19_ch 16 Mechanical Design budynas_SM_ch16

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online