Round and Round

When the object is at the beginning at equilibrium. The tension force in the spring is equal to the weight of the object. Therefore:

$Tension\quad =\quad Weight\\ \\ Tension\quad =\quad Spring\quad Constant\quad \times \quad Extension\\ \\ Weight\quad =\quad Spring\quad Constant\quad \times \quad Extension\\ \\ 5\quad =\quad K\quad \times \quad \frac { 14.8-13 }{ 100 } \\ \\ K\quad =\quad \frac { 2500 }{ 9 } \quad (keep\quad as\quad is)\\ \\$

Now when the object has rotates 180 degrees, the length is again 13 cm. Which means there is no extension and since Tension = KX where X is the extension (which is zero) then the tension here is zero. The only force acting on the object is its weight perpendicularly downward. So by setting an equation using F = ma we get:

$Force\quad Resultant\quad =\quad Centripetal\quad Force\\ \\ Force\quad resultant\quad =\quad Weight\\ \\ Weight\quad =\quad Centripetal\quad force\\ \\ Centripetal\quad force\quad \quad =\quad Mass\quad \times \quad (Angular\quad momentum)^{ 2 }\quad \times \quad Radius\\ \\ Weight\quad =\quad Mass\quad \times \quad (Angular\quad momentum)^{ 2 }\quad \times \quad Radius\\ \\ 5\quad =\quad \frac { 5 }{ 9.81 } \quad \times \quad \omega ^{ 2 }\quad \times \quad \frac { 13 }{ 100 } \\ \\ \omega ^{ 2 }\quad =\quad \frac { 981 }{ 13 } \quad (keep\quad as\quad is)$

Next when the object is finally at the bottom (after 360 degrees), let us call its length (L). by setting again another formula we get:

$Extension\quad =\quad L\quad -\quad \frac { 13 }{ 100 } \\ \\ Spring\quad constant\quad =\quad K\quad =\quad \frac { 2500 }{ 9 } \quad (derived\quad earlier)\\ \\ (Angular\quad velocity)^{ 2 }\quad =\quad \omega ^{ 2 }\quad =\quad \frac { 981 }{ 13 } \\ \\ Tension\quad Force\quad -\quad Weight\quad Force\quad =\quad Centripetal\quad Force\\ \\ T\quad -\quad W\quad =\quad F_{ c }\\ \\ T\quad =\quad Kx\quad =\quad K(L\quad -\quad \frac { 13 }{ 100 } )\\ \\ W\quad =\quad 5N\\ \\ F_{ c }\quad =\quad Mass\quad \times \quad \omega ^{ 2 }\quad \times \quad Radius\\ \\ Writing\quad this\quad again\quad we\quad get:\\ \\ T\quad -\quad W\quad =\quad F_{ c }\\ \\ K(L\quad -\quad \frac { 13 }{ 100 } )\quad \quad -\quad \quad 5\quad \quad =\quad \quad Mass\quad \times \quad \omega ^{ 2 }\quad \times \quad Radius\\ \\ \frac { 2500 }{ 9 } (L\quad -\quad \frac { 13 }{ 100 } )\quad \quad -\quad \quad 5\quad \quad =\quad \quad \frac { 5 }{ 9.81 } \quad \times \quad \frac { 981 }{ 13 } \quad \times \quad L\\ \\ \frac { 28000 }{ 117 } L\quad =\quad \frac { 370 }{ 9 } \\ \\ L\quad =\quad \frac { 481 }{ 2800 } \\ \\ a\quad =\quad 481,\quad b\quad =\quad 2800\\ \\ a\quad +\quad b\quad =\quad 3281\\ \\ Answer\quad =\quad 3281\\ \\$

The answer is 3281