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View History Bookmarks Window Help use11fLtheexpertta.com 0 and Mastering Module 8: Uniform Circular Motion and Satellites and Kepler's Laws: An Argum. The Expert TAI Human-like Grading, Au... Class Management Help W 8 Begin Date: 10/2/2017 11:00:00 AM- Due Date: 10/16/2017 11:00:00 AM End Date: 10/16/2017 11:00:00 AM (10%) Problem 7: A bobsled turn banked at 76° is taken at 33.65 m/s. 33% Part (a) what is the radius of the bobsled turn in m, assuming it is ideally banked and there s no friction between the ice and the bobsled? Grade Summary Potential sinO coso cotanO asinacos0 atanO acotanO sinh0 cosh0 tanh0 cotanhO Submissions Attempts remainin (5% per attempt) detailed view 1 2 3 0 O Degrees Radians Submit Hint I give up Hints:--deduction per hint. Hints remaining: 4 Feedback: 1 deduction per feedback 33% Part (b) Calculate the centripetal acceleration in m/s2. 33% Part (c) Calculate the centripetal acceleration as a multiple of g. 16

Explanation / Answer

given, banking angle, theta = 76 deg

speed v = 33.65 m/s

a. let the radius of turning be r and the coefficient of kinetic friction = 0

then from force balance

mv^2/r = Nsin(theta)

mg = Ncos(theta) ( where N is the normal reaction force of the ropad on the car)

tan(theta) = v^2/rg

tan(76) = 33.65^2/r*9.81

r = 28.778 m

b. centripital acceleration = v^2/r = 33.65^2/28.778 = 39.34576 m/s/s

c. centripital acceleration = 39.34576*g/9.81 = 4.01078g

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