Circular Motion & Highway Curves

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Circular Motion & Highway Curves

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Circular Motion & Highway Curves. Sect. 5-3: Highway Curves: Banked & Unbanked. - PowerPoint PPT Presentation

Transcript of Circular Motion & Highway Curves

Page 1: Circular Motion & Highway Curves

Circular Motion & Highway Curves

Page 2: Circular Motion & Highway Curves

Sect. 5-3: Highway Curves: Banked & UnbankedCase 1 - Unbanked Curve: When a car rounds a curve, there MUST be a net force toward the circle center (a Centripetal Force) of which the curve is an arc. If there weren’t such a force, the car couldn’t follow the curve, but would (by Newton’s 2nd Law) go in a straight line. On a flat

road, this Centripetal Force is the static friction force.

“Centripetal Force” No static friction?

No Centripetal Force The Car goes straight!

There is NEVER a “Centrifugal Force”!!!

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Example 5-6: Skidding on a curveA car, mass m = 1,000 kg car rounds a curve on a flat road of radius r = 50 m at a constant speed v = 14 m/s (50 km/h). Will the car follow the curve, or will it skid? Assume: a. Dry pavement with the coefficient of static friction μs = 0.6. b. Icy pavement with μs = 0.25.

Newton’s 2nd Law: ∑F = max: ∑Fx = max Ffr = maR = m(v2/r)

y: ∑Fy = may = 0 FN - mg = 0; FN = mg

The maximum static friction is Ffr = μsFN

Free BodyDiagram

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If the friction force isn’t sufficient, the car will tend to move more nearly in a straight line (Newton’s 1st Law) as the skid marks show.

As long as the tires don’t slip, the friction is static. If the tires start to slip, the friction is kinetic, which is bad in 2 ways!!

1. The kinetic friction force is smaller than the static friction force.

2. The static friction force points toward the circle center, but the kinetic friction force opposes the direction of motion, making it difficult to regain control of the car & continue around the curve.

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Case 2- Banked CurveBanking curves helps keep cars from skidding. For every banked curve, there is one speed v at which the entire Centripetal Force is supplied by the horizontal component of the normal force FN, so that no friction is required!!

Newton’s 2nd LawTells us what speed v this is: x: ∑Fx = max FNx = m(v2/r) or

Also y: ∑Fy = may = 0 FN cosθ - mg = 0 FNcosθ = mg

So FN = (mg/cosθ)Put this into the x equation:

g(sinθ/cosθ) = (v2/r) or tanθ = (v2/gr)

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Example 5-7: Banking anglea. For a car traveling with speed v around a

curve of radius r, find a formula for the angle θ at which a road should be banked so that no friction is required.

b. Calculate this angle for an expressway off-ramp curve of radius r = 50 m at a design speed of v = 14 m/s (50 km/h).

Newton’s 2nd Lawx: ∑Fx = max FNx = m(v2/r) or FNsinθ = m(v2/r) (1)

y: ∑Fy = may = 0 FNcosθ - mg = 0 or FNcosθ = mg (2)

Dividing (2) by (1) gives: tanθ = [(v2)/(rg)]

Putting in the given numbers tanθ = 0.4 or θ = 22º