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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

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

View Full DocumentRight Arrow Icon
(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
Background image of page 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.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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 = 2 BC = 200 in / sec So angular velocity of bar CD is 3 = v C I C = 200 32 = 6 . 25 3 =-6 . 25 k rad / sec v D = 3 r I D = 6 . 25(33 . 6) = 210 in / sec v D = 210 i in / sec So angular velocity of bar DE is 3 4 = v D r DE = 210 15 = 14 4 =-14 k rad / sec 4...
View Full Document

Page1 / 4

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

This preview shows document pages 1 - 4. Sign up to view the full document.

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