Online Homework Answers 12 (Nov. 30, 2006)

# Online Homework Answers 12 (Nov. 30, 2006) - Hood Charles...

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Hood, Charles – Homework 12 – Due: Nov 30 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A frictionless roller coaster is given an ini- tial velocity of v 0 at height h = 10 m . The radius of curvature of the track at point A is R = 32 m. The acceleration of gravity is 9 . 8 m / s 2 . R 2 3 h h h 0 v 0 A B Find the maximum value of v 0 so that the roller coaster stays on the track at A solely because of gravity. Correct answer: 15 . 7565 m / s. Explanation: let : h = 10 m and R = 32 m . At point A , the weight of coaster must be just large enough to supply the centripetal acceleration. Thus m v 2 A R = m g , or v 2 A = R g . Applying conservation of mechanical energy from the start to point A , 1 2 m v 2 0 + m g h = 1 2 m v 2 A + m g 2 h 3 v 2 0 = v 2 A - 2 g h 3 = R 2 g 2 - 2 g h 3 v 0 = r R g - 2 g h 3 = s g R - 2 h 3 = s (9 . 8 m / s 2 ) 32 m - 2 (10 m) 3 = 15 . 7565 m / s . 002 (part 2 of 2) 10 points Using the value of v 0 calculated in question 1, determine the value of h 0 that is necessary if the roller coaster just makes it to point B. Correct answer: 22 . 6667 m. Explanation: If the speed of the coaster is to be zero at point B, conservation of mechanical energy from the start to point B gives 0 + m g h 0 = 1 2 m v 2 0 + m g h = 1 2 m g R - 2 h 3 + m g h h 0 = R 2 + 2 h 3 = 32 m 2 + 2 (10 m) 3 = 22 . 6667 m .

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