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Unformatted text preview: wolz (cmw2833) HW #9 Antoniewicz (57420) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 4.0 points As a result of friction, the angular speed of a wheel c hanges with time according to d dt = e t , where and are constants. The angular speed changes from an initial angular speed of 2 . 67 rad / s to 1 . 27 rad / s in 3 . 92 s . Determine the magnitude of the angular acceleration after 1 . 39 s. Correct answer: 0 . 388884 rad / s 2 . Explanation: Let : = 2 . 67 rad / s , t = 0 , 2 = 1 . 27 rad / s , t 2 = 3 . 92 s , and t 3 = 1 . 39 s . The equation of motion is = e t , so 2 = e t 2 ln parenleftbigg 2 parenrightbigg = t 2 = ln parenleftbigg 2 parenrightbigg t 2 = ln parenleftbigg 1 . 27 rad / s 2 . 67 rad / s parenrightbigg 3 . 92 s = 0 . 189557 s 1 . Thus the angular acceleration at t 3 is ( t 3 ) = d dt = ( ) e t 3 = (2 . 67 rad / s) (0 . 189557 s 1 ) e (0 . 189557 s 1 ) (1 . 39 s) = . 388884 rad / s 2 bardbl vector ( t 3 ) bardbl = . 388884 rad / s 2 . 002 (part 2 of 3) 3.0 points How many revolutions does the wheel make after 2 . 91 s ? Correct answer: 0 . 950465 rev. Explanation: Let : t f = 2 . 91 s . = integraldisplay t f e t dt = integraldisplay t f e t ( dt ) = e t vextendsingle vextendsingle vextendsingle t f = ( 1 e t f ) = 2 . 67 rad / s . 189557 s 1 bracketleftBig 1 e (0 . 189557 s 1 ) (2 . 91 s) bracketrightBig = 5 . 97195 rad , so the number of revolutions is n = 2 = 5 . 97195 rad 2 = . 950465 rev . 003 (part 3 of 3) 3.0 points Find the number of revolutions it makes be fore coming to rest. Correct answer: 2 . 24178 rev. Explanation: = ( 1 e t f ) wolz (cmw2833) HW #9 Antoniewicz (57420) 2 For t = , this reduces to end = = 2 . 67 rad / s . 189557 s 1 1 rev 2 rad = 2 . 24178 rev . 004 10.0 points The net work done in accelerating a propeller from rest to an angular speed of 204 rad/s is 1286 . 8 J. What is the moment of inertia of the pro peller? Correct answer: 0 . 0618416 kg m 2 . Explanation: Let : W net = 1286 . 8 J and f = 204 rad / s . Since = 0 , W net = KE = 1 2 I ( 2 f 2 i ) = 1 2 I 2 f I = 2 W net 2 f = 2 (1286 . 8 J) (204 rad / s) 2 = . 0618416 kg m 2 ....
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 Spring '08
 Turner
 Physics, Friction

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