Suspend a typical light airplane from its wingtips, and it will probably be OK. Suspend a helicopter from its rotor tips, and you’ll probably break the rotors. Rotors get most of their rigidity from the outward force caused by their rotation.
So… How great is the force? How do you define/quantify the rigidity of the rotating blade? IANA engineer, so I’m not even sure how to ask the questions. Basically I’m wondering about the engineering behind making a rotor blade.
For a starting point, let’s assume a Robinson R22. It has a rotor diameter of about 25 feet. I don’t know the composition of the blades, but I assume they are aluminum honeycomb with aluminum skins. Tip speed is probably around 400 mph. Gross weight is 1,370 pounds.
Lifting an airplane by its wingtips is sort of chancy. Simple statics will show that the bending moment at the wing root would be at least doubled, and could be considerably more than that depending on the planform of the wing. I forget just what the safety factor of an airliner is, but that’s getting awfully close to the breaking point.
Of course, an airplane is designed for maximum gust loads, and I think these are more than twice the static load, but as I said, don’t remember the exact details. Bottom line; if the airplane were lifted by its wingtips very slowly and carefully, it might survive. But don’t jolt it.
According to a not-so-reliable source, a typical velocity for a helicopter rotor is 500 rpm. Assuming that a 25 foot diameter rotor spins at this rate, there would be over 1,000 Gs of outward force at the rotor tips. I can’t find the mass of the blades so I can’t convert that to force, but it would surely be a lot.