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Unformatted text preview: armington (kma786) – hw0210 – fiete – (57165) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A space station in the form of a large wheel, 341 m in diameter, rotates to provide an “ artificial gravity ” of 3 . 9 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 . 44425 rev / min. 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 (3 . 9 m / s 2 ) 2 π 2 (341 m) · 60 sec 1 min = 1 . 44425 rev / min . 002 10.0 points A mass slides with negligible friction on a circular track of 1 m radius oriented vertically. Its speed at the position shown in the figure is 3.13 m/s. v At the position shown in the figure, which arrow best represents the direction of the ac- celeration of the mass? The acceleration of gravity is 9.8 m/s 2 and no external forces act on the system. 1. 2. 3. 4. 5. 6. The mass is traveling at a constant veloc- ity, so it has no acceleration. 7. 8. 9. correct Explanation: The magnitude of the centripetal accelera- tion is a r = v 2 r = (3 . 13 m / s) 2 1 m ≈ 9 . 8 m / s 2 , acting inward ( − ˆ r ) with gravity acting down- ward ( − ˆ k ). a a r g 003 10.0 points What angular speed (in revolutions per minute) is needed for a centrifuge to produce armington (kma786) – hw0210 – fiete – (57165) 2 an acceleration of 1260 times the gravitational acceleration 9 . 8 m / s 2 at a radius of 5 . 22 cm ? Correct answer: 4644 . 45 rev / min. Explanation: The centripetal acceleration in the cen- trifuge, a c = ω 2 r , has to be equal to 1260 times the free-fall acceleration g = 9 . 8 m / s 2 . Therefore 1260 g = ω 2 r , so ω = radicalbigg a c r = radicalbigg 1260 g r = radicalBigg 12348 m / s 2 5 . 22 cm · 1 m 100 cm = 486 . 366 rad / s = 60 s / min 2 π rad...
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This note was uploaded on 02/21/2011 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas.

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solution_pdf - armington(kma786 – hw0210 – fiete...

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