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Unformatted text preview: sousse (res2468) – HW 7 – Kleinman – (58225) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This HW covers Ch. 12. 001 10.0 points A large wheel is coupled to a wheel with half the diameter as shown. r 2 r How does the rotational speed of the smaller wheel compare with that of the larger wheel? How do the tangential speeds at the rims compare (assuming the belt doesn’t slip)? 1. The smaller wheel has four times the ro tational speed and the same tangential speed as the larger wheel. 2. The smaller wheel has twice the rotational speed and the same tangential speed as the larger wheel. correct 3. The smaller wheel has twice the rotational speed and twice the tangential speed as the larger wheel. 4. The smaller wheel has half the rotational speed and half the tangential speed as the larger wheel. Explanation: v = r ω The tangential speeds are equal, since the rims are in contact with the belt and have the same linear speed as the belt. The smaller wheel (with half the radius) rotates twice as fast: parenleftbigg 1 2 r parenrightbigg (2 ω ) = r ω = v 002 10.0 points A figure skater begins spinning counterclock wise at an angular speed of 3.0 π rad/s. Dur ing a 4.2 s interval, she slowly pulls her arms inward and finally spins at 8.7 π rad/s. What is her average angular acceleration during this time interval? Correct answer: 4 . 26359 rad / s 2 . Explanation: Let : ω 1 = +3 . π rad / s , Δ t = 4 . 2 s , and ω 2 = +8 . 7 π rad / s . α avg = ω 2 ω 1 Δ t = 8 . 7 π rad / s 3 π rad / s 4 . 2 s = 4 . 26359 rad / s 2 . 003 10.0 points How long does it take the second hand of a clock to move through 4.13 rad? Correct answer: 39 . 4386 s. Explanation: Let : Δ θ s = 4 . 13 rad , Δ θ = 2 π rad , and Δ t = 60 s . ω avg = Δ θ s Δ t s = 2 π rad 60 s Δ t s = Δ θ s (60 s) 2 π rad = (4 . 13 rad) (60 s) 2 π rad = 39 . 4386 s . keywords: sousse (res2468) – HW 7 – Kleinman – (58225) 2 004 (part 1 of 3) 10.0 points An athlete swings a 4 . 7 kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0 . 692 m at an angular speed of . 683 rev / s. What is the tangential speed of the ball? Correct answer: 2 . 96966 m / s. Explanation: Let : m = 4 . 7 kg , r = 0 . 692 m , and w = 0 . 683 rev / s . v t = r ω = (0 . 692 m) (0 . 683 rev / s) · parenleftbigg 2 π rad 1 rev parenrightbigg = 2 . 96966 m / s . 005 (part 2 of 3) 10.0 points What is its centripetal acceleration? Correct answer: 12 . 7441 m / s 2 . Explanation: a c = v 2 r = r ω 2 = (0 . 692 m) (0 . 683 rev / s) 2 · parenleftbigg 2 π rad rev parenrightbigg 2 = 12 . 7441 m / s 2 ....
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This note was uploaded on 04/08/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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