Force on Ball bearing

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1. Nov 19, 2014

IEatSouls

I'm trying to work out the free body diagram for a router tool (which is basically used to cut wood sideways and create grooves and interesting shapes). I need to work out the free body diagram in order to find out the bearing load but I'm having difficulty with the aforesaid.

The shaft that holds the cutting tool is spun by using an AC motor. So the torque is provided by the motor. My question is whether I should include the force that creates the torque. Since the shaft is spinning at a constant speed during the cutting process, there will be no torque and hence no force. The only external force is from the user having to push the tool and this I will include.

So basically what I want to know is whether I should include the Torque created by the motor despite it moving at constant speed.

2. Nov 20, 2014

billy_joule

Constant speed does not mean zero torque.
The motor torque is what provides the cutting force (ever tried to use a router when it's switched off?) so a FBD without it would not achieve much.

3. Nov 20, 2014

IEatSouls

Thank you for your reply. I understand that, but the cutting torque required is less than the torque provided by the motor. The cutting occurs at constant speed as well and thus under load, the shaft still rotates at constant speed.

Torque by definition requires force and to have a force, we need acceleration. F=ma, if it moves at constant speed then the force is 0 which in turn makes the torque 0.

If something is at constant speed, it is still considered to be in static equilibrium. I don't mean when the router is switched off :p Sorry for the misunderstanding. So at the end of the day, the only force I think that the ball bearing has to take is the force used to push the router by the user. I just need confirmation.

Correct me if I'm wrong and again, Thank you for your reply.

4. Nov 20, 2014

Dr.D

If the motor provided more torque than required for cutting (and friction/windage), then the router would accelerate to a higher speed. I suspect what you are confused on here is the max torque available versus the torque required for cutting. Cutting would rarely ever get up to the max torque available.

If there is any unbalance (and there is always some, but we hope not much), this will also need to appear in the FBD. Remember that the unbalance force varies with speed squared, and at router speeds, even a little bit can be catastrophic.

If you push hard enough to bend the cutter, even a well balanced system will momentarily appear unbalanced.

5. Nov 20, 2014

IEatSouls

The router doesn't have a speed adjustment option and thus can only cut at one speed. So assuming it is at maximum speed at the start, there will be an increase in torque under load due to deceleration when you begin cutting. However this is only momentarily. The shaft will then proceed to rotate at constant speed with that excess torque (Or am I mistaken here?) which means the torque is 0 during cutting due to the constant speed. (Ya it doesn't really make sense to me now, I've confused myself too much).

If what you are saying is correct, than I should include the excessive torque into my FBD?

6. Nov 20, 2014

Dr.D

Here again, I think you are mistaken. When cutting, the router speed is somewhat (not a lot) less than the no load speed. It depends somewhat on the type of motor in use (I think most routers use universal motors), but unless it is a synchronous motor, there will be some speed drop under load. (On a 60 Hz line, the fastest synchronous motor wold only turn 3600 rpm, too slow for a router.) You certainly cannot see the speed drop, but you can usually hear it.

You think my name is funny, but yours clearly indicates you are satan himself.

7. Nov 20, 2014

IEatSouls

I think you misunderstood what I said or I wasn't clear enough. Probably the latter, anyway, I acknowledge that the speed would drop thus a deceleration from the starting speed when the router is under load. This would in turn mean that the torque will increase as speed and torque are inversely proportional.

The shaft would then proceed to accelerate till the excess force is countered by drag and friction.

So let's say, the excess force is

Motor Force - Cutting Force = Excess Force.

The excess force will be balanced eventually by the friction and drag and the shaft will rotate at constant speed. Or in this case, should the excess force be balanced by the bearing to get that constant speed and thus no decrease in applied torque?

Furthermore, when calculating for bearings, do we consider all the stages of the motor and select the bearing on which one gives the highest forces or just directly go to the dynamic condition? or constant speed static condition?

Thank you again! The RPM is well above 30,000 for the router and Aahahah :p Ya...I have no clue how I even came up with this name, I use it for all my gaming accounts.

Thanks again!

Last edited: Nov 20, 2014