19_ch 16 Mechanical Design budynas_SM_ch16

# 19_ch 16 Mechanical Design budynas_SM_ch16 - I G 2 to the...

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414 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (2) Equivalent energy (1 / 2) I 2 ω 2 2 = (1 / 2)( I 2 ) 1 ( w 2 1 ) ( I 2 ) 1 = ω 2 2 ω 2 1 I 2 = I 2 n 2 Ans. (3) I G I P = ± r G r P ² 2 ± m G m P ² = ± r G r P ² 2 ± r G r P ² 2 = n 4 From (2) ( I 2 ) 1 = I G n 2 = n 4 I P n 2 = n 2 I P Ans. (b) I e = I M + I P + n 2 I P + I L n 2 Ans. (c) I e = 10 + 1 + 10 2 (1) + 100 10 2 = 10 + 1 + 100 + 1 = 112 reﬂected load inertia reﬂected gear inertia Ans. pinion inertia armature inertia 16-27 (a) Reﬂect I L ,
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Unformatted text preview: I G 2 to the center shaft Reect the center shaft to the motor shaft I e = I M + I P + n 2 I P + I P n 2 + m 2 n 2 I P + I L m 2 n 2 Ans . I P I M n 2 I P I P m 2 I P I L m 2 n 2 I P I G 1 I M n I P m 2 I P I L m 2 r P r G I L...
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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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