Homework4asol.fall10

Homework4asol.fall10 - ME 560 Kinematics Fall Semester 2010...

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1 ME 560 Kinematics Fall Semester 2010 Homework No. 4 Wednesday, September 15th Consider the sliding-block linkage shown in Example 2.5, see Figure 2.24, page 77. The position solution for the input angle o 2 135 θ= is presented on pages 78 and 79. The answers, which can be obtained from the law of cosines and the law of sines, are 34 4 R O A 12.59 inches == and . o 3 434 104.64 Assume that the input link 2 is rotating with an angular velocity 2 5rad s ω= / clockwise and an angular acceleration 2 2 10 rad s α= / counterclockwise. The length of link 4; i.e., 4 OC 2 0 i n c h e s = and the origin of the XY reference frame is chosen to be coincident with the ground pin O 2 as shown in Figure 1. Use the method of kinematic coefficients to determine the following : (i) the relative velocity and acceleration between links 3 and 4 (denoted as 34 R ± and 34 R ±± ). (ii) the magnitude and direction of the angular velocity and angular acceleration of links 3 and 4. (iii) the velocity and acceleration of point C. Give the magnitude and direction of each vector. (iv) the unit tangent and normal vectors of point C. Show the vectors on Figure 1. (v) the radius of curvature of the path of point C. Show on Figure 1. (vi) the X and Y coordinates of the center of curvature of the path of point C. Show on Figure 1. Figure 1. The Sliding-Block Linkage.
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2 Solution to Homework Set 4. (A). The vectors for the sliding-block linkage is shown in Figure 2. Figure 2. The Vector Loop for the Sliding-Block Linkage. The vector loop Equation (VLE) for the sliding-block linkage is ?? 0 123 4 I RRR √√ + −= (1) The X and Y components of Equation (1) are 11 2 2 3 4 3 4 cos cos cos 0 RR R θ θθ + (2a) 3 4 sin sin sin 0 R + (2b) Differentiating Equations (2) with respect to the input position θ 2 gives 2 2 34 34 34 34 34 sin sin cos 0 R −+ = (3a) 2 2 34 34 34 34 34 cos cos sin 0 R (3b)
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3 Equations (3) can be written in matrix form as 34 34 34 34 22 34 34 34 34 sin θ cos θθ sin θ cos θ sin θ R cos θ R R R R ⎡⎤ = ⎢⎥ −− ⎣⎦ (4) The determinant of the coefficient matrix is 34 34 34 34 34 34 34 34 34 34 34 sin θ cos θ sin θ cos θ cos θ sin θ R DET R R R R == = (5) Using Cramer’s rule, the first-order kinematic coefficient for link 3 can be written as 3 4 3 4 2 2 34 2 2 34 34 34 sin cos cos sin sin sin cos cos R R RR DET R θ which can be written as 3 4 34 34 cos ( ) R R ′ = (6a) The first-order kinematic coefficient for links 3 and 4 can be written as 34 34 2 2 34 34 2 2 2 34 2 34 2 34 2 34 34 34 sin sin cos cos cos sin sin cos R DET R −+ which can be written as 23 4 2 3 4 34 34 sin ( ) R R ′ = (6b) Differentiating Equations (3) with respect to the input position θ 2 gives 2 2 2 34 34 34 34 34 34 34 34 34 34 34 cos 2 sin cos sin cos 0 R R R ′′ + + = (7a) 2 2 2 34 34 34 34 34 34 34 34 34 34 34 sin 2 cos sin cos sin 0 R R R + = (7b) Equations (7) can be written in matrix form as 2 34 34 34 34 2 2 34 34 34 34 34 34 2 34 34 34 34 2 2 34 34 34 34 34 34 sin θ cos cos 2 sin cos cos θ sin θ R sin 2
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Homework4asol.fall10 - ME 560 Kinematics Fall Semester 2010...

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