Online Homework Answers 12 (Nov. 30, 2006)

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

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Unformatted text preview: 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 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 v A B Find the maximum value of v 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 µ v 2 A R ¶ = mg , or v 2 A = Rg . Applying conservation of mechanical energy from the start to point A , 1 2 mv 2 + mg h = 1 2 mv 2 A + mg µ 2 h 3 ¶ v 2 = v 2 A- 2 g h 3 = R 2 g 2- 2 g h 3 v = r Rg- 2 g h 3 = s g µ 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 calculated in question 1, determine the value of h 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 + mg h = 1 2 mv 2 + mg h = 1 2 mg µ R- 2 h 3 ¶ + mg h h = R 2 + 2 h 3 = 32 m 2 + 2(10 m) 3 = 22 . 6667 m ....
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This note was uploaded on 03/22/2009 for the course PHYS 105 taught by Professor Ken during the Fall '08 term at NJIT.

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

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