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10_InstSolManual_PDF_Part7 - Du 5 1 25.0 N 21 4.40 m 21 898...

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10.17. Set Up: The free-body diagram for a log is given in Figure 10.17. Use the coordinates shown and take counterclockwise rotation to be positive. Figure 10.17 Solve: (a) Apply to the motion of the center of mass: Apply to the rotation about the center of mass: and Combine these two equations to eliminate (b) No slipping means the friction is static. 10.18. Set Up: where must be in Solve: 10.19. Set Up: (a) Let the direction of rotation of the merry-go-round be positive. Apply to the merry- go-round to calculate and then use a constant acceleration equation to calculate after 20.0 s. (b) Calculate from a constant acceleration equation. (c) Solve: (a) gives (b) (c) Reflect: The power applied when at the end of the 20.0 s time interval, is At and and hence P increase linearly in time, so This agrees with our answer in part (c). P av 5 P f 2 P i 2 5 9.88 3 10 3 W 2 5 4.94 3 10 3 W. v P 5 0. v 5 0 t 5 0, P 5 tv 5 1 25.0 N 21 4.40 m 21 89.8 rad / s 2 5 9.88 3 10 3 W. v 5 89.8 rad / s, P av 5 W D t 5 9.88 3 10 4 J 20.0 s 5 4.94 3 10 3 W W 5 t Du 5
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Unformatted text preview: Du 5 1 25.0 N 21 4.40 m 21 898 rad 2 5 9.88 3 10 4 J u 2 u 5 v t 1 1 2 a t 2 5 1 2 1 4.49 rad / s 2 21 20.0 s 2 2 5 898 rad v 5 v 1 a t 5 1 1 4.49 rad / s 2 21 20.0 s 2 5 89.8 rad / s a 5 g t I 5 FR I 5 1 25.0 N 21 4.40 m 2 24.5 kg # m 2 5 4.49 rad / s 2 . g t 5 I a P av 5 W D t . Du 5 u 2 u W 5 t Du . v a g t 5 I a P 5 1 4.30 N # m 21 503 rad / s 2 5 2.16 3 10 3 W 5 2.90 hp 1 hp 5 746 W. 1 4800 rev / min 2 1 2 p rad 1 rev 21 1 min 60 s 2 5 503 rad / s. rad / s. v P 5 tv , f 5 1 2 Ma cm 5 1 2 1 575 kg 21 4.86 m / s 2 2 5 1.40 3 10 3 N. a cm 5 g sin f 1.5 5 1 9.80 m / s 2 2 sin 48.0° 1.5 5 4.86 m / s 2 Mg sin f 2 1 2 Ma cm 5 Ma cm f : f 5 1 2 Ma cm . fR 5 A 1 2 MR 2 B 1 a cm R 2 fR 5 I cm a g t 5 I cm a Mg sin f 2 f 5 Ma cm . g F x 5 ma x y f n Mg x a cm a f 1 a cm 5 R a . R 5 0.600 m. I cm 5 1 2 MR 2 . Dynamics of Rotational Motion 10-7...
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