ME 242 Exam 1 Solutions

ME 242 Exam 1 Solutions - ME 242 Dynamics March 6 2009...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ME 242 Dynamics March 6, 2009 Coordinate Transformation i j k 0 -1 Eric Wang b1 -sin cos 0 b2 cos sin b3 0 0 Problem #5 Problem #6 P vA = vB + vA /B vA /B = vA - vB b1 v A = -v A i v B = v B b1 b2 j er i e FBD = IRD R P P 6 unknowns, 6 eqns mar Re + N j = m( - r 2 )e r + m(2 + r )e r r m( - r 2 )e r + m(2 + r )e = mi r r x ma N P v p = re r + r e = x i m x Re + N j = mar e r + ma e = mi x 1 Alternative solution (fastest) Re + N j = m( - r 2 )e r + m(2 + r )e r r r= 0.5 = .577m cos Re + N j = m( - r 2 )e r + m(2 )e r r m( - r 2 )e r + m(2 )e = mi r r x v p = re r + r e = x i i er e sin cos j cos -sin 0.5 tan = r tan cos r 2 = r tan + r = r 2 + 2 tan r cos2 r= m( - r 2 )e r + m(2 )e = m(sin e r + cos e ) r r x e = x (sin e r + cos e ) v p = re r + r v p = re r + r e = x (sin e r + cos e ) x (sin ) = r x (cos ) = r r = r tan (cos ) = 2 x r r = r tan R P m( - r 2 )e r + m(2 )e = m(sin e r + cos e ) r r x (sin ) = - r 2 x r (cos ) = 2 x r 2r 2 tan cos 2r 2 tan F = m i cos 2r 2 tan R cos = m cos 2 R = 2mr tan e = x N 2 ...
View Full Document

{[ snackBarMessage ]}

Page1 / 2

ME 242 Exam 1 Solutions - ME 242 Dynamics March 6 2009...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online