with v= 60(1000/3600) = 17 m/s and R= 200 m. The banking angle is therefore θ= 8.1°. Now we consider a vehicle taking this banked curve at v'= 40(1000/3600) = 11 m/s. Its (horizontal) acceleration is 2/avR′′=, which has components parallel the incline and perpendicular to it: 2| |2coscossinsin.vaaRvR⊥′′==′′These enter Newton’s second law as follows (choosing downhill as the +xdirection and away-from-incline as +y):||sincossNmgfma
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.