Practice 09 Solutions - practice 09 GADHIA, TEJAS Due: Mar...

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Unformatted text preview: practice 09 GADHIA, TEJAS Due: Mar 21 2007, 4:00 am 1 Question 1 part 1 of 2 10 points A record has an angular speed of 31 . 7 rev / min. What is its angular speed? Correct answer: 3 . 31962 rad / s (tolerance 1 %). Explanation: 1 rev = 6 . 28319 rad, and 1 min = 60 s , Therefore, 1 = (31 . 7 rev / min) parenleftbigg 2 rad rev parenrightbiggparenleftbigg 1 min 60 s parenrightbigg = 3 . 31962 rad / s . Question 2 part 2 of 2 10 points Through what angle, in radians, does it rotate in 1 . 27 s? Correct answer: 4 . 21592 rad (tolerance 1 %). Explanation: = t = (3 . 31962 rad / s) (1 . 27 s) = 4 . 21592 rad . Question 3 part 1 of 2 10 points A racing car travels on a circular track of radius 240 m. The car moves with a constant linear speed of 38 . 2 m / s. Find its angular speed. Correct answer: 0 . 159167 rad / s (tolerance 1 %). Explanation: The linear speed v and the angular speed are related by, v = R = v R . Question 4 part 2 of 2 10 points Find the magnitude of its acceleration. Correct answer: 6 . 08017 m / s 2 (tolerance 1 %). Explanation: If the car is moving at a constant speed, there is no tangential acceleration, thus the acceleration is purely radial, a r = v 2 R . Question 5 part 1 of 1 10 points A horizontal disk with a radius of 17 m ro- tates about a vertical axis through its center. The disk starts from rest and has a constant angular acceleration of 3 . 1 rad / s 2 . At what value of t will the radial and tan- gential components of the linear acceleration of a point on the rim of the disk be equal in magnitude? Correct answer: 0 . 567962 s (tolerance 1 %). Explanation: The tangential and radial accelerations are given by a t = r a r = 2 r Equating the two gives us the expression 2 = = (1) Now consider the equation = - t (2) practice 09 GADHIA, TEJAS Due: Mar 21 2007, 4:00 am 2 Plug the Eq. (1) into (2) and solve for t , where = t = 0. t = = = 1 = 1 radicalbig (3 . 1 rad / s 2 ) = 0 . 567962 s Question 6 part 1 of 1 10 points The tub of a washer goes into its spin- dry cycle, starting from rest and reaching an angular speed of 7 . 9 rev / s in 10 . 9 s . At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 16 . 1 s . Through how many revolutions does the tub turn? Assume Constant angular acceleration while it is starting and stopping. Correct answer: 106 . 65 rev (tolerance 1 %). Explanation: We will break the motion into two stages: (1) an acceleration period and (2) a deceleration period. While speeding up, 1 = t = 0 + 2 t 1 = 1 2 t 1 = 1 2 (7 . 9 rev / s) (10 . 9 s) = 43 . 055 rev ....
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Practice 09 Solutions - practice 09 GADHIA, TEJAS Due: Mar...

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