A 50kg grindstone is a solid disk .26 in radius. You press an ax down on the rim with a normal force of 160N. The coefficient of kinetic friction between the blade and the stone is .6, and there is a constant friction torque of 6.5 NM between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle .5 m long to bring the stone from rest to 120 rev/min in 9 seconds? (b) 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? (c) 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?
for solid disk: I = (1/2)MR^2
The Attempt at a Solution
120 rev/min = 12.56 rad/s
i found the acceleration would have to be 1.4 rad/s^2 and it would travel 56.7 radians
I = 1.69
i tried this:
t - 6.5Nm(from torqu friction) - torque friction from ax = I(Arad)
but im not entirely sure how to get torque friction from ax. is it just Uk(Normal)(r) ?
and after i get total torque required im not sure how to convert it into tangential force with the handle.