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Torque and force

  1. Apr 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A 50kg grindstone is a solid disk .26 in radius. You press an ax down on the rim with a normal force of 160N. The coefficient of kinetic friction between the blade and the stone is .6, and there is a constant friction torque of 6.5 NM between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle .5 m long to bring the stone from rest to 120 rev/min in 9 seconds? (b) After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev/min? (c) How much time does it take the grindstone to come from 120 rev/min to rest if it is acted on by the axle friction alone?


    2. Relevant equations
    for solid disk: I = (1/2)MR^2


    3. The attempt at a solution
    120 rev/min = 12.56 rad/s
    i found the acceleration would have to be 1.4 rad/s^2 and it would travel 56.7 radians
    I = 1.69

    i tried this:

    t - 6.5Nm(from torqu friction) - torque friction from ax = I(Arad)

    but im not entirely sure how to get torque friction from ax. is it just Uk(Normal)(r) ?

    and after i get total torque required im not sure how to convert it into tangential force with the handle.
     
    Last edited: Apr 25, 2008
  2. jcsd
  3. Apr 25, 2008 #2
    think about the defining equation for torque...
     
  4. Apr 25, 2008 #3
    ok torque is FD so i guess Uk(normal)(2pie)(r) i'll try that
     
  5. Apr 25, 2008 #4
    ok nevermind i found all the answers. im just tired and Physics hw takes sooooo long lol + i have 2 more assignments to do after this in phyiscs :(
     
  6. Apr 26, 2008 #5
    first work out what tangential force the normal force will produce... you are given the coefficient of friction. You then know what this tangential force is and how far it is from the centre, so you can work out the torque
     
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