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

In summary, the grinder motor we currently use has a torque of 2.21 hp with a coefficient of friction of .75.
  • #1
clearwater304
89
0
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-5yc1v/R-202938803/h_d2/ProductDisplay?catalogId=10053&langId=-1&keyword=cup+grinder&storeId=10051
 
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  • #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.
 
  • #3
This yields a result in ft-lb when you solve for Torque, by the way
 
  • #4
Is their a relationship between slip friction and torque?
 
  • #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.
 
  • #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.
 
  • #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)
 
  • #8
I get about .43 hp, but I'm not sure how accurate that is. Shouldn't cross sectional area come in somewhere?
 
  • #9
Your analysis needs a few items that need looking at.

The grinded surface has a:
Coefficient of Friction: 0.75

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?

What are the variables in this equation: F = Fn * .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.

Distance= Circumfrence X Revolutions Per Second (Guessing)

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.
 
  • #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:

1. What is torque and why is it important for a grinder motor?

Torque is a measure of the rotational force or twisting power of an object. In the case of a grinder motor, torque is important because it determines the ability of the motor to rotate the blades and effectively grind materials.

2. How do I determine the minimum torque needed for a grinder motor?

The minimum torque needed for a grinder motor can be calculated by considering the speed and power requirements of the grinder. The formula for torque is torque = power / (2 x pi x speed). By knowing the power and speed requirements of the grinder, you can plug them into this formula to determine the minimum torque needed.

3. What factors can affect the minimum torque needed for a grinder motor?

The minimum torque needed for a grinder motor can be affected by several factors including the type and hardness of the material being ground, the size and shape of the grinder blades, and the desired speed and fineness of the grind.

4. Can I use a motor with higher torque than the minimum requirement for a grinder?

Yes, using a motor with higher torque than the minimum requirement can provide additional power and efficiency for the grinder. However, it is important to make sure that the motor is compatible with the grinder and does not cause any damage or overheating.

5. What are some common units of measurement for torque in grinder motors?

The most common units of measurement for torque in grinder motors are Newton-meters (Nm), foot-pounds (ft-lb), and inch-pounds (in-lb). These units represent the amount of force needed to produce a rotational acceleration of one meter, foot, or inch per second squared, respectively.

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