I am working on a project and the image you see is a 12 inch diameter flywheel that weighs 27.006 pounds. The RPM's will be 3600. My questions will be about the very large values I get for the acceleration and force. Since I am most familiar with ft/sec and pounds, those are the units I will use. Time period for 1 revolution = .01666 V=2∏r/T= 2x3.14159x.5(ft)/.01666(sec)= 188.570ft/sec A=V2/r=188.572/.5= 71118.0347ft/sec2 centripetal force=mr(4∏2/t2)=27.006x.5x4x3.141592/.016662=1,920,613.646(lbf) I have checked the math and that force value seems to be very high. What I would like to know is how much force it takes to spin the flywheel at 3600RPM's. Any help would be greatly appreciated. Thanks as I kept working on this I divided 1,920,613.646 by 3600=533.5037. Is that the amount of force per revolution?