Torque and angular accelerationfinding coeff. of friction

[offbeatjumi]

Homework Statement

The problem states: grindstone in shape of solid disk with diameter .52m and mass 52 kg rotates at 850 rev/min. You press an ax against the rim with normal force 160 N and grindstone comes to rest in 7.5 s. Find coefficient of friction between ax and grindstone.

Homework Equations

The sum of all torques t = I*alpha (angular accel). = alpha*mass*radius^2
avg.angular.accel = (change in angular velocity)/(change in time)
850 rpm = 89 rad/s

The Attempt at a Solution

Using Newton's second law I get the sum of external forces = m*a(tangential) = m*r*alpha = (mu)_k*n
(mu)_k = (mass*radius*alpha)/n = 0.964

The answer in the book is half of the answer I got, 0.482. Where did I miss this? Thanks so much =)

 

[godtripp]

Easy, it seems like your moment of inertia is wrong. If I'm not mistaken you're currently using I=MR^2. Which is the moment of inertia for a thin hoop, not a disk.

A disk has the moment of inertia I= (MR^2)/2... half of what you're using =)

 

[offbeatjumi]

Thanks, that was a silly mistake.
I feel completely stupid asking this now but i returned to the question and I don't see how I was using moment of inertia to answer my question.
It just seems that what I did was solve for ang.accel by taking the change in ang.vel. over change in time to get 11.87 rad/s. Then I equated the sum of ext.forces = f(k) = ma(tangential) = m*r*ang.accel.
since f(k) = mu_k*n... so mu_k = (m*r*ang.accel)/n .
Thank you so much, I'm just having a massive mental block.

 

[godtripp]

Don't worry about it, angular momentum can be a hard concept but I recommend working hard at it. It will come a lot in physics from now on =)