Path of a coin placed on a frictionless rotating disk 
The torque of the disk is through its Center of mass
A coin is placed on the disk in r distance.
First when the disk has friction, the system seems like this-

If the disk don’t have any friction, would not the coin go toward the center?
 A: The major thing to understand here is that centripetal force is not a real force.
It is an effect of the resultant forces.It is not created on rotating the disk, but the forces acting towards the center provide the force.
Centripetal force is the effect not the cause. It is not a separate force but merely the sum of the components of all the forces acting towards the center
In absence of friction, there is no horizontal component of force.
There is only gravity and normal reaction which both act along an axis perpendicular to the coin.
Friction is hence required to move the coin from initial position.
A: In the situation with no friction, the only forces acting on it are the normal force and gravity
which do not provide torque about the axis of the disc.
Also, the disc is in translational equilibrium.
Therefore the coin will stay in its original place and will not go towards the center.
The coin will stay in its original place or move with its initial velocity or slide off the disc if the disc is not aligned horizontally. – user1079505
Are also possibilities depending on the initial conditions of the coin and the disc.
A: When the rotating disk has friction, the friction between the coin and the disk is the centripetal force, it is what is opposing the coin's centrifugal force to keep it from sliding outwards. Your friction arrow should be labeled centrifugal force. If the disk has no friction it cannot affect the movement of the coin except for the normal force that keeps it from falling through the disk. If the frictionless disk is not perfectly horizontal the coin will only slide downhill.
