What is the Ideal banking angle (in degrees) for a gentle turn of 2.00 km radius

Question

What is the Ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 115 km/h speed limit (about 71 m/h), assuming everyone traveling at the limit? ____ degree If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding towards the inside of the curve (a real problem on icy maintain roads.) (a) Calculate the ideal speed in (m/s) to take a 120 m radius curve banked at 15 degree. ____ m/s (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 30.0 km/h? ____

Explanation / Answer

weight component due to gravity = component of centrifugal force mg sin = (mv²/r) cos v = sqrt[gr tan] v = sqrt[(9.8 m/s²)(120 m) tan15°] v = 17.75 m/s (b) (30 km/h)(1000 m/km)(h / 3600 s) = 8.333 m/s weight component due to gravity = component of centrifugal force + friction mg sin = (mv²/r) cos + [(mv²/r) sin + mg cos] = [g sin - (v²/r) cos] / [(v²/r) sin + g cos] = [(9.8 m/s²) sin15° - ({8.333 m/s}² / {120 m}) cos15°] / [({8.333 m/s}² / {120 m}) sin15° +     (9.8m/s²) cos15°] = 0.179

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