Answer
The normal force at the lowest point is $~~1110~N$
Work Step by Step
At the top, the net downward force on the person is $(667~N - 556~N)$ which is $111~N$
This force of $F = 111~N$ provides the centripetal force to keep the person moving in a circle.
We can find an expression for the person's speed:
$F = \frac{mv^2}{r}$
$v^2 = \frac{F~r}{m}$
$v = \sqrt{\frac{F~r}{m}}$
We can find the centripetal force when the speed is doubled:
$F' = \frac{m(2v)^2}{r}$
$F' = \frac{m(2\sqrt{\frac{F~r}{m}})^2}{r}$
$F' = \frac{4m~F~r}{m~r}$
$F' = 4F$
$F' = 444~N$
We can find the normal force at the lowest point:
$F_N = 444~N+667~N = 1110~N$
The normal force at the lowest point is $~~1110~N$
