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Unformatted text preview: homework 04 – ASH, BEN – Due: Feb 14 2008, 11:00 pm 1 Question 1, chap 4, sect 5. part 1 of 1 10 points A space station in the form of a large wheel, 286 m in diameter, rotates to provide an “ artificial gravity ” of 5 . 6 m / s 2 for people located at the outer rim. What is the frequency of the rotational mo tion for the wheel to produce this effect? Correct answer: 1 . 88972 rev / min (tolerance ± 1 %). Explanation: d = v t And the frequency (where T is the period) is f = 1 T = v π d . Since bardbl vectora r bardbl = v 2 r = 2 v 2 d or v = radicalbigg a r d 2 Therefore, we have f = radicalbigg a r 2 π 2 d = radicalBigg (5 . 6 m / s 2 ) 2 π 2 (286 m) · 60 sec 1 min = 1 . 88972 rev / min . Question 2, chap 4, sect 5. part 1 of 2 10 points The radius of the Earth is about 6 . 37 × 10 6 m. a) What is the centripetal acceleration of a point on the equator? Correct answer: 0 . 0336877 m / s 2 (tolerance ± 1 %). Explanation: Basic Concept: a c = v t 2 r = r ω 2 Given: r equator = 6 . 37 × 10 6 m ω = 2 π rad / day Solution: a c = (6 . 37 × 10 6 m)(2 π rad / day) 2 · parenleftbigg 1 day 24 h parenrightbigg 2 · parenleftbigg 1 h 3600 s parenrightbigg 2 = . 0336877 m / s 2 Question 3, chap 4, sect 5. part 2 of 2 10 points b) What is the centripetal acceleration of a point at the North Pole? Correct answer: 0 m / s 2 (tolerance ± 1 %). Explanation: Solution: The Pole is 0 m from the axis of rotation, so a c = (0 m) (2 π rad / day) 2 = 0 m / day 2 = 0 m / s 2 Question 4, chap 4, sect 5. part 1 of 1 10 points Consider a toosmall space habitat that consists of a rotating cylinder of radius 4m. If a man standing inside is 2m tall and his feet are at 1 g , what is the g force at the elevtion of his head? (Do you see why projections call for large habitats?) 1. 2 g 2. 0.5 g correct 3. 0.25 g 4. 4 g Explanation: Centripetal force (and weight and g in the rotating habitat) is directly proportional to radial distance from the hub. At half the homework 04 – ASH, BEN – Due: Feb 14 2008, 11:00 pm 2 radial distance, the g force will be half that at his feet. The man will literally be light headed . (Gravitational variations of greater than 10 percent headtotoe are uncomfort able for most people.) Question 5, chap 4, sect 6. part 1 of 1 10 points A bee flies to a flower 531 m due south of its hive. The bee’s speed in still air is 0 . 69 m / s, and there is a wind blowing toward the south at 0 . 23 m / s. How long will it take the bee to travel to the flower and back to the hive? Correct answer: 1731 . 52 s (tolerance ± 1 %). Explanation: The wind helps the bee fly faster on the downwind trip than on the return trip. The time to go to the flower is given by t with = s with v with = s v bee + v wind The wind slows the bee on the return trip, so the time for the return trip is t against = s against v against = s v bee − v wind The total time is t = t with + t against Question 6, chap 4, sect 6....
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This note was uploaded on 09/29/2008 for the course PHY 303 taught by Professor Erskine/tsoi during the Spring '08 term at University of Texas at Austin.
 Spring '08
 ERSKINE/TSOI
 Gravity, Work

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