{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

19_ch 16 Mechanical Design budynas_SM_ch16

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

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

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 m G m P = r G r P 2 r G r P 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
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 Reﬂect 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

{[ snackBarMessage ]}

Ask a homework question - tutors are online