# Circular Motion & Highway Curves

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Circular Motion & Highway Curves

Sect. 5-3: Highway Curves: Banked & Unbanked
Case 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”!!!

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 is Ffr = μsFN Free Body Diagram Figure Caption: Example 5–14. Forces on a car rounding a curve on a flat road. (a) Front view, (b) top view. Solution: The normal force equals the weight, and the centripetal force is provided by the frictional force (if sufficient). The required centripetal force is 4500 N. The maximum frictional force is 5880 N, so the car follows the curve. The maximum frictional force is 2450 N, so the car will skid.

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. Figure Caption: Race car heading into a curve. From the tire marks we see that most cars experienced a sufficient friction force to give them the needed centripetal acceleration for rounding the curve safely. But, we also see tire tracks of cars on which there was not sufficient force—and which unfortunately followed more nearly straight-line paths.

Case 2- Banked Curve Newton’s 2nd Law
Banking 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 Law Tells us what speed v this is: x: ∑Fx = max  FNx = m(v2/r) or Figure Caption: Normal force on a car rounding a banked curve, resolved into its horizontal and vertical components. The centripetal acceleration is horizontal (not parallel to the sloping road). The friction force on the tires, not shown, could point up or down along the slope, depending on the car’s speed. The friction force will be zero for one particular speed. 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)

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). Answer: a. Set FN = mg in previous equation. Find tan θ = v2/rg. b. Tan θ = 0.40, so θ = 22°. Newton’s 2nd Law x: ∑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