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Homework Help: Torque and Forces of grindstone

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Young University Physics 12e: Mastering Physics Problem 10.53

    A 70.0-kg grindstone is a solid disk 0.540 m in diameter. You press an ax down on the rim with a normal force of 150 N. The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50 Nm between the axle of the stone and its bearings.

    Part A
    How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 9.00 s ?

    Part 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?

    Part 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

    Torque = Radius(R) x Force(F) or Moment of Inertia(I) x Angular Acceleration(alpha)


    3. The attempt at a solution

    Not even sure how to do part A

    Do I need to include the negative torque done by the ax along with the friction on the bearings in my calculation for part A?
     
    Last edited: Apr 28, 2008
  2. jcsd
  3. Apr 29, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi NinjaIQ!

    You do know what a crank handle is, don't you?

    You push tangentially on one end of the handle, the other end is a pivot (so the handle moves in a circle), and so you produce a torque at the pivot. That torque is transferred directly to the grindstone.

    In other words: torque produced by crank handle equals torque received by grindstone. :smile:

    oh, and yes … the whole purpose is to sharpen the axe … so include axe and bearings! :smile:
     
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