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Torque and angular accelerationfinding coeff. of friction

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant 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


    3. 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 =)
     
  2. jcsd
  3. Nov 26, 2009 #2
    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 =)
     
  4. Nov 29, 2009 #3
    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.
     
  5. Nov 29, 2009 #4
    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 =)
     
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