Homework 7, rotation 18-19-solutions.pdf

Homework 7, rotation 18-19-solutions.pdf - litzner(bnl645...

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litzner (bnl645) – Homework 7, rotation 18-19 – marder – (OlsonM302K1819 6) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A car accelerates uniformly from rest and reaches a speed of 15 . 6 m / s in 10 . 9 s. The diameter of a tire is 34 . 9 cm. Find the number of revolutions the tire makes during this motion, assuming no slip- ping. Correct answer: 77 . 5436 rev. Explanation: Let : v f = 15 . 6 m / s , r = 17 . 45 cm = 0 . 1745 m , and t = 10 . 9 s . Since v o = 0, the distance the car traveled during the time interval is s = ¯ v t = v o + v f 2 t = 1 2 v f t , so from rolling without slipping, θ = s r = v f t 2 r = (15 . 6 m / s) (10 . 9 s) 2 (0 . 1745 m) · 1 rev 2 π rad = 77 . 5436 rev . 002 10.0points A figure skater begins spinning counterclock- wise at an angular speed of 4.0 π rad/s. Dur- ing a 4.9 s interval, she slowly pulls her arms inward and finally spins at 8.0 π rad/s. What is her average angular acceleration during this time interval? Correct answer: 2 . 56457 rad / s 2 . Explanation: Let : ω 1 = +4 . 0 π rad / s , Δ t = 4 . 9 s , and ω 2 = +8 . 0 π rad / s . α avg = ω 2 - ω 1 Δ t = 8 π rad / s - 4 π rad / s 4 . 9 s = 2 . 56457 rad / s 2 . 003 10.0points Earth orbits the sun once every 365 . 25 days. Find the average angular speed of Earth about the sun. Correct answer: 1 . 99102 × 10 7 rad / s. Explanation: Let : Δ θ = 1 rev = 2 π rad and Δ t = 365 . 25 days . ω avg = Δ θ Δ t = 2 π rad 365 . 25 days · 1 day 24 h · 1 h 3600 s = 1 . 99102 × 10 7 rad / s . keywords: angular speed, orbital speed 004(part1of2)10.0points A disk with a radius of 0.1 m is spinning about its center with a constant angular speed of 10 rad/sec. What are the speed and magnitude of the acceleration of a bug clinging to the rim of the disk? 1. 10 m/s and 10 m/s 2 2. 1 m/s and 10 m/s 2 correct 3. 1 m/s and 0 m/s 2 (Disk spins at constant speed.) 4. 0.1 m/s and 1 m/s 2 Explanation:
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litzner (bnl645) – Homework 7, rotation 18-19 – marder – (OlsonM302K1819 6) 2 The speed of the bug is v = R ω = (0 . 1 m)(10 1 / s) = 1 m / s and its acceleration is a r = R ω 2 = (0 . 1 m)(10 1 / s) 2 = 10 m / s 2 . 005(part2of2)10.0points Let the turntable spin faster and faster, with constant angular acceleration vectorα . Which sketch qualitatively shows the direc- tion of the acceleration vector vectora of the bug? 1. vectora 2. vectora 3. vectora correct 4. vectora Explanation: The bug has an acceleration vectora r directed toward the center of the circle, and an accel- eration vectora t in the same direction as its velocity (counterclockwise); the net acceleration is the vector sum of vectora r and vectora t . 006 10.0points The speed of a moving bullet can be deter- mined by allowing the bullet to pass through two rotating paper disks mounted a distance 84 cm apart on the same axle. From the angular displacement 18 . 7 of the two bul- let holes in the disks and the rotational speed 997 rev / min of the disks, we can determine the speed of the bullet.
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