HW1Solutions_Winter2012

HW1Solutions_Winter2012 - ENME361 Vibrations Controls and...

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ENME361: Vibrations, Controls and Optimization I January 6, 2012 Winter 2012 Solutions for ENME361 HW 1 1.5 In Figure E1.5, a slider of mass M r is located on a bar whose angular displacement in the plane is described by the coordinate . The motion of the slider from the pivot point is measured by the coordinate r 1 . The acceleration due to gravity acts in a direction normal to the plane of motion. Assume that the point O is fixed in an inertial reference frame and determine the absolute velocity and absolute acceleration of the slider. Solution Choose unit vectors e 1 and e 2 fixed to the slider as shown in the figure. Then, the position vector from point O to the slider is 11 m r re The absolute velocity is then given by     1 1 1 1 1 1 1 1 1 1 m dd r r r r r dt dt e V e e e e ( t ) O 2 e r 1 Noting that k , where k is the unit vector that points out of the plane, we find that 1 1 1 1 1 1 1 2 m r r r r  V e k e e e The absolute acceleration is found from           12 1 1 1 2 1 1 1 1 2 1 2 1 1 1 1 1 1 2 1 2 1 2 2 1 1 1 1 1 2 2 m d d d r r r r r r r dt dt dt r r r r r r r r r  ee a e e e e e e e e e e  1.10 Determine the linear momentum for the system shown in Figure E1.5 and discuss if it is conserved. Assume that the mass of the bar is bar M and the distance from the point O to the center of the is bar L . Solution The linear momentum of this system is given by bar slider  p p p where 2 bar bar bar bar bar M M L  p V e
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where bar V is the velocity of the center of mass of the bar. Making use of the slider velocity determined in the solution to Exercise 1.5, we arrive at   1 1 1 2 slider r M r r   p e e Thus,   1 1 2 r bar bar r M r M L M r p e e Because reaction forces exist at point O , we conclude that the linear momentum is not conserved. 1.14 For the system shown in Figure E1.13, construct the system kinetic energy. Solution We make use of Eq. (1.23) to determine the system kinetic energy. The position vector to the center of mass G can be written as:   sin cos m x l l  r i j And the velocity of the center of mass G can be obtained as,   cos sin m x l l V i j Then, from Eq. (1.23), the kinetic energy is l G m , J G O O j i     2 2 2 2 11 22 2 cos m m G G T m J m x l xl J
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HW1Solutions_Winter2012 - ENME361 Vibrations Controls and...

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