rigid_kinematics

rigid_kinematics - the angular velocity and angular...

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KINEMATICS OF RIGID BODIES Reference : Any undergraduate text in dynamics. Suppose that you need to relate the velocity and acceleration of a point B on a rigid body to a second point A on the SAME rigid body. A B r B/A Let ϖ and α represent the angular velocity and angular acceleration, respectively, of the rigid body. Then if v A and a A are the velocity and acceleration of point A, the velocity and acceleration of point B are given by: v B = v A + ϖ × r B/A a B = a A + α × r B/A + ϖ × ϖ × r B/A ( 29 where r B/A is the position vector FROM point A TO point B, as shown in the figure above.
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Example Point A is traveling horizontally with a speed of 3 ft/sec and has a vertical acceleration of 4ft/ sec 2 , as shown in the figure below. At the position shown, the bar to which point A is attached is at an angle of θ = 30° with θ increasing at a rate of ˙ θ= 0.2rad/sec with ˙ θ decreasing such that ˙ ˙ θ= - 0.1rad/ sec 2 . 2 ft θ A B v A a A i j With this information, find the velocity and acceleration of point B. Solution Since the angle θ represents a rotation of the bar about the z-axis, we can write
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Unformatted text preview: the angular velocity and angular acceleration of the bar as: = k = 0.2k = k = -0.1k Using the above information we can now write: v B = v A + r B/A = 3i + 0.2k ( 29 2cos i + 2sin j ( 29 = 3i + 0.2 ( 29 2cos ( 29 j + 0.2 ( 29 2sin ( 29-i ( 29 = 3-0.4sin30 ( 29 i + 0.4cos30 ( 29 j = 2.80i + 0.346j ( 29 ft/ sec a B = a A + r B/A + r B/A ( 29 = 4j +-0.1k ( 29 2cos i + 2sin j ( 29 + 0.2k ( 29 0.2k ( 29 2cos i + 2sin j ( 29 [ ] = 4j +-0.1 ( 29 2cos ( 29 j +-0.1 ( 29 2sin ( 29-i ( 29 + 0.2k ( 29 0.2 ( 29 2cos ( 29 j + 0.2 ( 29 2sin ( 29-i ( 29 [ ] = 4 + 0.2sin30 ( 29 i +-0.2cos30 ( 29 j + 0.2 ( 29 2 2cos ( 29-i ( 29 + 0.2 ( 29 2 2sin ( 29-j ( 29 = 4 + 0.2sin30 -0.2 ( 29 2 2cos ( 29 [ ] i +-0.2cos30 -0.2 ( 29 2 2sin ( 29 [ ] j = 4.03i-0.213j ( 29 ft/ sec 2...
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rigid_kinematics - the angular velocity and angular...

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