A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50

Question

A grindstone in the shape of a solid disk with diameter 0.520m and a mass of 50.0kg is rotating at 850rev/min . You press an ax against the rim with a normal force of 160N (see the figure below), and the grindstone comes to rest in 8.50s . Find the coefficient of kinetic friction between the ax and the grindstone.
Part A There is negligible friction in the bearings. ? = A grindstone in the shape of a solid disk with diameter 0.520m and a mass of 50.0kg is rotating at 850rev/min . You press an ax against the rim with a normal force of 160N (see the figure below), and the grindstone comes to rest in 8.50s . Find the coefficient of kinetic friction between the ax and the grindstone.
Part A There is negligible friction in the bearings. ? = A grindstone in the shape of a solid disk with diameter 0.520m and a mass of 50.0kg is rotating at 850rev/min . You press an ax against the rim with a normal force of 160N (see the figure below), and the grindstone comes to rest in 8.50s . Find the coefficient of kinetic friction between the ax and the grindstone.
Part A There is negligible friction in the bearings. ? = Part A There is negligible friction in the bearings. ? = ? = ? = ? =

Explanation / Answer

for a solid disk, radius of Gyration is K = R/sqrt 2

here it is R = 0.52 /2 = 0.26

so K = 0.26/1.414 = 0.1838m

Nowm applyn the formula for moemnt of inertia of disk as I = mk^2

I = 50 x 0.1838^2 = 1.689 kg.m^2

Ang velcoity W = = 850 * 2*3.14/60 = 89 rad/s

Deceleration a = (?i - ?f) /t = (89 - 0) / 8.5

a = 10.47 m/s^2

Now apply the formula of torque T = I ?

T = 1.689 * 10.47

T =   17.68 Nm

as Torque T =    force x radius

Ff = 17.68/0.26 = 68 N

Coefficient of friction = F/ R = 68/160

coeffcien to friction u =0.425

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