Doesn't rotational KE of a rolling marble change if there is no friction to provide torque? The question arise from the following situation:
A marble at the border of a uniform bowl begins rolling within it from rest. There is enough friction in the first half the bowl for the marble to not slip, but there's no friction in the other half. Find the height reached by the marble, measured from the bottom of the bowl.
I read the solution, but I can't undestand the following:
When marbel arives to the bottom, it has both translational an rotational KE. The solution says when marble stops at the maximun height, its rotational KE is still the same as when it was at the bottom (because there's no friction to provide torque). Does it mean that marble is still rolling? Could anyone explain why would it be wrong if I propose that RKE is zero at the maximun height?
 A: Here's the marble in its starting position:

During its descent on the side with friction, potential energy $U$ is converted to kinetic energy $K$. Assuming there's enough friction for rolling without slipping ($v=\omega R$) then kinetic energy is partitioned into translational and rotational energy, so:
$$U=K$$
$$mgy=\frac12mv^2+\frac12I\omega^2$$
$$mgy=\frac12mv^2+\frac12\frac{I}{R^2}v^2$$
If $I$ is known, like in the case of a marble, $v$ can be calculated from that equation. For example for a solid sphere like a marble, $I=\frac25 mR^2$, so:
$$v^2=\frac{10}{7}gy$$
Now when the marble starts travelling upwards in the friction-free zone, its rotational state of motion remains unaltered, as per Newton's law, because no friction also means no torque that can affect $\omega$. Effectively, rotational kinetic energy remains constant during the climb.
The kinetic energy available for conversion to potential energy is thus only:
$$K_{trans}=mgy-\frac12\frac{I}{R^2}v^2$$
This energy is converted back to potential energy, acc.:
$$mgy_2=mgy-\frac12\frac{I}{R^2}v^2$$
So:
$$y_2=y-\frac12\frac{I}{mgR^2}v^2$$
Reworked with the above we get:
$$y_2=\frac57 y$$
Note that there's no loss of energy: part of the initial kinetic energy remains owned by the marble due to its rotation.
A: If there is no friction, there is no torque to change the rotation of the marble.
Normally, the marble would use its rotational KE to "climb up" the side of the bowl, but in this instance, it has no grip.
When the linear motion of the marble stops, it is still spinning (slipping without friction). This means you can only use the linear KE from the motion at the bottom of the bowl to estimate how high it will climb up the side.
A: The marble begins to slip instead of roll; thus it continues to move upwards until gravity converts all of the kinetic energy the linear motion into potential energy - then it will begin to reverse course.
A: By suggesting that the RKE at the top is zero, you are suggesting that there is a change in RKE from the bottom and the top. However, just like you need a net force to change kinetic energy (i.e. do Work) you need a net torque in order to change the rotational kinetic energy. In this situation, because this part of the ramp is frictionless, there is no torque on the ball throughout this part of the ramp, so there can be no change in RKE. Therefore the RKE at the top must be the same as the RKE on the bottom, i.e. non-zero.
A: Perhaps it is easier to start with the following example?
When in contact there is no friction between a marble and a surface.  
A marble is set spinning about a horizontal axis through its centre and dropped onto the surface.
What will happen?
As there is no frictional force on the marble its centre of mass does not accelerate and since the centre of mass started at rest it will stay at rest.
As there is no frictional force on the marble there is no torque on the marble about its centre of mass and so there is no angular acceleration.  The marble continues to spin indefinitely.

In the example you gave, as the marble goes down it gains both rotational and translational kinetic energy.
Assume that there is no slipping.
There are three forces acting on the marble:
Weight of marble acting downwards
Normal reaction of surface on the marble
A frictional force acting up along a tangent to the surface.  
Since the weight and the normal reaction act through the centre of mass of the marble it it only the frictional force which applies a torque on the marble about its centre of mass.
That torque produces an angular acceleration which makes the marble spin faster as it goes down the slope.
So going down the slope the marble loses gravitational potential energy and gains translational kinetic energy and rotational kinetic energy.
On the other side there is no friction and so there are only two forces acting on the marble:
Weight of marble acting downwards
Normal reaction of surface on the marble  
Since the weight and the normal reaction act through the centre of mass of the marble and there is no frictional force there is no torque on the marble about its centre of mass and so no angular acceleration about the centre of mass of the marble.  
The marble travels up the slope this time gaining gravitational potential energy and losing translational energy.
When the centre of mass of the marble comes to a stop it will still have some rotational kinetic energy and so the marble's centre of mass has not risen as far as it had fallen.
An interesting addition to the problem is to ask what happens when the marble makes the return journey.
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Later
To make the mathematics easier suppose that the slopes are flat and inclined at an angle $\theta$ to the horizontal.  The FBDs for going down and then going up are shown below.

Again to make like easier suppose that the marble starts from rest and does one complete revolution down the slope with friction before going up the slope without friction.
If the radius of the marble is $r$ then to undergo one revolution with slipping the centre of mass of the marble must have moved $2 \pi r$ and the marble must have rotated though an angle $2 \pi$ radians.
The work done by the net force down the slope is $(mg \sin \theta – F) \cdot 2 \pi r$ and this is equal to the gain in translational kinetic energy of the centre of mass of the marble.
If a torque $\tau$ rotates through an angle $\phi$ the work done by the torque is $\tau \phi$.
So in this example when the marble is going down the slope the torque about the centre of mass of the marble is $Fr$ and a rotation of $2 \pi$ means that the work done by the torque is $Fr 2 \pi$.  This work increases the rotational kinetic energy of the marble.  
So the total work done on the marble is $(mg \sin \theta – F) \cdot 2 \pi r + Fr 2 \pi  = mg \sin \theta \cdot 2 \pi r$ which is the work done by the gravitational field on the marble.
Now going up the frictionless slope it is the marble which is doing the work and assume that the marble travels a distance $s$ up the slope before it stops.
The work done by the marble is $mg \sin \theta \cdot x$ which must be less than the work done on the marble when going down the slope by an amount $Fr 2 \pi$ which is the rotational kinetic energy the marble still possesses when its centre of mass stops moving.
So s < 2\pi r$ and the marble does not rise up as fre on the frictionless slope as it fell on the slope with friction.
