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# You have a horizontal grindstone (a disk) that is 85 kg, has a 0.39 m radius, is turning at 85 rpm (in the positive direction), and you press a steel axe against the edge with a force of 19 n in the radial direction. 50% part (a) assuming that the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone in rad/s2.

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m = mass of disk = 85 kg r = radius of disk = 0.39 mmoment of inertia of the disk is given as I = (0.5) m r2I = (0.5) (85) (0.39)2 = 6.5 kgm2F = normal to the applied force on the edge = 19 N = coefficient of friction = 0.20 frictional force is, therefore, given by f = Ff = 0.20 x 19 = 3.8 N = angular accelerationTorque due to the force of friction is given as = f rTorque is also given as = I so I = f r, the insertion of the values (6.5) = (3.8) (0.39) = 0.23 rad/s2