Kinematic analysis using intermediate frames.pptx
- 2. Calculation of velocity and acceleration when two intermediate frames are used. Point P is defined in the intermediate reference frame 𝑜2𝑥2𝑦2𝑧2 , which is moving w.r.t another intermediate reference frame 𝑜1𝑥1𝑦1𝑧1. Frame 𝑜1𝑥1𝑦1𝑧1 is rotating at the angular velocity 𝑚1 and angular acceleration 𝑚 ˙ 1 w.r.t the fixed reference frame OXYZ. Frame 0 Frame 2 Frame 1
- 3. Case – 1 The angular motions of 𝑜2𝑥2𝑦2𝑧2, 𝑚2 and 𝑚˙2, are defined w.r.t frame OXYZ. Case – 2 The angular motions of 𝑜2𝑥2𝑦2𝑧2, 𝑚2 and 𝑚˙2, are defined w.r.t frame 𝑜1𝑥1𝑦1𝑧1.
- 4. Case 1: The angular motions of 𝑜2𝑥2𝑦2𝑧2, 𝑚2 and 𝑚˙2, are defined w.r.t frame OXYZ. Question: Find the motion of P w.r.t OXYZ. Solution: Find the motion of 𝑜2 w.r.t OXYZ and then the motion of P w.r.t OXYZ.
- 5. Motion of 𝑜2 (defined in 𝑜1𝑥1𝑦1𝑧1) w.r.t OXYZ. 𝑑 𝑡 𝑟𝑒 𝑙 Velocity of P: 𝒗 = 𝑑𝑹0 + 𝒗 + 𝑚 × 𝒓 𝑑𝑡2 2 Acceleration of P: 𝒂 = 𝑑 𝑹0 + 𝒂𝑟𝑒 𝑙 + 2𝑚 × 𝒗𝑟𝑒 𝑙 + 𝑚˙× 𝒓 + 𝑚 × 𝑚 × 𝒓
- 6. 𝑹0 = 𝒓1 𝒓 = 𝒓2 𝑚 = 𝑚1 The time derivatives: 𝑑𝑹0 𝑑𝑡 = 𝒓 ˙1 𝑑2𝑅0 𝑑𝑡2 = 𝒓¨1 𝒗𝑟𝑒𝑙 = 𝒓 ˙2 𝒂𝑟𝑒𝑙 = 𝒓¨ 2 𝑚 ˙ = 𝑚 ˙ 1
- 7. 𝒗𝑜2(𝑂𝑋𝑌𝑍) 𝑑𝑡 𝑟𝑒 𝑙 𝑑𝑹0 = + 𝒗 + 𝑚 × 𝒓 = 𝒓˙1 + 𝒓˙2 + 𝑚1 × 𝒓2 𝒂𝑜2 𝑂𝑋𝑌𝑍 = 𝑑2𝑹0 𝑑𝑡2 + 𝒂𝑟𝑒𝑙 + 2𝑚 × 𝒗𝑟𝑒𝑙 + 𝑚˙× 𝒓 + 𝑚 × 𝑚 × 𝒓 = 𝒓¨1 + 𝒓¨2 + 2𝑚1 × 𝒓˙2 + 𝑚˙1 × 𝒓2 + 𝑚1 × 𝑚1 × 𝒓2 Velocity and acceleration of point 𝑜2 w.r.t OXYZ.
