If I've done my math right, assuming 3,000ft/sec, they would be rotating at 270,000 RPM and 135,000 RPM respectively. I'm not sure how much that rotational force difference translates into the bullet opening, but that would seem to be a significant force. Out of curiosity, I wonder how much the force is different for a larger diameter bullet given the same rotation rate? Ex: a .243 bullet vs. a .338 spinning at the same rate. The outer edge of the .338 is further from the center of rotation.
For the hell of it I calculated the centrifugal force for you just to see what would happen. Being pressed for time this evening I didn't bother to look into actual rpm values for a 243 and 338, just plugged and chugged with those rpm numbers.
For a 95 grain .243 bullet:
270k rpm = 3,414 lbf
135k = 853 lbf
For 250 gr .338 bullet:
270k rpm = 12,497 lbf
135k rpm = 3,124 lbf
So yes, both bigger bullet and higher rpms yields substantially higher centrifugal force.
