Potential energy is energy of a position or orientation relative to other positions or orientations.
For example, if the hamster wheel is on a table, it will have more gravitational potential energy than if it is on the floor upon which the table is standing.
If the hamster wheel is rotationally symmetric about the axis of rotation, and the axis is not moving, then gravitational potential energy will stay constant (at least to the extent gravity is constant).
There could be potential energy associated with outward stretching of the wheel as it rotates (analogous to potential energy of a stretched spring), but this should be very small.
You assume that the force which acts on the car is the same one that acts on the passengers, which is not correct since the car and the passengers are two different objects.
What you actually feel when the car accelerates forward is the chair pushing you forwards and you pushing the chair backwards. When there is no contact with the chair your body tends to remain in equilibrium (remains at the same velocity). As soon as your body makes contact with the chair, the chair acts with forward force on the body which you can feel.
Consider a different example which is a bit more intuitive - a passenger on roller skates in an empty wagon that accelerates at $a$ with respect to earth. The roller skates are introduced to cancel the friction force as much as possible.
The passenger and the wagon are two different objects, which means that force acting on the wagon is not the same force acting on the passenger. When there is a net force applied to the wagon, the wagon starts accelerating relative to a stationary (inertial) reference frame $A$. Since no force is applied to the passenger, the passenger remains in equilibrium as seen from $A$, which follows directly from the first Newton's law.
In the wagon (non-inertial) reference frame $B$, which moves together with the wagon in the same direction, the passenger appears to be accelerating at $-a$ (going backwards) although no force has been applied to them. This clearly violates the first Newton's law which says
An object acted on by no net external force remains in equilibrium, i.e. has a constant velocity (which my be zero) and zero acceleration.
This is why we must introduce a force that acts on the passenger in order to make the reference frame $B$ satisfy the first and second Newton's laws. Since this force actually does not exist, it is called a pseudo-force or fictitious force.
To describe this with equations
$$a_{P/A} = a_{P/B} + a_{B/A}$$
where $a_{P/A}$ is the passenger acceleration in $A$, $a_{P/B}$ is the passenger acceleration in $B$, and $a_{B/A}$ is the wagon acceleration in $A$. Since no external force acts on the passenger $a_{P/A} = 0$, the above equation becomes
$$a_{P/B} = -a_{B/A}$$
Suddenly, the passenger has some acceleration although there is no external force acting on them. Introduce a fictitious force to make the first and second Newton's laws work in $B$ and problem solved. However, this clearly violates the third Newton's law since the fictitious force has no reaction pair, i.e. there is no force that acts from the passenger to the wagon. This is why non-inertial reference frames are not suitable for the Newton's laws.
Best Answer
Both the equations are equivalent. By definition of angular velocity, $$\omega=v/r$$
You substitute this in $m \omega^2 r$ to get $m v^2/r$.
NOTE THAT $\omega$ is not constant if you change radius, as you claim!
And for the mass to radius relation, its straight forward. Take, $$F_c = m v^2 /r$$ and ask what you get if you keep $F_c$ constant while increasing the radius.