A 50.0-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 (?Fig. P10.54?). The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of between the axle of the stone and its bearings. (a) 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? (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?

Solution 57P Introduction First we have to calculate the acceleration of the from the given initial and final angular speed and time. Then we have to calculate the required torque. Now from the torque we can calculate the applied force. Step 1 The mass of the disc is m = 50 kg The diameter of the disc is d = 0.520 m Hence the radius of the disc is r = d/2 = (0.520 m)/2 = 0.260 m Hence the moment of inertia of the disc is Step 2 The normal force is N = 160 N Coefficient of kinetic friction is = 0.26 Hence the frictional force is f = N = (0.26)(96 N) = 96 N So the torque due to friction is = rF = (0.260 m)(96 N) = 24.96 Nm f f Step 3 The torque due to friction of axle and bearing is b = 6.50 Nm Step 4 The grindstone starts from the rest and reaches at speed of 120 rev/min. Hence we have Initial angular speed is 0 = 0 Final angular speed is = 120 rev/min = 12.56 rad/s Time taken to reach this speed si t = 9.00 s Hence the acceleration is