{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

315 HW_6 - 6 θ = F t J ¨ θ k‘ 6 17 k‘ 2 36 θ = F t...

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

View Full Document Right Arrow Icon
North Carolina State University Department of Mechanical and Aerospace Engineering MAE 315-1 Dynamics of Machines Summer I, 2009 Homework #6: Due date June 17, 2010 Rao’s Textbook problems: 5.28, 5.29, 5.34, 5.41, 5.48, 5.49, 5.57. Answers: 5.28: ω 1 = 0 . 5176 q k t 1 J 0 2 = 1 . 9319 q k t 1 J 0 r 1 = 1 . 3661 ,r 2 = - 0 . 3661. 5.29: ω 2 1 2 2 = k 2 2 m + k 1 + k 2 m o r k 2 2 m + k 1 + k 2 m o · 2 - 2 k 1 k 2 mm o . 5.34: ω 1 = 0 . 5904 q k m 2 = 1 . 4668 q k m . 5.41: m ¨ x + 3 kx + k‘
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6 θ = F ( t ) J ¨ θ + k‘ 6 + 17 k‘ 2 36 θ = F ( t ) ‘ 3 where J = m‘ 2 12 . 5.48: X 1 =-(40 . 004 + 0 . 0192 j ) × 10-4 in,X 2 = (0 . 922 + 0 . 295 j ) × 10-4 in . 5.49: k 2 m 2 = ω 2 . 5.57: ω 1 = 0 ,ω 2 = q m 1 k + m 2 k + kJ o /r 2 m 1 m 2 + m 1 J /r 2 . 1...
View Full Document

  • Summer '08
  • WU
  • University of North Carolina at Chapel Hill, North Carolina State University, Mechanical and Aerospace Engineering, 0.0192j, 0.295j

{[ snackBarMessage ]}

Ask a homework question - tutors are online