09_InstSolManual_PDF_Part4

09_InstSolManual_PDF_Part4 - a u 5 u 1 v t 1 1 2 a t 2 v 5...

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Then, with and the equation gives Reflect: The angular acceleration is negative because the wheel is slowing down. 9.18. Set Up: Let the direction of rotation of the flywheel be positive. Solve: (a) gives At the start of the interval the flywheel was rotating at in the opposite direction. (b) gives 9.19. Set Up: Let the direction of rotation of the flywheel be positive. Solve: gives Reflect: At the end of the 4.00 s interval, which checks. 9.20. Set Up: Eq. 9.7 is Eq. 9.11 is Solve: so and 9.21. Set Up: Solve: Reflect: requires that be in 9.22. Set Up: where must be in Solve: (a) (b) v5 v r 5 0.600 m / s 0.0400 m 5 15 rad / s 5 143 rpm v 5 1 0.0750 m 21 64.9 rad / s 2 5 4.87 m / s 620 rpm 5 64.9 rad / s. 1 rpm 5 2 p 60 rad / s. rad / s. v v 5 r v , rad / s. v v 5 r v v 5 r v5 1 6.5 3 10 2 3 m 21 130.9 rad / s 2 5 0.85 m / s v5 1250 rpm 5 130.9 rad / s r 5 6.5 mm 5 6.5 3 10 2 3 m. v 2 2v 0 2 5 2 a 1 u2u 0 2 . u5u 0 1 1 2 a 1 v2v 0 21 v1v 0 2 5u 0 1 1 2 a 1 v 2 2v 0 2 2 . u5u 0 1v 0 1 v2v 0 a 2 1 1 2 a 1 v2v 0 a 2 2 5u 0 1 1 2 a 1 v2v 0 21 2 v 0 1v2v 0 2 t 5 v2v
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Unformatted text preview: a u 5 u 1 v t 1 1 2 a t 2 . v 5 v 1 v t . u 2 u 5 1 v 1 v 2 2 t 5 1 10.5 rad / s 1 19.5 rad / s 2 2 1 4.00 s 2 5 60.0 rad, v 5 v 1 a t 5 19.5 rad / s. v 5 u 2 u t 2 1 2 a t 5 60.0 rad 4.00 s 2 1 2 1 2.25 rad / s 2 21 4.00 s 2 5 10.5 rad / s. u 2 u 5 v t 1 1 2 a t 2 a 5 v 2 v t 5 108 rad / s 1 27 rad / s 4.00 s 5 33.8 rad / s 2 . v 5 v 1 v t 27.0 rad / s, v 5 2 1 u 2 u 2 t 2 v 5 2 1 162 rad 2 4.00 s 2 108 rad / s 5 2 27.0 rad / s u 2 u 5 1 v 1 v 2 2 t u 2 u 5 1 v 1 v 2 2 t 5 1 8.33 rev / s 1 2 2 1 75.0 s 2 5 312 rev. t 5 v 2 v a 5 2 8.33 rev / s 2 0.111 rev / s 2 5 75.0 s. v 5 v 1 v t v 5 8.33 rev / s v 5 9-4 Chapter 9...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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