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How do I calculate the minimum torque needed for a grinder motor

  1. Jul 20, 2012 #1
    I'm working as an intern trying to automate this grinder system. I would like to find the minimum torque needed for the grinder motor.

    The grinder motors we currently use have:
    1.25 HP
    RPM between 8,000-11,000
    13 amps
    120 volt AC

    The grinded surface has a:
    Coefficient of Friction: 0.75

    We grind the surface with braided cup grinders like the one seen here:

    http://www.homedepot.com/h_d1/N-5yc...3&langId=-1&keyword=cup+grinder&storeId=10051
     
  2. jcsd
  3. Jul 20, 2012 #2
    In general, HP = [itex] \frac{2*pi*Torque*RPM}{33000}[/itex]
    However this may not accurately describe the requirements of your grinder during operating conditions.
     
  4. Jul 20, 2012 #3
    This yields a result in ft-lb when you solve for Torque, by the way
     
  5. Jul 20, 2012 #4
    Is their a relationship between slip friction and torque?
     
  6. Jul 20, 2012 #5
    What I mean to say is:
    Is there a way I could calculate the energy loss due to friction yet still keep the motor running at a minimum RPM. For instance, if I put 5 pounds of force on the grinder, and the surface has a friction of .75, what would the energy loss be, and how does that translate to HP.
     
  7. Jul 20, 2012 #6
    If your motor remains spinning at the same rpm (which I doubt it will) then your torque is just the motor torque calculated previously plus the torque required to overcome the surface friction.

    T = F x r
    F = Fn * .75
    r = radius of grinding wheel

    I find it difficult to believe you'll find an accurate measurement for the kinetic friction for the cutter going through a piece of material as you are not just dealing with the contact point, but also lateral forces on the blade when it enters the cutting area at a not-quite-perpendicular angle (like how you see jig saw blades bend), as well as surface friction on the top and bottom of the wheel from the cut surfaces.
     
  8. Jul 20, 2012 #7
    Grinding cups abrade the surface, they don't cut it. What are the variables in this equation: F = Fn * .75

    Can I use the work done due to friction to calculate energy loss?
    Work done=Friction Force X Distance
    Friction Force= Coefficient of Friction X Normal Force
    Distance= Circumfrence X Revolutions Per Second (Guessing)
     
  9. Jul 20, 2012 #8
    I get about .43 hp, but I'm not sure how accurate that is. Shouldn't cross sectional area come in somewhere?
     
  10. Jul 20, 2012 #9
    Your analysis needs a few items that need looking at.

    A coefficient of friction is that which is listed or obtained between 2 surfaces. The 0.75 you quote represents what - steel to wood, steel to steel, steel to copper, ?? from where did you obtain this 0.75?

    One variable is the 0.75. A polished surface will have a coefficien different than unpolished. Rust, paint, oil, surface irregularities, edges, etc will all change the coefficient. You are removing material so it really is not a coefficient of friction per se.

    Do you intend to apply the tool only at the circumference of the cup or the whole flat surface of the bristles, in which case you do an integration of the flat surface of the cup mating with the material. As you press more down on the tool, some energy is taken up by the flex of the bristles. If you press down hard enough your tool may stall - you might want to take that into account. If you are automating the procedure you certainly do not want your tool to stall and burn out the motor while it is unattended.
     
  11. Jul 23, 2012 #10
    I assume the equation from this thread would be the closest thing I could get to a correct formula.

    2. angular velocity w= (rpm*2*pi)/60
    angular accel a= w/time
    inertia I=mr^2
    Torque T=Ia

    https://www.physicsforums.com/showthread.php?t=412001

    I'm not really sure how they put in the coefficient of friction. I'm also assuming the Inertia of a cup grinder is defined as:

    I=(1/2)m(r^2+R^2)
    Where r and R represent the inner and outer radius.

    Without putting in the cooefficient of friction I get a Torque of about 4.46 newton meters. This is using 3500 rpm, time at 1 second, inner and outer radius of .075 and .05 meters, and a mass of 3 kg on the system. 4.46 N*m at 3500 rpm translates to 2.21 hp. (Again, without putting in the coefficient of friction)

    This doesn't quite add up as I was able to run the grinder on a 1/30 hp motor with about 1 kg of force at 3200 rpm. I'm going to experiment with a larger motor until I get the right size.
     
    Last edited: Jul 23, 2012
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