Circular Motion & Orbits
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
Anything moving in a circle accelerates toward the center: ac = v²/r. Real forces (tension, normal, friction, gravity) must add up to a net inward force mv²/r. There is no extra “centripetal force” to draw.
1. Find what points inward
2. Vertical circles and orbits
Top of a loop: gravity and the normal force (or tension) both point down, toward the center. At the minimum speed, N = 0 and gravity alone provides mv²/r.
Bottom: tension points up, gravity down: T − mg = mv²/r, so the tension is largest there.
Orbits: gravity provides the centripetal force. Farther out means slower: v ∝ 1/√r. The period follows Kepler’s third law: T² ∝ r³.
Use r from the planet’s center, not its surface.
3. Example, then try it
A road curve has radius 90 m. At what bank angle can cars take it at 15 m/s without needing friction?
tan θ = v²/(rg) = 225 ÷ 900 = 0.25, so θ ≈ 14°.
A 0.50 kg ball swings in a vertical circle on a 0.80 m string. At the top its speed is 4.0 m/s. Find the tension at the top, its speed and tension at the bottom, and the minimum speed at the top. (g = 10 m/s²)
Top: T + mg = mv²/r → T = (0.50)(16) ÷ 0.80 − 5.0 = 10 − 5.0 = 5.0 N.
Bottom speed (energy): v² = 16 + 2g(2r) = 16 + 32 = 48, so v ≈ 6.9 m/s.
Bottom tension: T − mg = mv²/r → T = 5.0 + (0.50)(48) ÷ 0.80 = 5.0 + 30 = 35 N.
Minimum at the top: T = 0, so v = √(gr) = √8.0 ≈ 2.8 m/s.
Answer: 5.0 N at the top; 6.9 m/s and 35 N at the bottom; at least 2.8 m/s at the top.
4. Common mistakes
A cart goes around the inside of a vertical loop. At the very top, which forces point toward the center?
At the top, the center is below the cart. Gravity points down, and the track pushes down on the cart too, so both point toward the center.
Ready to practice?
5 questions on Circular Motion & Orbits, each with a hint and an explanation for every wrong answer.