A grindstone in the shape of a solid disk with diameter ? and a mass of is rotat

Question

A grindstone in the shape of a solid disk with diameter ?and a mass of is rotating at ?. You press an ax against the rim with a normal force of F , and the grindstone comes to rest in ?t

my answer is (m((d/2)omega)/delta t)/F. I don't know why it's wrong. I got my answer by using a=v/t, but v=(d/2)*omega, then plugging all that in for a in F=ma and since the grindstone stops, the friction force must equal the grindstone force so F=uN, but N uses the variable F. Sorry if my explanation is confusing....

Explanation / Answer

Here d = diameter in m
m = mass in kg
r = radius = D/2
rev = revolution in rev/min
F = Force in N
t = time in s

Now the rotational inertia of a solid circular disk :

I = [m*(D/2)^2]/2

the initial angular velocity is = [rev*2*pi/60] rad/s

Angular acceleration :

omega = alpha*t

alpha = omega/t = [rev*2*pi/60] / t

The torque is Ff*r but it is also I*alpha

Ff*(D/2) = u*F*(D/2) = I*alpha

u = I*alpha/(F*D/2)

u = { [m*(D/2)^2]/2 * [rev*2*pi/60] / t } / F(D/2)

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