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# test2asol.fall10 - ME 560 Kinematics Class Test 2 Friday...

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ME 560 Kinematics. Class Test 2. Fall Semester 2010 Friday, October 1st Name of Student : ________________________________ OPEN BOOK AND OPEN NOTES Problem (50 points). For the inverted slider-crank mechanism in the position shown in Problem 3.15, see Figure P3.15, page 160, the angular velocity and the angular acceleration of the input link 2 are 2 60 rad s ω= / counterclockwise and 2 2 5rad s α= / clockwise, respectively. Using the method of kinematic coefficients , determine: (i) The angular velocity and the angular acceleration of the coupler link 3. Give the magnitude and the direction of each vector. (ii) The velocity and the acceleration of the coupler point B. Give the magnitude and the direction of each vector. (iii) The unit tangent vector and the unit normal vector to the path of point B. (iv) The radius of curvature of the path of point B. (v) The X and Y coordinates of the center of curvature of the path of point B.

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2 Solution to Class Test 2. The known position variables are o 2 150 , θ= o 1 0, 1 R 125 mm, = and 2 R7 5 m m . = The remaining position variables can be determined from trigonometry. Using the law of cosines gives 22 2 2 34 4 1 2 1 2 4 2 R( A O ) R R 2 R R c o s O O A == + (1) Therefore 2 34 R (125 mm) (75 mm) 2(125 mm)(75 mm)cos150 =+ ° (2a) that is 34 R 193.62 mm = (2b) Using the law of sines gives 34 1 24 4 2 R R sin O AO sin O O A = ∠∠ (3a) Rearranging this equation gives 1 4 2 34 R (125 mm) sin O AO sin O O A sin150 R (193.62 mm) ∠= ° (3b) that is O AO 18.83 (3c) Therefore 33 4 330 18.83 348.83 θ =θ = °+ °= ° (3d) or 4 (30 18.83 ) 11.17 θ=θ =− °− °=− ° (3e) (i) The angular velocity and the angular acceleration of link 3. The vectors for the mechanism are shown in Figure 1. Figure 1. The vectors for the inverted slider-crank mechanism. The vector loop equation (VLE) for the mechanism can be written as 1 I? ? RR R 0 23 4 √√ + −= (4) Note that 34 R is the vector from point A (which is fixed in link 3) to the ground pin 4 O . Therefore, the time rate of change of the magnitude of 34 R is the relative velocity between these two points and the time rate of change of the direction (angle) is the angular velocity of link 3 (and link 4).
3 The X and Y components of Equation (4) are 22 3 43 11 cos cos cos 0 RR R + −= θ θθ (5a) and 3 sin sin sin 0 R + (5b) Differentiating Equations (5a) and (5b) with respect to the input position θ 2 gives 3 3 3 4 3 sin sin cos 0 R −− + = (6a) and 2 2 34 3 3 34 3 cos cos sin 0 R ++ = (6b) Then writing Equations (6) in matrix form gives 34 3 3 3 34 3 3 34 sin cos sin cos sin cos R R R ⎡⎤ = ⎢⎥ ⎣⎦ (7) The determinant of the coefficient matrix in Equation (7) can be written as 34 3 3 34 34 3 3 sin cos 193.62 mm cos sin == = R DET R R (8) Using Cramer’s rule, the first-order kinematic coefficient of link 3 and 4 can be written as 3 3 3 3 34 sin cos cos sin cos ( ) R R R DET R (9a) or as 3

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test2asol.fall10 - ME 560 Kinematics Class Test 2 Friday...

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