- 8. Motion of P (defined in 𝑜2𝑥2𝑦2𝑧2) w.r.t OXYZ. 𝑹0 = 𝒓1 + 𝒓2 𝑑𝑹0 = 𝒗 𝑑𝑡 𝑜2(𝑂𝑋𝑌𝑍) 𝑑2𝑹0 𝑑𝑡2 = 𝒂𝑜2(𝑂𝑋𝑌𝑍) 𝒓 = 𝒓3 𝒗𝑟𝑒𝑙 = 𝒓 ˙3 𝒂𝑟𝑒𝑙 = 𝒓¨ 3 𝑚 = 𝑚2 𝑚 ˙ = 𝑚 ˙ 2
- 9. 𝒗𝑃(𝑂𝑋𝑌𝑍 ) 𝑑𝑡 𝑟𝑒 𝑙 𝑑𝑹0 = + 𝒗 + 𝑚 × 𝒓 = 𝒗𝑜2(𝑂𝑋𝑌𝑍) + 𝒓˙3 + 𝑚2 × 𝒓3 = 𝒓˙1 + 𝒓˙2 + 𝒓˙3 + 𝑚1 × 𝒓2 + 𝑚2 × 𝒓 3 Velocity of point P w.r.t OXYZ.
- 10. 𝒂𝑃 𝑂𝑋𝑌𝑍 = 𝑑2𝑹0 𝑑𝑡2 + 𝒂𝑟𝑒𝑙 + 2𝑚 × 𝒗𝑟𝑒𝑙 + 𝑚˙× 𝒓 + 𝑚 × 𝑚 × 𝒓 = 𝒂𝑜2(𝑂𝑋𝑌𝑍) + 𝒓¨3 + 2𝑚2 × 𝒓˙3 + 𝑚˙2 × 𝒓3 + 𝑚2 × 𝑚2 × 𝒓3 = 𝒓¨1 + 𝒓¨2 + 𝒓¨3 + 2𝑚1 × 𝒓˙2 + 2𝑚2 × 𝒓˙3 + 𝑚 ˙ 1 × 𝒓2 + 𝑚 ˙ 2 × 𝒓 3+ 𝑚1 × 𝑚1 × 𝒓2 + 𝑚2 × 𝑚2 × 𝒓3 Acceleration of point P w.r.t OXYZ.
- 11. Case – 2 The angular motions of 𝑜2𝑥2𝑦2𝑧2, 𝜔2 and 𝜔˙ 2, are defined w.r .t frame 𝑜1𝑥1𝑦1𝑧1. Question: Find the motion of P w.r.t OXYZ. Solution: Find the motion of P w.r.t 𝑜1𝑥1𝑦1𝑧1 and then the motion of P w.r.t OXYZ.
- 12. Motion of point P (defined in 𝑜2𝑥2𝑦2𝑧2) w.r.t 𝑜1𝑥1𝑦1𝑧1. 𝑹0 = 𝒓2 𝑑𝑹0 𝑑𝑡 = 𝒓 ˙2 𝑑2𝑹0 𝑑𝑡2 = 𝒓¨2 𝒓 = 𝒓3 𝒗𝑟𝑒𝑙 = 𝒓˙ 3 𝒂𝑟𝑒𝑙 = 𝒓¨ 3 𝑚 = 𝑚2 𝑚 ˙ = 𝑚 ˙ 2 𝒗𝑃(𝑜1𝑥1𝑦1𝑧1) = 𝒓˙2 + 𝒓˙3 + 𝑚2 × 𝒓3 𝒂𝑃(𝑜1𝑥1𝑦1𝑧1) = 𝒓¨2 + 𝒓¨3 + 2𝑚2 × 𝒓˙3 + 𝑚˙2 × 𝒓3 + 𝑚2 × (𝑚2 × 𝒓 3 )
- 13. Motion of point P (defined in 𝑜1𝑥1𝑦1𝑧1) w.r.t OXYZ. 𝑹0 = 𝒓1 𝑑𝑹0 𝑑𝑡 = 𝒓 ˙1 𝑑2𝑹0 𝑑𝑡2 = 𝒓¨1 𝒓 = 𝒓2 + 𝒓3 𝒗𝑟𝑒𝑙 = 𝒗𝑃(𝑜1𝑥1𝑦1𝑧1) 𝑚 = 𝑚1 𝑚 ˙ = 𝑚 ˙ 1 𝒂𝑟𝑒𝑙 = 𝒂𝑃(𝑜1𝑥1𝑦1𝑧1)
- 14. 𝒗𝑃(𝑂𝑋𝑌𝑍) = 𝒓˙1 + 𝒗𝑃(𝑜1𝑥1𝑦1𝑧1) + 𝑚1 × = 𝒓˙1 + 𝒓˙2 + 𝒓˙3 + 𝑚1 × 𝒓2 + 𝒓2 + 𝒓3 𝑚1 + 𝑚2 × 𝒓3 𝒓2 + 𝒓3 + 𝑚1 𝒂𝑃(𝑂𝑋𝑌𝑍) = 𝒓¨1 + 𝒂𝑃(𝑜1𝑥1𝑦1𝑧1) + 2𝑚1 × 𝒗𝑃(𝑜1𝑥1𝑦1𝑧1) + 𝑚 ˙ 1 × 𝑚1 × 𝒓2 + 𝒓3 = 𝒓¨1 + 𝒓¨2 + 𝒓¨3 + 2𝑚1 × 𝒓˙2 + 2 𝑚1 + 𝑚2 × 𝒓3̇ + 𝑚 ˙ 1 × 𝒓2 + 𝑚˙1 + 𝑚˙2 × 𝒓3 + 𝑚2 × 𝑚2 × 𝒓3 + 2𝑚1 × (𝑚2 × 𝒓3) + 𝑚1 × 𝑚1 × 𝒓𝟐 + 𝒓3
- 15. Bird on mobile. An artistic mobile structure is modeled as sketched in Figure 1(a), consists of a large + and four smaller Y’s, one each attached to a tip of the +. The + is horizontal at all times and rotates at a constant angular velocity 𝜔1 (w.r.t ground) about