A grindstone in the shape of a solid disk with diameter 0.520m and a mass of 50.0kg is rotating at 850rpm . You press an ax against the rim with a normal force of 160N (see the figure below), and the grindstone comes to rest in 8.00sec . Find the coefficient of kinetic friction between the ax and the grindstone. (negligible friction in the bearings)
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inertia = 1/2mr^2
0.5 x 50 x 0.26^2 = 1.69 kg.m^2
rotational KE = 1/2Iω^2
850 rpm / 9.55 = 89 rad/s
0.5 x 1.69 x 89^2 = 6693.24 J
1/2Iω^2 = work done by torque
work done by torque = torque x distance in radians
distance in radians = (89 + 0)/2 x 8 = 356
6693.24/356 = torque = 18.8 N.m
torque of ax = 160 x 0.26 = 41.6
18.8/41.6 = coefficient of kinetic friction = 0.452
0.5 x 50 x 0.26^2 = 1.69 kg.m^2
rotational KE = 1/2Iω^2
850 rpm / 9.55 = 89 rad/s
0.5 x 1.69 x 89^2 = 6693.24 J
1/2Iω^2 = work done by torque
work done by torque = torque x distance in radians
distance in radians = (89 + 0)/2 x 8 = 356
6693.24/356 = torque = 18.8 N.m
torque of ax = 160 x 0.26 = 41.6
18.8/41.6 = coefficient of kinetic friction = 0.452
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