This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 1 ME 242 Dynamics January 28, 2009 Eric Wang Todays Key Concepts Cartesian Coordinates 2D motion Coordinate transformation array Position Vector r P / O = xi + y j Velocity v P = d dt XYZ r P / O = dx dt i + x d dt XYZ i + dy dt j + y d dt XYZ j + dz dt k + z d dt XYZ k r P / O = xi + y j + zk 3D Kinematics Position Velocity Acceleration v P = d dt XYZ r P / O = x i + y j + z k a P = d dt XYZ v P = x i + y j + z k r P / O = xi + y j + zk Given: V =11.31 ft/s Find: 2 Solution: V =11.31 m/s v x = v cos v y = v sin v x v y Final Position: tan30 = 0.97 x 1 x 1 = 1.68 ft x 1 = y 1 x = y = Constant Acceleration v x = 11.31cos v y = 11.31sin x = y = x 1 = 1.68 y 1 = 0.97 v 1 x 2 = v x 2 + 2 a x ( x 1 x ) v 1 y 2 = v y 2 + 2 a y ( y 1 y ) Constant Acceleration v x = v x + a x t x =...
View Full Document
This note was uploaded on 09/19/2009 for the course ME 242 taught by Professor Kam during the Spring '06 term at Nevada.
- Spring '06