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Hw 7 - Bell Evan Homework 7 Due Mar 2 2005 11:00 pm Inst...

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Bell, Evan – Homework 7 – Due: Mar 2 2005, 11:00 pm – Inst: Furlan 1 This print-out should have 34 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 record has an angular speed of 39 . 4 rev / min. What is its angular speed? Correct answer: 4 . 12596 rad / s. Explanation: 1 rev = 6 . 28319 rad, and 1 min = 60 s , Therefore, ω 1 = (39 . 4 rev / min) 2 π rad rev 1 min 60 s = 4 . 12596 rad / s . 002 (part 2 of 2) 10 points Through what angle, in radians, does it rotate in 1 . 42 s? Correct answer: 5 . 85887 rad. Explanation: θ = ω t = (4 . 12596 rad / s) (1 . 42 s) = 5 . 85887 rad . 003 (part 1 of 3) 10 points A ball at the end of a string of length 2 . 1 m rotates at a constant speed in a horizontal circle. It makes 5 . 7 rev / s. What is the period of the ball’s motion? Correct answer: 0 . 175439 s. Explanation: The period T is the time for one complete revolution and in this case the ball makes 5 . 7 rev / s complete revolutions per second, so T = 1 rev 5 . 7 rev / s = 0 . 175439 s . 004 (part 2 of 3) 10 points What is the frequency of the motion? Correct answer: 5 . 7 Hz. Explanation: f = 1 0 . 175439 s = 5 . 7 Hz . 005 (part 3 of 3) 10 points What is the ball’s angular velocity? Correct answer: 35 . 8142 rad / s. Explanation: ω = 2 π rad T = 2 π 0 . 175439 s = 35 . 8142 rad / s . 006 (part 1 of 3) 10 points A(n) 54 . 5 kg child stands at the rim of a merry-go-round of radius 1 . 51 m, rotating with an angular speed of 1 . 44 rad / s. The acceleration of gravity is 9 . 8 m / s 2 . What is the child’s centripetal accelera- tion? Correct answer: 3 . 13114 m / s 2 . Explanation: a r = r ω 2 = (1 . 51 m)(1 . 44 rad / s) 2 = 3 . 13114 m / s 2 . 007 (part 2 of 3) 10 points What minimum force between her feet and the floor of the carousel is required to keep her in the circular path? Correct answer: 170 . 647 N. Explanation: F = m a r = (54 . 5 kg)(3 . 13114 m / s 2 ) = 170 . 647 N . 008 (part 3 of 3) 10 points
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Bell, Evan – Homework 7 – Due: Mar 2 2005, 11:00 pm – Inst: Furlan 2 What minimum coefficient of static friction is required? Correct answer: 0 . 319504 . Explanation: We know the centripetal force is equal to the force of friction. Therefore, m a r = μ m g Also, the normal force, N , is equal to m g . Thus, μ = a r g = 3 . 13114 m / s 2 9 . 8 m / s 2 = 0 . 319504 . 009 (part 1 of 1) 10 points A small wheel of radius 1 . 8 cm drives a large wheel of radius 12 . 9 cm by having their circumferences pressed together. If the small wheel turns at 410 rad / s, how fast does the larger one turn? Correct answer: 57 . 2093 rad / s. Explanation: Basic Concept: Linear and angular ve- locity are related by v = r ω , where ω is in radians per unit time. Solution: The rim of the small wheel moves at v 1 = r 1 ω 1 . Since their circumferences are in contact, the rim of the larger one is forced to travel at the same linear speed, and v 2 = r 2 ω 2 r 1 ω 1 = r 2 ω 2 . Thus ω 2 = r 1 ω 1 r 2 = (1 . 8 cm) (410 rad / s) (12 . 9 cm) = 57 . 2093 rad / s . 010 (part 1 of 3) 10 points A wheel has a radius of 4 . 77 m.
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