OK, physics graduate here.
v0n you are confusing gravity to angular momentum. Changing the rotational speed of a body has no effect on it's gravitational pull which entirely a function of mass:
f = G x (m1 X m2)/r^2 Where G is the gravitational constant. Notice how there is no speed or angular speed values in there.
Calculating it out, the force on a 100kg man at the surface is:
f = 6.67x10^-11 x (100kg X 5.9x10^24kg)/6378000M^2 = 967.4N
As for the speed of the planet rotating causing a large shift in the perceived gravitational pull at the surface. Let's calculate it.
f = (mv^2)/r where v = d/t
v = 40,000,000M / 86164s = 464.2 Ms^-1
Assuming a 100kg person
f = (100 x 464.2^2)/6,378,000M = 3.4N
So, the rotational speed of the earth causes a 0.35% change in the felt gravitational force at the surface.
Let's more than quadruple the rotational speed of the earth:
v = 40,000,000M / 20000s = 2000 Ms^-1
f = (100 x 2000^2)/6,378,000M = 62.7N which is a 6.5% change in the felt gravitational force.
Nice. Thank you.
Now if we half the radius, it needs to be 4 times faster just to have the same effect it has now right?