ME 242 Exam 1 Solutions

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

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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 ...
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## This note was uploaded on 09/19/2009 for the course ME 242 taught by Professor Kam during the Spring '06 term at Nevada.

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

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