Circular Motion Dynamics: Banked Curves, Loops & Conical Pendulums
- Apply Newton's second law in the radial direction to circular-motion problems
- Analyze banked curves, vertical loops and conical pendulums with free-body diagrams
- Determine minimum speeds and critical conditions for circular motion
The method never changes
Every circular-motion dynamics problem is solved the same way. Draw a free-body diagram with real forces only. Choose the radial direction as positive toward the center. Write ΣF_radial = mv²/r. Write ΣF = ma separately for any perpendicular direction, usually with a = 0. Solve. The difficulty in these problems is never the physics; it is correctly identifying which forces have components pointing toward the center.
The banked curve
On a frictionless banked turn, only gravity and the normal force act. The normal force is perpendicular to the road surface, so it tilts inward: its horizontal component N sin θ supplies the centripetal force while its vertical component N cos θ supports the weight. Dividing one equation by the other eliminates both N and m and gives tan θ = v²/(rg). The design speed depends only on the bank angle and radius — a fully loaded truck and a motorcycle negotiate the same turn at the same ideal speed.
The vertical loop and the minimum speed
At the top of a vertical loop, gravity and the normal force both point downward, toward the center, so N + mg = mv²/r. As the speed decreases, N decreases; the critical case is N = 0, where gravity alone supplies exactly the required centripetal force. That gives mg = mv²/r and v_min = √(gr). Below that speed the required centripetal force exceeds what gravity can supply and the object leaves the track. Note that N = 0 is the condition of apparent weightlessness — the rider feels nothing from the seat.
Find the ideal bank angle for a curve of radius 120 m designed for 25 m/s. Separately, find the minimum speed at the top of a vertical loop of radius 8.0 m.
- 1.Banked curve: tan θ = v²/(rg) = (25)²/(120 × 9.8) = 625/1176 ≈ 0.531.
- 2.θ = arctan(0.531) ≈ 28°.
- 3.Vertical loop at minimum speed: N = 0, so mg = mv²/r and v = √(gr).
- 4.v = √(9.8 × 8.0) = √78.4 ≈ 8.9 m/s.
At the very top of a vertical circular loop, a car travels at the minimum speed for maintaining contact. The normal force from the track is —
Never add mv²/r to a free-body diagram as though it were a force. It is the result the real forces must produce — the right-hand side of ΣF = ma, not a term on the left. Putting it on both sides is the single most common error on this topic.
On a frictionless banked curve, the ideal speed for negotiating the turn depends on —
Answer the 2 checkpoints as you read.
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