PHY302K-Sai-HW4-solutions

PHY302K-Sai-HW4-solutions - Oslund (hno56) Homework 4 sai...

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Oslund (hno56) – Homework 4 – sai – (58015) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 3) 10.0 points An athlete swings a 4 . 21 kg ball horizontally on the end oF a rope. The ball moves in a circle oF radius 0 . 824 m at an angular speed oF 0 . 663 rev / s. What is the tangential speed oF the ball? Correct answer: 3 . 43258 m / s. Explanation: Let : m =4 . 21 kg , r =0 . 824 m , and w =0 . 663 rev / s . v t = r ω = (0 . 824 m) (0 . 663 rev / s) · ± 2 π rad 1 rev ² = 3 . 43258 m / s . 002 (part 2 oF 3) 10.0 points What is its centripetal acceleration? Correct answer: 14 . 2993 m / s 2 . Explanation: a c = v 2 r = r ω 2 = (0 . 824 m) (0 . 663 rev / s) 2 · ± 2 π rad rev ² 2 = 14 . 2993 m / s 2 . 003 (part 3 oF 3) 10.0 points IF the maximum tension the rope can with- stand beFore breaking is 111 N, what maxi- mum tangential speed can the ball have? Correct answer: 4 . 66105 m / s. Explanation: Let : T = 111 N . The tension supplies the centripetal accel- eration, so T = m v t 2 r v t = ³ rT m = ´ (0 . 824 m) (111 N) 4 . 21 kg = 4 . 66105 m / s . keywords: 004 10.0 points A space station in the Form oF a large wheel, 324 m in diameter, rotates to provide an “ar- tifcial gravity” oF 7 m / s 2 For people located on the outer rim. ±ind the rotational Frequency oF the wheel that will produce this e²ect. Correct answer: 1 . 98501 rpm. Explanation: We want the centripetal Force to equal mg a F r = mg a = m v 2 D 2 . Solving For v , v = ³ Dg a 2 . The time in seconds to complete one revolu- tion is T = circumFerence speed = π D ³ g a D 2 . Hence, the number oF revolutions per second is f = 1 T = ³ g a D 2 π D = ³ g a 2 D π 2 = ´ 7m / s 2 2(324 m) π 2 =0 . 0330835 rps .
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Oslund (hno56) – Homework 4 – sai – (58015) 2 And the number of revolutions per minute is f = (0 . 0330835 rps) 60 s min =1 . 98501 rpm . 005 (part 1 of 2) 10.0 points At the Six Flags Great America amusement park in Gurnee, Illinois, there is a roller coaster that incorporates some of the latest design technology and some basic physics. Each vertical loop, instead of being circu- lar, is shaped like a teardrop (shown in the ±gure). The cars ride on the inside of the loop at the top, and the speeds are high enough to ensure that the cars remain on the track. The biggest loop is 35 m high, with a maxi- mum speed of 31 m / s (nearly 69 . 44 mph) at the bottom. Suppose the speed at the top is 15 m / s and the corresponding centripetal acceleration is 2 . 8 g (where g is the free-fall acceleration). What is the radius of the arc of the teardrop at the top? Correct answer: 8 . 19971 m. Explanation:
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PHY302K-Sai-HW4-solutions - Oslund (hno56) Homework 4 sai...

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