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Torque/ find the coefficient of friction

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data

    A grindstone in the shape of a solid disk with diameter 0.520m and a mass of 50.0kg is rotating at 800rev/min . You press an ax against the rim with a normal force of 160N, and the grindstone comes to rest in 8.50s .
    Find the coefficient of kinetic friction between the ax and the grindstone.


    2. Relevant equations



    3. The attempt at a solution

    The concept of torque has been giving me alot of problems, mostly UNDERSTANDING what the equation means, and when to use them.

    So here is how I started out,

    I=1/2MR^2

    850rev/min= 85pi rad/3s

    So my logic is I have the inertia, and I have enough information to solve for acceleration which means I can find the torque.

    I solve for the acceleration

    wz-woZ/t=-11.9rad/s^2

    Here is where I get a bit confused, where I was going with this problem is I was going to solve for the torque using Ʃτz= Iαz

    Then when I aquired the torque I was going to use

    ƩFext=Macm

    to find the coefficient of friction.

    The solution manual sets Ʃτz= μkNR=Iαz, and solves for μk.

    How do we know when to stay at Ʃτz=Iaz or when we have to switch over to ƩFext=Macm?

    Thank you

    Higgenz
     
  2. jcsd
  3. Apr 9, 2012 #2

    BruceW

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    In the problem statement, you say it is originally rotating at 800rev/min, but in the attempt at the solution, you are using 850rev/min... was this a typing error?

    About the torque: the torque is due to a force at the rim of the grindstone. There is an equation to calculate the torque due to a force at a point, but the equation is not ƩFext=Macm
     
  4. Apr 9, 2012 #3
    Typing error is is originally 850 rev/min and 7.50s apologies for that.
     
  5. Apr 9, 2012 #4

    BruceW

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    good, I get the same answer of 11.9rad/s^2 for the angular acceleration.

    So now you can go to the next step, "I was going to solve for the torque using Ʃτz= Iαz" And then after that, what equation relates torque to the force and the position that force is applied at?
     
  6. Apr 9, 2012 #5
    the lever arm R?
     
  7. Apr 9, 2012 #6

    BruceW

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    yep, and generally it also involves the angle between the force and the lever arm, but this case happens to be simple, because of what?

    Edit: and so, what is the equation between torque and applied force? You've already got a big clue, since you've seen the solution manual
     
  8. Apr 9, 2012 #7
    Would it be because we are dealing with angular, and not a mix between angular and linear motion, and the equation bewteen torque and applied force would be torque= FR
     
  9. Apr 10, 2012 #8

    BruceW

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    You are right that it is simpler because it's purely angular motion. But I was talking about the general equation between force and torque:
    [tex]torque = RF sin(\theta)[/tex]
    And you're right that we can use the simpler version torque=FR, but why are we allowed to say this? (I'm just trying to make sure you've thought through each step properly).

    And then what is the next step?
     
  10. Apr 10, 2012 #9
    That is where I get a big confused there are so many equations that can give you the same exact thing.

    Torque=FL
    Torque=RF

    I also dont understand why they use sin(theta) on some problems it would require you to find another angle, when it would be easier to use cos theta instead.
     
  11. Apr 10, 2012 #10

    BruceW

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    I'm not sure what you mean about finding another angle... But in this problem, we have the equation torque = RF sin(theta), or you might know it better as Force times perpendicular distance. Anyway, the question tells us that there is a frictional force at the rim. So what is the direction of the frictional force, and therefore what is theta? (and so what is the torque?)
     
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