What would happen if the Earth was hollow and somebody rode on the "inside shell"? I'm having a difficult time understanding basic centripetal force, forgive me. On the earth as it currently is, if you fall off a cliff you will fall closer to the core. However on a cylindrical amusement park ride, as it spins faster you are "glued" to the wall by centripetal force. My question is, does a rotating cylinder exert force to keep somebody "glued" to both the interior and exterior? If we could hollow out the Earth and ride along the inside of the crust like an amusement park ride, what force would stop us from falling to the core that doesn't exist with the non-hollow Earth. (I guess it's necessary to state that the necessary adjustments would have to be made to keep the angular speed of the Earth constant despite hollowing it out)
 A: Your intuition of gravity inside the hollowed-out earth seems off. Gravity inside a spherical body only comes from the mass that is "below" you, as the mass above cancels out entirely. If you're inside a hollow spherical shell, there is no net gravitational force at all, regardless of your position (see the Shell Theorem). If you're inside a hollowed-out earth, you are in a zero-G environment and don't fall in any direction.
If you were inside the hollowed-out earth with non-zero velocity, you'll eventually hit one of the inside walls. The interior wall would then carry you along in a circle as the planet rotates, preventing you from flying off in the straight line trajectory you would have had otherwise. The centripetal force in this case comes from the structural integrity of the inner wall of the earth-shell, and gravity plays no role whatsoever. This situation is functionally identical to the rotating amusement park ride - the only force comes from the wall pushing you toward the center.
For someone on the outside of the earth-shell, the centripetal force comes from gravity (assuming the hollowed-out earth somehow has the same surface gravity).
A: A rotating cylinder doesn't keep someone glued to the exterior. It tries to throw you off.
If the Earth were hollow and somehow maintained structural integrity, you would likely be thrown off the surface since the gravitational acceleration is insufficient to overcome the tangential velocity imparted to you by the surface beneath your feet dragging you as it rotates. As soon as the surface drags your feet to a sufficiently high tangential velocity, your feet leave the ground since you now just following the linear path of that tangent away from the Earth rather than the curved path that the earth's surface is following as it rotates.
As for being on the inside, the surface beneath your feet imparts the same tangential velocity under your feet as if you were standing on the outside.  The difference here is that if you keep traveling along that straight, tangential path you end up hitting more surface since it is curved. The planet's rotation keeps trying to throw you tangentially outward, but since you are on the inside of its curved surface you keep hitting the planet whenever it tries to do so. This effect is not as strong the closer you get to the axis of rotation (i.e. near the poles)
It's the flip side (literally) how of orbiting something is moving so fast that you keep falling over the edge. Instead of falling over the edge you keep hitting the planet.
A: If we were to assume that your hollow Earth somehow kept its current gravity, then you would still fall, because the gravity would be by far still stronger than the fictitious centrifugal force (tangential velocity technically). In fact, a little playing around with the equations for gravity and centrifugal force and you will soon find that the distance from Earth at which the two forces equal out is around 36,000km away from the surface, or 43,000km from the core (assuming that you are still somehow connected to Earth to take advantage of its' spin, via an impossibly strong tower, and on the equator, where the spin is fastest). If your Earth does not keep its gravity, then you will indeed be pushed to the side and spun around by centrifugal force (tangential velocity), and could 'ride it' like a carousel as you suggested. In fact, as it currently is, you weigh about 1.2% less on the equator than at the poles due to this.
