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Wheels and rotational motion

  1. Nov 24, 2006 #1
    Hi. I did a 5 part problem and I just want to know if I did it right or if I'm on the right track. Please help. Thank you. The problem is as follows:

    Two identical grinding wheels of mass m and radius r are initially spinning about their centers. Wheel A has an intial angular speed of w, while wheel B has an initial angular speed of 2w. Both wheels are being used to sharpen tools. For both wheels the tool is being pressed against the wheel with a force F directed toward the center of the wheel, and the coefficient of kinetic friction between the wheel and the tool is uk. You are holding the tool firmly so it does not move tangentially to the wheel.

    (a) If it takes wheel A a time T to come to a stop, how long does it take for wheel B to come to a stop?

    My work: p=Ft, so m(wf-wi)=uFt, so t=(m(wf-wi)/umg)= (mwi)/(umg)=TA, so since B=2w then the time for B would be 2TA.

    (b) Find an expression for T in terms of teh variables specified above.

    My work: T=mwi/umg, so T=wi/ug

    (c) If wheel A rotates through an angle @ before coming to rest, through what angle does wheel B rotate before coming to rest?

    My work: wf^2-wi^2=2a@, so @=(wf^2-wi^2)/(2a) so @=wi^2/2a...So the @B would be 2wi^2/2a, so @B is 4@A. (@ is angle and a is alpha)

    (d) Find an expression for @ in terms of teh variables specified above.

    My work: @= (wi^2/2a) = (wi^2/2(Ffric/m)) = (wi^2/2(uF/m)) = (wi^2/2(umg/m)) = (wi^2/2ug). So @= (wi^2)/(2ug)

    (e) If you doubled the value of uk, how would that affect the time required to stop wheel A?

    My work: Doubling the value of uk doubles the friction, which would reduce the time to stop wheel A by 1/2. Double uk would cause 1/2TA.

    Thank you for reading through this!
     
  2. jcsd
  3. Nov 26, 2006 #2

    Fermat

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    You have uF = umg. But F is simply the force of the tool against the wheel and has no relation to the wheel's mass. Just leave F is it is, F.
    So, check your answers for parts a), b), and d).
    Also, in d), a (alpha) should be a = (2Fr)/(mr). {I_disc = (1/2)mr²}
    All your other working looks OK.

    Edit: corrected F = umg to uF = umg
     
    Last edited: Nov 27, 2006
  4. Nov 26, 2006 #3
    Ok so if F can't be simplified then for (a) it is mwi/F so TB would still be double Ta...for (b) it T= mwi/F, and for (d) I'm a little confused...
    is it wi^2/(2Fr/mf) = (wi^2)/(2F/m) = (mwi^2)/(2F)? Right? Thanks for looking over all this.
     
  5. Nov 26, 2006 #4

    Fermat

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    correct.

    correct, but use w , rather than wi, since you are asked to use the variables specified above.

    I'm not sure how you got alpha = Fr/mf, but ...

    F = normal reaction of disc against tool face.
    μ is coefft of (kinetic) friction
    Fr = frictional force on rim of wheel
    and
    Fr = μF

    Decelerating torque on wheel is T = Fr x r
    and
    T = Iα, where I = (1/2)mr²
    Therefore,
    (1/2)mr² x α = Fr x r
    (1/2)mr x α = μF
    α = 2μF/(mr)

    Now, you should have

    θ = ω²/(2α)
    θ = ω²/(4μF/(mr))
    θ = mrω²/(4μF)
     
  6. Nov 26, 2006 #5
    So for the entire problem I don't have to factor in the coefficient of kinetic friction between the wheel and the tool? That's why I used the umg in each of the steps
     
  7. Nov 27, 2006 #6

    Fermat

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    You have Fr = uF = umg, but F = mg is wrong.

    Also, you have P = Ft, which is also wrong. I assumed that was a typo. It should be P = Fr x t.

    So, you can't use umg in each of the steps.
     
  8. Nov 27, 2006 #7
    wait so,

    (a) (mwi)/F ?
    (b) (mw)/F ?

    wouldnt you have to factor in uk?
     
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