an axis through its center. Also, each four Y’s remains at all times in a vertical plane and rotates at a constant angular velocity 𝜔2 (w.r.t its + tip) about an axis through its center. At the instant shown, a bird of mass 𝑚 is on a leg of one of the Y’s, which is oriented as indicated in Figure 1(b). Relative to the Y, the bird is running with a velocity 𝑣0 and an acceleration 𝑎0, at the instant shown. At the same instant, a gust of wind exerts a force 𝐹𝑤 on the bird in the X direction. Find the velocity and acceleration of the bird, which may be modeled as a point, at the instant shown.
- 16. Figure 1(a): Mobile structure Figure 1(b): Detail of Y and bird. Figure 1: Sketch of artistic mobile structure with bird running along horizontal leg of Y.
- 17. 1. Motion of bird (defined in 𝑜2𝑥2𝑦2𝑧2) w.r.t 𝑜1𝑥1𝑦1𝑧1. 2. Motion of bird (defined in 𝑜1𝑥1𝑦1𝑧1) w.r.t OXYZ.
- 18. Robot manipulating work piece. A robot named JT is rolling w.r.t the shop floor at a constant speed of 0.5 m/s and carrying a work piece 1 m long, as sketched in Figure 2. each of the links of the robot arm is 0.75 m long, and the second link has an end gripper that holds the work piece which ,ay be considered as rigid. At the instant shown, the link AB is rotating a 𝜔1 (1 rev/3 s), and link BD is rotating at 𝜔2 (1 rev/2 s) and 𝜔˙2 (0.5 rad/s2), all w.r.t the shop floor. At the instant shown, find the velocity and acceleration of the center of the work piece, labeled point C, as sketched in Figure 2.
- 19. Figure 2: JT manipulating work piece.
- 20. 1. Using the frame 𝐴𝑥1𝑦1𝑧1 as an intermediate frame, find the motion of point B w.r.t the fixed reference frame 𝑂𝑋𝑌𝑍. 2. Using the frame 𝐵𝑥2𝑦2𝑧2 as an intermediate frame, find the motion of point C w.r.t the fixed reference frame 𝑂𝑋𝑌𝑍.