Question

(10%) Problem 10: Modem roller coasters have vertical loops like the one shown in the figure. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. Otheexpertta.com * 33% Part (a) What is the speed of the roller coaster in m/s at the top of the loop if the radius of curvature there is 19 m and the downward acceleration of the car is 1.5 g? v - 24.6 Deductions 29 Potential Grade Summary 989 E cos t an) cotan) asin() T acos() atan) acotano) sinh cosh() tanh cotanho Degrees Radians L ( 7 8 9 HOME 4 5 6 1012131 Submissions Attempts remaining: (2% per attempt) detailed view V BACKSPACE CLEAR Hint Submit Hints: 4% deduction per hint. Hints remaining: 2 give up! Feedback: 5% deduction per feedback. HA 33% Part (b) The beginning of this roller coaster is at the top of a high hill. If it starte from rest at the top of this hill, how high, in meters, above the top of the loop is this initial starting point? You may assume there is no friction anywhere on the track. HA 33% Part (c) If it actually starts 7.5 m higher than your answer to the previous part (y still reaches the top of the loop with the same velocity), how much energy, in joules, did it lose to friction? Its mass is 1700 kg.

Answer 1

Step a centripetal acco=a= losg a g also, Centripetal accla - lisy = v= (1.59 XR) 2 = (1.5x9.81x 190.5 - 16.7 ms Stepa In ragh- v2 vo V29h - 167= √2x9.8 xh h= 16.72 a = 14.23 m Ans 2X lops in friction = mgh = 1700x9.81X 75 – 1.25x1055 and

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