HW #9-solutions - wolz (cmw2833) HW #9 Antoniewicz (57420)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: wolz (cmw2833) HW #9 Antoniewicz (57420) 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 (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 ....
View Full Document

Page1 / 7

HW #9-solutions - wolz (cmw2833) HW #9 Antoniewicz (57420)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online