A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50.0 kg is rotating at 850 rev/min. You press an ax against the rim with a normal force of 160 N (Fig. P10.54), and the grindstone comes to rest in 7.50 s. Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

Solution 59P Step 1: Data given Diameter d = 0.520 m Radius R = d/2 = 0.26 m Mass m = 50.0 kg Angular velocity = 850 rev/min = 89 rad /sec Normal force F = 160 N Time t = 7.50 s We need to find the friction coefficient We shall find the radius of gyration It is given by K = R/2 Substituting values we get K = 0.26 m/2 K = 0.18 m Step 2 : We need to find the moment of inertia of the wheel It is given by I = mK 2 Substituting values we get 2 I = 50.0 kg × (0.18 m) I = 1.62 kg m 2 Step 3 : We shall find the acceleration of the wheel while reducing the speed It is given by = ( )/t i f Substituting values we get Here = f as the wheel is stopping = (89 rad /sec 0 )/7.50 s = 11.86 rad/s 2