Circular Motion on a Banked Curve
A car of mass 1200 kg rounds a frictionless banked curve of radius 45 m at 20 m/s. The banking angle is set to 25°. A designer claims that because the curve is frictionless, the car will stay on the road at any speed as long as the banking angle is steep enough, since the normal force provides a horizontal component. On a rainy day the designer proposes increasing the banking angle to 40° so the car can safely take the curve at 30 m/s. The car's tyres are treated as providing no friction on the wet surface.
1. Calculate the design speed for the 25° bank and compare it with the car's actual speed of 20 m/s.
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For a frictionless bank, the centripetal force is provided entirely by the horizontal component of the normal force: N sinθ = mv²/r and N cosθ = mg. Dividing gives tanθ = v²/(rg).
Design speed: v = √(rg tanθ) = √(45 × 9.8 × tan25°) = √(45 × 9.8 × 0.466) ≈ √205.6 ≈ 14.3 m/s.
The car travels at 20 m/s, faster than the 14.3 m/s design speed, so on a frictionless bank it would tend to slide up and outward. Without friction it cannot maintain the curve at 20 m/s.
2. Evaluate the designer's proposal to bank at 40° for 30 m/s, and explain the physical reason the 'any speed' claim is wrong.
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Required condition: tanθ = v²/(rg) = 30²/(45 × 9.8) = 900/441 ≈ 2.04, so θ ≈ 63.9°.
A 40° bank gives design speed v = √(45 × 9.8 × tan40°) ≈ √370 ≈ 19.2 m/s, far below 30 m/s. The proposal fails.
The 'any speed' claim is wrong because for a frictionless bank the required centripetal force mv²/r grows with v², while the available horizontal normal component is fixed by θ and mg. Only one speed satisfies both force balances simultaneously; at other speeds there is a net radial force the road cannot supply without friction. Steepening the bank only shifts the single design speed upward; it does not remove the constraint.