Yeah, you'll get flung off due to the faster-than-escape-velocity speeds at the surface. You need to land near the pole.
Check it out.
I was thinking about it being possible to land near the equator if you just had enough friction with the ground.
The rotation is not pushing you upwards, but rather sideways, so with enough landing legs planted it should be possible to stay stuck to the surface.
Then again, it might turn out you'd need millions of legs or something.
Let g be the planet's gravitation at your current position, ω be the planet's rotation speed (in radians per time unit), R be the distance from the center of mass to your landing position, α be the global latitude, and β be the local latitude (the angle between the equator and your local surface normal).
To keep staying on the ground without hooks, you need an acceleration of at least ωēRcos(α) to the axis of rotation. Let's put this in a coordinate system so we can use 2D vectors. The acceleration required is a1 = (-ωēRcos(α), 0), and the planet's gravity provides a2 = (-gcos(α), -gsin(α)). The difference a1 - a2 = (gcos(α)-ωēRcos(α), gsin(α)) needs to be covered by surface friction. For this to be possible, we need tan(ß) to be greater than (ωēR/g - 1)cot(α). For α=ß (landing on a horizontal surface), we get tanē(α) > ωēR/g - 1, or cosē(α) < g/ωēR.
In case of Inaccessable, g = 1.18 m/sē, ω=0.0143, R = 15000, so you can only stay on horizontal surfaces in latitudes greater than around 52°, although you'll need to go at least above 75° unless you have a buttload of friction to spare.
/physics