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Torque Grindstone problem

  1. Nov 4, 2014 #1
    1. The problem statement, all variables and given/known data
    A 60.0-kg grindstone is a solid disk 0.550m in diameter. You press an ax down on the rim with a normal force of 140N(Figure 1) . The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50N⋅m between the axle of the stone and its bearings.

    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 8.00s?

    I then have the questions:
    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?
    and
    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?

    2. Relevant equations

    T = RFsin(theta)

    3. The attempt at a solution
    I set up an equation, shown below, which sets the Torque net to all of the variables that add up to get that net force, which is equal to I * alpha.. I found alpha to be 1.57 using the formula w = wo + alpha * t. I is 1/2MR^2.

    Tnet = Fr - Taxle - Taxe = I * alpha
    r being the handle length, Taxle being RF, or (.275) * (60) * (9.8), Taxe being Mk * Fn or (.6) * (140).

    After doing this out, I got 155.51 Nm. The system is telling me I'm wrong. Help please, I'm dying.
     
    Last edited: Nov 4, 2014
  2. jcsd
  3. Nov 4, 2014 #2

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    Wait, I'm confused. [itex] T_{axle} [/itex] is given to you in the problem statement as a constant 6.50 N⋅m. No calculations necessary on this torque. Where are the 0.275, 60, and 9.8 figures coming from?

    (0.6)(140 N) gives you the tangential force from the axe. But you still need to calculate the torque attributed to that force.
     
  4. Nov 4, 2014 #3
    So, taking the piece of information, my recalculated answer comes to 162.124. What do you think? I figured I was going to be wrong somewhere related to that piece of information, but I didn't know exactly what it meant so i didn't know where to put it.

    Btw, before I put this in to my hmwk, since it is online, I had put in prior to this, 162.5, which pearson said was wrong. Does that mean this new answer is still wrong as well?
     
  5. Nov 4, 2014 #4

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    I did not end up with 162.124 or even 162.5 Newtons. The answer that I came up with was orders of magnitude different. Wait, hold on, let me recalculate ...
     
  6. Nov 4, 2014 #5

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    I recalculated, but still did not come near your answers. (Although it is within the same order of magnitude.)
     
  7. Nov 4, 2014 #6
    My answers were 162.52, 155.51, and 142.51. All wrong so far. I only have two chances left, could you share your answer?
     
  8. Nov 4, 2014 #7

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    Sorry, no. But if you show us how you arrived at your answer(s), we might be able to help point out what went wrong.
     
  9. Nov 4, 2014 #8
    I thought I already did, what else do you need to know?
     
  10. Nov 4, 2014 #9

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    You set up the problem correctly, stating earlier that

    [tex] \tau_{net} = F r_{handle} - \tau_{axle} - \tau_{axe} = I \alpha, [/tex]

    where [itex] I = \frac{1}{2} m r_{stone}^2. [/itex]

    So far so good.

    But you didn't show us how you came up with the answers you did. There must be a calculation error in there somewhere. Show us the calculations and we might be able to help more.
     
  11. Nov 4, 2014 #10
    Ok, I'll add in the variables where I felt they should go.

    T_net = F(.5) - 6.5 - (.6)(140) = 1/2(60)(.275)^2(1.57)

    To this my answer was: 162.124
    Thoughts?
     
  12. Nov 4, 2014 #11

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    (0.6)(140 Newtons) is a force, not a torque (it's the tangential, frictional force from the axe). You need to throw the radius of the wheel in there somewhere to get the corresponding torque.
     
  13. Nov 4, 2014 #12
    I see what you mean, I'm starting to get it a little more. I put in the radius of the wheel in to that part of the equation and it looks like my final answer became 40.32
     
  14. Nov 4, 2014 #13

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    Wait, 40.32 Newtons? That's still different from my answer.
     
  15. Nov 4, 2014 #14
    I have no idea then, I feel like everything is correct, I don't know whats wrong.

    Am I close to yours at least?
     
  16. Nov 4, 2014 #15

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    What did you come up with for the torque attributed to the axe friction? The rest of your numbers look okay.
     
  17. Nov 4, 2014 #16
    23.1

    What radius did you use for the axe? I used the grindstone, .275.
     
  18. Nov 4, 2014 #17

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    That's correct. [itex] \tau_{axe} = \mu_k F_{N} r_{stone} = \mathrm{(0.6) (140 \ [N])(0.275 \ [m]) = 23.1 \ [N \cdot m]} [/itex]

    Yet when I do the calculations, I still come up with something different for the final answer.

    Did you remember to convert the torque of the handle to the force applied to the handle? [Edit: then again, that alone does not account for the discrepancy.]

    [Edit: Maybe you are subtracting something when calculating your final answer when you should be adding?]
     
  19. Nov 4, 2014 #18
    These are my calculations:
    F(.5) - 6.5 - (.6)(140)(.275) = 1/2(60)(.275)^2(1.57)
    f(.5) -16.6 = 3.56194
    f(.5) = 20.1619
    f = 40.3239
     
  20. Nov 4, 2014 #19

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    -6.5 - 23.1 ≠ -16.6
     
  21. Nov 4, 2014 #20

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    In other words, take another look on the parts that I highlighted in red below:

     
    Last edited: Nov 4, 2014
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