I assume that when you say "set it in circular motion" you mean thrown like a discus, or perhaps if you're holding a child's hands and you start rotating around and swing them higher and higher, thus appearing "lighter" (at least compared to lifting them up stiff with their hands outstretched).

The reason for this is due to torque. Recall that now this problem exists in three dimensions, and there is a torquing force pulling the object downwards, quite far from your own center of mass.

In truth, when you swing them around you are still pulling on them harder than if you simply just wanted to hold it off the ground. An experiment you can do to demonstrate a basic principle of torque exerted on your body is to try to lift a heavy object up at arms length, and them try to lift them up closer to your body. You will find that you exert less force the closer the object is to your center of mass. This is because in order to stay upright you require several other applied forces (such as friction on your feet, etc) to prevent the object from torquing your upper body towards the ground. This requires other torques on your body from other places (your feet, your arms, etc) to balance it all out.

To illustrate a less complicated system, it is easier to see from a regular lever (see-saw) problem, where the net force upwards or downwards is zero. And the pivot would only have to support the net weight of the system. However the individual forces involved in balancing could be quite great, depending on the length of the lever and where the forces are applied (try a lever with a box, a force pushing down on the same side as the box, and another force counter balancing. Then show that the net torques are zero, the net force is zero, however if you have the counterforce at a much shorter radius this force can become quite large).

Now, back to your original question: if you set the object in rotational motion, it becomes easier to lift simply because there is much less torque being applied on you. Most of the necessary force is being applied along your arms as opposed to perpendicular to them (recall torque is equal to F*r*sin (theta) where r is the distance from the pivot point and theta is the angle between where the force is applied and the radius) and thus you feel much less torque. (If a force is being applied along the radius, the torque is zero according to that formula).

Let us do a thought experiment to see that the centripetal case is actually more difficult. To do this, we have to imagine a situation where we eliminate the torque upon the body for balancing. Perhaps the easiest example would be to imagine yourself hanging over a cliff holding onto a person that is about to fall off. If you just want to hold them at arms length, it is much easier to do so when they are not swinging. As soon as they start swinging (going into rotational motion) they become much harder to hold onto, especially at the low point of the swing. this type of situation would make the torques required to balance your body come from the forces that the environment (i.e. the ground and your own weight) places upon you instead of forces your body would have to provide (your legs and shoes providing friction as well as lots of other muscles counterbalancing each other).

As you swing the hammer in a circle, then the rotating system (the hammer plus your arm) pick up angular momentum, which is a conserved quantity. As long as the system holds together (i.e. you maintain your grip on the hammer), the angular inertia will act to prevent the hammer from slowing down, including actively resisting the force of gravity on the upswing. Of course, when you release the hammer, it flies off at a tangent. Still, the hammer's mass is the same, and the force with which the hammer is accelerating downward when it is on the downswing is greater than its weight at rest.

Do you think this might explain your airplane experiment?

You correctly point out that when the plane is flying in circular motion, there is an additional force required to keep the plane at a fixed radius. This force is always perpendicular to the instantaneous direction of motion of the plane. As you spin the plane faster and faster, the force required to keep it at this fixed distance will increase, which of course is why you observed the spring scale to increase. I wouldn't necessarily think of this as due to the plane 'weighting more', instead, this is fundamentally due to Newton's 1st Law. This law states that an object will continue traveling along a straight line unless acted on by an unbalanced force. Essentially, the plane wants to travel in a straight line, and as soon as you let go of the string, it will. The reason I think it feels lighter when you throw it, or just let go of the string, is because you are not supplying the additional centripetal force required to keep it in circular motion.