10_InstSolManual_PDF_Part29

10_InstSolManual_PDF_Part29 - 2 B1 a / R c 2 2 5 1 4 1 m p...

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Figure 10.74 Solve: (a) Treating both marbles as a single object, gives axis at P and counter- clockwise positive gives gives (b) For the bottom marble, gives and Or, gives and which checks. 10.75. Set Up: (a) The pulley has moment of inertia with and The cylinder has moment of inertia with and Apply conservation of energy. Take downward and let at the final position of the bucket. and (b) and (c) The free- body-diagram for the bucket is given in Figure 10.75. Figure 10.75 Solve: (a) Conservation of energy gives For the bucket, and and so the conservation of energy expression becomes This gives We can rewrite the rotational kinetic energies in terms of a : Then and v 5 Å 4 m b gy i 2 m b 1 m p 1 m c 5 Å 4 1 3.00 kg 21 9.80 m / s 2 21 2.00 m 2 6.00 kg 1 2.00 kg 1 5.00 kg 5 4.25 m / s m b gy i 5 1 4 1 2 m b 1 m p 1 m c 2 v 2 1 2 I p v p 2 1 1 2 I c v c 2 5 1 2 A 1 2 m p R p 2 B1 a / R p 2 2 1 1 2 A 1 2 m c R c
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Unformatted text preview: 2 B1 a / R c 2 2 5 1 4 1 m p 1 m c 2 a 2 m b gy i 5 1 2 I p v p 2 1 1 2 I c v c 2 1 1 2 m b v 2 . U i 5 K f . U f 5 K i 5 y f 5 0. y i 5 2.00 m U i 1 K i 5 U f 1 K f . x y m b g T b a a 5 R c a c . a 5 R p a p y 5 1 y R c 5 0.400 m. m c 5 5.00 kg I c 5 1 2 m c R c 2 , R p 5 0.200 m. m p 5 2.00 kg I p 5 1 2 m p R p 2 , n 5 F B 2 w cos 30 5 1.47 N 2 0.735 N cos 30 5 0.848 N, w 1 n cos 30 F B 5 g F y 5 n 5 F A sin 30 5 0.848 N. F A 5 n sin 30 g F x 5 F A 5 F C 5 0.424 N. g F x 5 F C 5 w 2 cos 30 5 0.424 N. F C 1 2 R cos u 2 2 wR 5 0. t 5 0, F B 5 2 w 5 1.47 N. g F y 5 w C F C F B F A n n R w P A B u Dynamics of Rotational Motion 10-29...
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