K this is the last problem for me in the Rotation/Torque section of my homework and its nuts. I've put it off after working on it for an hour and now I have to do it: A 55.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on the rim with a normal force of 160 N . The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50 N*m between the axle of the stone and its bearings. for a picture go here, trust me it helps alot: http://members.cox.net/brianjw/wheel.JPG [Broken] It has 3 questions: 1)How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 9.00 s? For this one I know I need to get the sum of the torques. But the crank and the way the axe is being onto the stone is confusing me. Also, what formula do I use that has time in it? Can't seem to find any from this section. The next 2 I can't really solve cause the first part is stumping me. 2)After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev/min? 3)How much time does it take the grindstone to come from 120 rev/min to rest if it is acted on by the axle friction alone? Thanks for any input, even where to start would be greatly appreciated. I don't expect someone to give me an answer, just a nudge in the right direction!