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Circular Motion & Highway Curves

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

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Circular Motion & Highway Curves

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  1. Circular Motion & Highway Curves

  2. Sect. 5-3: Highway Curves: Banked & Unbanked Case 1 - Unbanked Curve:When a car rounds a curve,there MUSTbe 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”!!!

  3. Example 5-6: Skidding on a curve A 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 = ma x: ∑Fx = max Ffr =maR = m(v2/r) y: ∑Fy = may = 0  FN -mg = 0; FN =mg The maximum static friction isFfr =μsFN Free Body Diagram

  4. 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.Thekinetic friction force is smaller than the static friction force. 2.Thestatic friction forcepoints 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.

  5. Case 2- Banked Curve Banking curves helps keep cars from skidding. For every banked curve, there is one speed v at which the entire Centripetal Forceis supplied by the horizontal component of the normal force FN, so that no friction is required!! Newton’s 2nd Law Tells 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)

  6. Example 5-7: Banking angle a. 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 Law x: ∑Fx = max FNx = m(v2/r) or FNsinθ=m(v2/r) (1) y:∑Fy = may = 0  FNcosθ -mg = 0 orFNcosθ=mg (2) Dividing (2) by (1) gives:tanθ = [(v2)/(rg)] Putting in the given numberstanθ = 0.4 or θ = 22º

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