44_ch 04 Mechanical Design budynas_SM_ch04

44_ch 04 Mechanical Design budynas_SM_ch04 - W 2 and the...

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Chapter 4 113 4-82 Choose t ± = 0 at the instant of impact. At this instant, v 1 = (2 gh ) 1 / 2 . Using momentum, m 1 v 1 = m 2 v 2 . Thus W 1 g (2 gh ) 1 / 2 = W 1 + W 2 g v 2 v 2 = W 1 (2 gh ) 1 / 2 W 1 + W 2 Therefore at t ± = 0, y = 0,and ˙ y = v 2 Let W = W 1 + W 2 Because the spring force at y = 0 includes a reaction to W 2 , the D.E. is W g ¨ y =− ky + W 1 With ω = ( kg / W ) 1 / 2 the solution is y = A cos ω t ± + B sin ω t ± + W 1 / k ˙ y =− A ω sin ω t ± + B ω cos ω t ± At t ± = 0, y = 0 A =− W 1 / k At t ± = 0, ˙ y = v 2 v 2 = B ω Then B = v 2 ω = W 1 (2 gh ) 1 / 2 ( W 1 + W 2 )[ kg / ( W 1 + W 2 )] 1 / 2 We now have y =− W 1 k cos ω t ± + W 1 ± 2 h k ( W 1 + W 2 ) ² 1 / 2 sin ω t ± + W 1 k Transforming gives y = W 1 k ³ 2 hk W 1 + W 2 + 1 ´ 1 / 2 cos( ω t ± φ ) + W 1 k where φ is a phase angle. The maximum deflection of
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Unformatted text preview: W 2 and the maximum spring force are thus W 1 ky y W 1 ± W 2 Time of release ² 0.05 ² 0.01 2 1 0.01 0.05 Time t ³ Speeds agree Inflection point of trig curve (The maximum speed about this point is 29.8 in/s.) Equilibrium, rest deflection During contact Free fall y max y...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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