Quadruple somersault

$Since\quad the\quad only\quad force\quad acting\quad on\quad him\quad (gravitaional)\quad is\\ acting\quad on\quad the\quad C.O.M,\quad angular\quad momentum\quad is\quad conserved\\ 1.\quad { I }_{ 1 }{ \omega }_{ 1 }={ I }_{ 2 }{ \omega }_{ 2 }\\ Let\quad the\quad time\quad taken\quad to\quad perform\quad \\ the\quad first\quad quater\quad +\quad last\quad quarter\quad =\quad { t }_{ 1 }\\ Number\quad of\quad revolutions\quad in\quad { t }_{ 1 }=\quad \frac { 1 }{ 2 } \\ Let\quad the\quad time\quad taken\quad to\quad perform\quad \\ the\quad rest\quad of\quad the\quad jump\quad =\quad { t }_{ 2 }\\ Number\quad of\quad revolutions\quad in\quad { t }_{ 2 }=\quad \frac { 7 }{ 2 } \\ 2.\quad \frac { 1 }{ 2 } rev=\frac { { I }_{ 2 }{ \omega }_{ 2 } }{ { I }_{ 1 } } \quad \times \quad { t }_{ 1 }\quad \quad \quad \quad \quad \\ ({ \omega }_{ 1 }=\frac { { I }_{ 2 }{ \omega }_{ 2 } }{ { I }_{ 1 } } \quad from\quad angular\quad momentum\quad conservation)\\ \\ 3.\quad \frac { 7 }{ 2 } rev=\quad { \omega }_{ 2 }\quad \times \quad { t }_{ 2 }\\ Dividing\quad the\quad above\quad two\quad equations,\quad we\quad get\\ 4.\quad { t }_{ 2 }-\frac { { 7I }_{ 2 } }{ { I }_{ 1 } } { t }_{ 1 }\quad =\quad 0\\ Given:\quad \\ 5.\quad { t }_{ 1 }\quad +\quad { t }_{ 2 }\quad =\quad 1.87\\ Solving\quad the\quad above\quad two\quad linear\quad equations,\quad we\quad get\\ { t }_{ 1 }=0.785\quad s\\ { t }_{ 2 }=1.085\quad s\\ Inserting\quad the\quad above\quad values\quad in\quad eq\quad 3,\quad we\quad get\\ { \omega }_{ 2 }=\boxed { \boxed { 3.23\quad rev\quad { s }^{ -1 } } } \\ \\$