Code Plus Math

Q. The collar A in Figure slides along the rotating rod OB. The angular position  of the rod is given by 𝜃 =2/3 *(𝜋𝑡)² rad and the distance of the collar from O varies as r =18t⁴+4 m, where time t is measured in seconds. Determine the velocity and acceleration vectors of the collar at t =0.4 s. Solution  :-     Given polar co-ordinates are      r = 18t⁴+4       and       𝜃 =2/3 ×(𝜋𝑡)²      r`= 72t³                        𝜃` =4/3 ×(𝜋𝑡)      r``= 216t²                    𝜃`` =4/3 ×(𝜋) At t=0.4 sec       r = 18(0.4)⁴+4 = 4.46m       r`= 72(0.4)³ = 4.608 m/a       r``= 216(0.4)² = 34.56 m/s²      𝜃 = 2×(3.14×0.4)²/3 =0.335 rad      𝜃`= 4×3.14×0.4/3   = 1.675 rad/sec      𝜃``= 4×3.14/3 = 4.188 rad/sec² The velocity vector is given by V = (r`) Er + (r×𝜃`) E𝜃     = 4.608 Er +4.46×1.675 E𝜃     = 4.608 Er + 7.47 E𝜃 Where Er - radial unit vector               E𝜃 -  angular unit vector Also acceleration vector is given by A = (r``- r𝜃`²)Er + (2r