obtain v²=GM/r. Substituting into the period relation gives P²=4π²r³/(GM), the circular-orbit form of Kepler’s third law. Radius is measured from the central body’s center. This simplified form assumes the orbiting mass is negligible relative to M; an isolated comparable-mass pair requires the total mass and its relative orbit.

Worked ratio: For two circular orbits around the same dominant body, an orbit at four times the radius has half the orbital speed and eight times the period. A longer orbit is traversed more slowly; comparing circumference alone would miss the speed change. Transfer: If radius is multiplied by 9, predict speed factor 1/3 and period factor 27 before using the equations. These are conditional circular-orbit statements, not numerical predictions of drag or atmospheric reentry.

Banking and conical motion: resolve real forces

For a frictionless banked road at bank angle θ above horizontal, the normal force tilts θ from vertical. Vertical balance gives N cosθ=mg and inward balance gives N sinθ=mv²/r. Dividing yields tanθ=v²/(rg). At r=20 m and assigned g=10 m/s², tanθ=0.5 supports v=10 m/s without friction. The normal force is larger than mg because only its vertical component supports weight.

Static friction can help at other speeds, but its direction is not fixed. Above the frictionless-design speed, the tendency is up and outward on the bank, so static friction acts down the bank; below that speed, it may act up the bank against the downward tendency. Treat these frictional banking cases qualitatively here and test the static bound before assuming a path is feasible.

For a conical pendulum, let α be the string angle from vertical and F_T its tension. The bob moves in a horizontal circle of radius r=L sinα for string length L. Resolve tension: F_T cosα=mg and F_T sinα=mv²/r, giving tanα=v²/(rg). At r=1.5 m, α=30° and g=10 m/s², v=√(15 tan30°)≈2.94 m/s. The bob has no vertical acceleration even though the string is angled.

Representation transfer: For tanα=0.5 and radius 0.8 m using g=10 m/s², derive v=2 m/s. Draw only tension and weight, with their components used in the equations, not additional force arrows. Explain why α is measured from vertical here whereas the road’s bank angle is measured from horizontal. The widget models a simpler horizontal tension circle; these worked and transfer cases explicitly extend the force construction without claiming to simulate banking or a full pendulum.

4. Predict and investigate

Predict first: Would doubling both speed and radius leave required tension unchanged? Use the square on speed before answering.

Find a feasible and an infeasible setting pair. Predict the tension ratio after changing only speed, then only radius. Derive the maximum permitted speed symbolically and compare with the ledger. Explain why an outward force arrow would be inappropriate in the inertial-frame free-body diagram. Use the lesson controls and record the settings with your evidence.

5. Transfer and explain

Try a new practice case

A 3 kg object moves in a horizontal circle of radius 2 m at speed 4 m/s. Find required inward net force.

Your numerical result (N) Check number

Reveal worked check (practice, not mastery)

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

If the connector broke, what is the instantaneous velocity direction? Distinguish the tangent direction from the inward acceleration that existed just before breaking.

Retain your first answer, force diagram, equations, units, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Supplied synthetic outputs do not certify actual laboratory skill.

Top-of-loop contact and combined acceleration

For an object on the inside of a vertical loop at its top, gravity and an inward normal force give mg+N=mv²/r. A unilateral contact can push but not pull, so N≥0 and the minimum contact speed is sqrt(gr). A taut string has the same inward-only constraint; a rigid bilateral linkage can push or pull and does not obey that same minimum-speed contact argument.

Predict and check: with m=1 kg,r=2 m,g=10 m/s², the limiting speed is sqrt20≈4.472 m/s. At 6 m/s, N=36/2−10=8 N; at 4 m/s the algebra asks for −2 N, so the assumed inside-contact path fails rather than creating an outward pulling normal. Quadrupling radius doubles the threshold speed.

When speed changes, radial and tangential accelerations are perpendicular components. A separate instantaneous packet has radial acceleration 4 m/s² inward and tangential acceleration 3 m/s² along the velocity. The net magnitude is 5 m/s², directed about 36.87° from inward toward the tangent, not 7 m/s². Reverse only the tangential component: net magnitude stays 5 but speed now decreases. This vector construction does not add a new interaction called centripetal force.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 2 topic mapping: 2.9. Checked 30 October 2026. This is a content cross-reference, not AP authorization.