圈上泽充大峰 Cycloidal Displacement Integrating acceleration gives velocity d Q= _=Csin d jh-jcs2r →v=-C cos 2π B +k 2π K:constant of integration Obtained using boundary conditions:v=0 at 0=0 k=CB 2π And 圆上清支大华 Cycloidal Displacement Integrating again to get displacement Ja-p s-CB0-cB +k2 2π B k2 Obtained using boundary conditions: >s=0at0=0 →k2=0 ▣For the constant C, >s=hate=B →C=2π- 1515  Integrating acceleration gives velocity               Csin 2 d dv a  K1:constant of integration.  Obtained using boundary conditions: v = 0 at  = 0    dv C  d            sin 2 1 cos 2 2 v C  k                 Cycloidal Displacement   2 k1  C                       1 cos 2 2 v C  And  Integrating again to get displacement 2 2 2 sin 2 2 4 s C C  k                         ds C d                          1 cos 2 2 Cycloidal Displacement  k2 Obtained using boundary conditions:  s = 0 at  = 0 0  k2   For the constant C,  s = h at  =  2 2   h  C 