 Using the information shown on the diagram on the other page, determine the correct answers to the following questions. 1 How much force is used to accelerate the train across the first 42.6 m? 2 How many g's does the rider feel as he is pushed by the initial accelerating force? 3 How fast is the car traveling at location A if it coasted down the back side of the first hill? 4 At location B the incline is 90° of a circle. How many g's does the rider feel as he enters the loop at location "B?" 5 How fast is the car traveling when it reaches location "C"? 6 In the banked turn the rider is traveling 21.6 m/s. What is the optimum angle of the banked curve? 7 How many g's does the rider feel as he enters the loop? 8 What is the height of the loop? 9 If the radius at the top of the loop is 7.45 m, then how many g's does the rider feel at this location? 10 How high does the coaster train coast at the end of the track? 1 Use energy relationships to solve ET(BEGINNING) + WORK = ET(BEGINNING) 0 + Fd = (1/2) mv2 F(42.6) = (1/2) 5634)(38.6)2 F(42.6) = 4197217.32 F = 98,526.22817 N F = 98,500 N 2 Use kinematics to solve vf2 = vo2 + 2ax 38.62 = 0 + 2a(42.6) ac = 17.48779 m/s2 3 Use energy relationships to solve ET(AFTER FORCE) = ET (at A) KE + PE = KE + PE (1/2)mv2 + mgh = (1/2)mv2 + 0 (1/2)v2 + gh = (1/2) v2 + 0 (1/2)(36.8)2 + 9.8(12.1) = (1/2)v2 863.56 = (1/2)v2 1727.12 = v2 v = 41.55863 m/s v = 41.6 m/s 4 v = 45.5 m/s, r = 70.3 m g's felt at the bottom = ac[in g's] + 1 g g's felt at the bottom = 3.0049787 + 1 g g's felt at the bottom = 4.00 g's 5 Use energy relationships and compare locations "B" with "C" ETB = ETC KE + PE = KE + PE (1/2)mv2 + 0 = (1/2)mv2 + mgh (1/2)v2 + 0 = (1/2)v2 + gh (1/2)(45.5)2 + 0 = (1/2)v2 + 9.8(61.3) 1035.125 = (1/2)v2 +600.74 868.77 = v2 v = 29.4749 m/s v = 29.5 m/s 6 Use banked curve equations v = 21.6 m/s, r = 42.3 m 7 Use circular motion equations v = 45.5 m/s, r = 35.3 m g's felt at the bottom = ac[in g's] + 1 g g's felt at the bottom = 5.984419 + 1 g g's felt at the bottom = 6.98 g's 8 Use energy relationships v(BOTTOM) = 45.5 m/s, V(TOP) = 15.2 m/s ET(TOP) = ET(BOTTOM) KE + PE = KE + PE (1/2)mv2 + mgh = (1/2)mv2 + 0 (1/2)v2 + gh = (1/2)v2 (1/2)(15.2)2 + (9.80)h = (1/2)(45.5)2 115.52 + 9.8(h) = 1035.125 h =93.8372 m h = 93.8 m 9 Use circular motion relationships v = 15.2 m/s, r = 7.45 m g's felt at the top of a loop = ac[in g's] + 1 g g's felt at the top of a loop = 3.164498 + 1 g g's felt at the top of a loop = 2.16 g's 10 Use energy relationships to solve. Velocity at the highest point will be zero, (apogee). ET(BOTTOM) = ET(TOP) KE + PE = KE + PE (1/2)mv2 + 0 = 0 + mph (1/2)v2 = gh (1/2)(45.5)2 + = (9.80)h h =105.625 m h = 106 m If you use or find this page useful or have any comments, please contact the author so maybe he'll do more. Author: Tony Wayne "ROLLER COASTER PHYSICS" TABLE OF CONTENTS ... PHYSICS PAVILION TABLE OF CONTENTS <--PREVIOUS SECTION ... NEXT SECTION --> A special thanks to VASTfor hosting our web site.