What real force causes outward acceleration in rotation? Consider a thick solid cylinder with a smooth groove along a diameter. If a block was placed somewhere in the groove and the cylinder was spun about an axis passing through its centre, I think the block would move outwards with respect to the cylinder because in the cylinders frame, the block would experience a centrifugal acceleration.
However I can't figure out what real force could be causing this as there is no friction and the normal reaction doesn't have any component in that direction.
 A: 
the normal reaction doesn't have any component in that direction.

The problem with this statement, and the resolution of the apparent paradox, is that "that direction" in this case is relative to position. When the block is due west of the center, and the normal force is due northward, a northward force has no inwards or outwards component, but that does not remain true as the block changes position with the resulting imparted velocity.
To simplify the situation, consider only a momentary impulse, rather than continuous acceleration.
The block's initial position is due west of the cylinder's center. A momentary impulse pushes due northward on the block, normal to the sides of the groove, giving it a northward velocity. At this initial point, the force/impulse and the resulting velocity are neither inward nor outward - inward would be due east, outward would be due west, and the force and velocity are due north.
Some time later, the block has moved some distance north, and is now northwest of the cylinder's center. Because of the block's new position relative to the cylinder, inward is now southeast, and outward is now northwest. The block is still moving due north, however. The block's straight line northward movement, which began as purely perpendicular to the groove, now has an outward component. The block's movement gained this outward component, not because the direction of its movement changed, but because the definition of "outward" changed due to the change in position.
A: Since there is no friction, the only horizontal force acting on the block is the normal force from the side walls of the groove. In the non-rotating (inertial) reference frame it is this force that makes the block move outwards along the groove. In the rotating (non-inertial) frame this force appears to makes the block move in a circle; we think we need an additional force to account for the block moving outwards along the groove, so we invent a fictitious centrifugal force.
You can see that the normal force from the groove side walls must be involved by thinking about what happens if there is no groove, and the block is simply resting on the smooth flat horizontal face of the cylinder. In this case the block will not move no matter how fast the cylinder rotates.
A: The real force when the block is at position $A$ is the normal force, $N$, on the block due to the groove wall.

This force accelerates the block towards position $B$ which is further from the centre of rotation, $X$, than position $A$.
This is demonstrated by Walter Lewin using a marble in a glass tube.
However the trajectory of the marble/block is not a straight line.
This is because the force on the block is always at right angles to the groove wall and thus there was no radial component I decided to use polar coordinates. I also decided not to consider motion in the rotating frame of the cylinder because I did not want to introduce pseudo forces although it would have made the analysis a little easier.
In polar coordinates the position of a particle is defined by $r$ and $\theta$.  Let the block at time $t = 0$ be at position $(R,0)$ and with a velocity $(0,R \dot \theta)$ where $\dot{\theta}$ is the angular speed of the cylinder and assumed to be constant.
The formula for acceleration in polar coordinates looks horrific,
$\vec{ \ddot{r}} = ( \ddot{r} - r {\dot {\theta}}^2) \hat{r} + (r \ddot{\theta} + 2 \dot{r}\dot{\theta})\hat{\theta}$
Applying Newton's second law with the normal reaction force having no radial component gives
$\vec F = F \hat{\theta} = m \vec{\ddot{r}} =  m(( \ddot{r} - r {\dot {\theta}}^2) \hat{r} + (r \ddot{\theta} + 2 \dot{r}\dot{\theta})\hat{\theta})$
So $\; \ddot{r} - r {\dot {\theta}}^2 = 0\;$  which is equivalent to the formula one would have obtained in the rotating frame of reference.
Solving this equation and applying the initial conditions gives $r = \frac R2e^{-\dot \theta t}(e^{2\dot\theta t}+1)$ and because $\dot{\theta}t = \theta$ this can be rewritten as $r = \frac R2e^{-\theta}(e^{2\theta}+1)$.
Here is the path taken by the block with $R=1$ and you can see that for the first part of the motion it is "almost" a straight line.

A: I think answer was given by Chris above but I like to add something to clarify more. The force is real or not it depends because some force moves it out of rotational motion. But that force is due to frame and cease to exist at rest, so force is apparent in universal context but limited to frame or motion only. This is experience in linear acceleration of a car also, passenger push back to seat or seat moves forward. In relative motion, things appear to move backward is also kind of apparent motion.
The block is at rest can be seen by putting value of acceleration caused by frame, centrifugal and coriolis in equation given by farcher above. The radial and tangential acceleration of the block is canceled by acceleration of the cylinder with or without any external acceleration. So in absence of external or actual acceleration, the block is at rest. As in linear acceleration, water in a glass spill backward, so centripetal force of frame cause centrifugal force outward and then tangentially as it gain some speed, spirally outward if diameter is large.
A: There is no outward acceleration. Now, your logic, although not stated explicitly, seems to be along the lines of "The object moves outward, it doesn't have initial outward velocity, therefore there must be outward acceleration". The flaw in that logic is that the direction that is labelled "outward" is not constant. When I say "There is no outward acceleration", what I mean that at any point in time, the component of the acceleration in the direction that is, at the time of the acceleration, in the direction that is "outward" at that time, is zero. What happens is that the object is accelerated in a direction that is, at the time, tangential, and that direction then becomes "outward" later. For instance, if the block starts on the Eastern side of the cylinder, and the cylinder rotates counterclockwise, then the block will start out being accelerated Northwards, which will, at that time, be tangential to the block's displacement from the center, and therefore will not be "outward". Once the block reaches the Northern side of the cylinder, however, the Northward velocity that it received at the beginning will now be "outward". But at that time, all of the acceleration will be Westward.
In the nonrotating frame of reference, the direction that is "outward" rotates, and so the angle $\theta$ between the initial acceleration and "outward" decreases, and so the component of that acceleration in the "outward" direction, which is given by $a_0 \cos (\theta)$, increases. In the rotating frame of reference, the velocity of the block rotates (Coriolis force), and so its initial tangential velocity is transformed into "outward" velocity.
A: There is no spoon.
Really, there is no real force, and there is no outward acceleration.
The only reason you think there IS is because you are thinking, and analyzing the scenario in an accelerated frame of reference.
None of the simplified laws of Physics work the same way unless they are applied in an unaccellerated (zero-G) frame of reference. (Or unless you compensate for the acceleration by adding in a fictitious force to compensate.) A rotating frame of reference is accelerating!
This is akin to asking what force is causing an apple to fall in a rocket ship out in space with the motor firing. The Apple isn't falling, you are accelerating upwards!
