# hw21 - Again using similarity x 8 = 12 5 x = 19 2 in And x...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 21 (3.47, 3.51) 3.47 100 rpm = 100 ± 2 π rad 60 sec ² = 10 . 471976 rad / sec 30 mph = 30 ± 88 60 ² f t / sec = 44 f t / sec (a) v p = v A + ω × r AP = 44 ˆ i + 10 . 471976 ˆ k × 4 ˆ j = 44 ˆ i - 41 . 88790 ˆ i = 2 . 11209 ˆ i f t / sec (b) I A P 44 ϖ With respect to the instantaneous center I ± , the center of the disk, has a tangential velocity of 44 f t / sec . ³ ³ ³ r I ± A ³ ³ ³ = v A ω = 44 10 . 471976 = 4 . 201 f t vertically above A This is equivalent to 0.201 ft vertically above P. 1

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(c) I 4 4.201 5.80 Q 4.201 4 5.80 I S I R 8.20 v Q = r I ± Q ω = 5 . 80(10 . 471976) ± 4 . 2 5 . 8 ˆ i - 4 5 . 8 ˆ j ² = 60 . 737 ± 4 . 2 5 . 8 ˆ i - 4 5 . 8 ˆ j ² v S = r I ± S ω = 5 . 80(10 . 471976) ± 4 . 2 5 . 8 ˆ i + 4 5 . 8 ˆ j ² = 60 . 737 ± 4 . 2 5 . 8 ˆ i + 4 5 . 8 ˆ j ² v R = r I ± R ω = 8 . 20(10 . 471976) ˆ i = 85 . 870 ˆ i 3.51 Since there is no slip, the speed for the discs at the point of contact is identical. r 1 ω 1 = r 2 ω 2 ω 2 = 2(100) / 10 = 20 rad / sec ω 2 = 20 ˆ k rad / sec 2
x B D I 3 12 5 3 4 C 10 8 6 8 x Using similarity of triangles and the slope (3/4) we can Fll in the required values in the Fgure.

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Unformatted text preview: Again using similarity x 8 = 12 5 x = 19 . 2 in And, x x + 6 = I ± C I ± C + BC I ± C = 32 in I ± D = 25 . 6 in Velocity of C is given by v C = ω 2 BC = 200 in / sec So angular velocity of bar CD is ω 3 = v C I ± C = 200 32 = 6 . 25 ω 3 =-6 . 25 ˆ k rad / sec ∴ v D = ω 3 r I ± D = 6 . 25(33 . 6) = 210 in / sec v D = 210 ˆ i in / sec So angular velocity of bar DE is 3 ω 4 = v D r DE = 210 15 = 14 ω 4 =-14 ˆ k rad / sec 4...
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hw21 - Again using similarity x 8 = 12 5 x = 19 2 in And x...

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