Question

A grindstone in the shape of a solid disk with diameter 0.510 m and a mass of m = 50.0 kg is rotating at ω = 830 rev/min . You press an ax against the rim with a normal force of F = 170 N (Figure 1) , and the grindstone comes to rest in 8.00 s .

Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

Answer 1

given,

m = 50 kg

radius, r = d/2

= 0.51/2

= 0.255 m

Moment of Inertia of wheeel, I = 0.5*m*r^2

= 0.5*50*0.255^2

= 1.616 kg.m^2

w1 = 830 rev/min = 830*2*pi/60 = 86.87 rad/s

w2 = 0

time taken, t = 8 s

angular acceleration, alfa = (w2 - w1)/t

= (0 - 86.87)/8

= -10.86 rad/s^2

let m is the coefficient of friction between the ax and the grindstone.

we know, Torque = I*alfa

F_tan*r*sin(90) = I*alfa

F*mue*r = I*alfa

mue = I*alfa/(F*r)

= 1.626*10.86/(170*0.255)

= 0.407 <<<<<<<<<<<<--------------------------Answer